2017.10高一月考卷(南昌十中)
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2016-2017学年江西省南昌十中高一(下)第一次月考数学试卷一、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,恰有一项是符合题目要求的).1.(5分)下列函数中,周期是的偶函数是()A.y=sin4x B..y=tan2xC.y=cos22x﹣sin22x D.y=cos2x2.(5分)数列1,,,,的一个通项公式a n是()A.B.C.D.3.(5分)下列各组向量中,可以作为基底的是()A.,B.,C.,D.,4.(5分)已知O是△ABC所在平面内一点,D为BC边中点,且,那么()A.B. C. D.5.(5分)在△ABC中,已知2sinAcosB=sinC,那么△ABC一定是()A.直角三角形B.等腰三角形C.等腰直角三角形 D.正三角形6.(5分)已知cos(α﹣)=,则sin2α等于()A.B.﹣C.D.﹣7.(5分)若非零向量,满足||=||,且(﹣)⊥(3+2),则与的夹角为()A.B.C. D.π8.(5分)如图是函数y=sin(ωx+φ)(ω>0,0<φ<)在区间[﹣,]上的图象,将该图象向右平移m(m>0)个单位后,所得图象关于直线x=对称,则m的最小值为()A.B.C.D.9.(5分)在△ABC中,若a=4,b=3,cosA=,则B=()A.B.C.D.10.(5分)在直角坐标系xOy中,分别是与x轴,y轴同向的单位向量,若直角三角形ABC中,,,则k的可能值有()A.4个 B.3个 C.2个 D.1个11.(5分)如图,从气球A上测得正前方的河流的两岸B,C的俯角分别为75°,30°,此时气球的高是60m,则河流的宽度BC等于()A.m B.m C.m D.m 12.(5分)设A1,A2,A3,A4是平面直角坐标系中两两不同的四点,若=λ(λ∈R),=μ(μ∈R),且+=2,则称A3,A4调和分割A1,A2,已知平面上的点C,D调和分割点A,B,则下面说法正确的是()A.C可能是线段AB的中点B.D可能是线段AB的中点C.C、D可能同时在线段AB上D.C、D不可能同时在线段AB的延长线上二、填空题:(本大题共4小题,每小题5分,共20分)13.(5分)在△ABC中,A=60°,AC=4,BC=2,则△ABC的面积等于.14.(5分)一质点受到平面上的三个力F1,F2,F3(单位:牛顿)的作用而处于平衡状态.已知F1,F2成60°角,且F1,F2的大小分别为2和4,则F3的大小为.15.(5分)如图,在△ABC中,H为BC上异于B,C的任一点,M为AH的中点,若=λ+μ,则λ+μ=.16.(5分)2002年在北京召开的国际数学家大会,会标是以我国古代数学家赵爽的弦图为基础设计的.弦图是由四个全等直角三角形与一个小正方形拼成的一个大正方形(如图).如果小正方形的面积为1,大正方形的面积为25,直角三角形中较小的锐角为θ,那么cos2θ的值等于.三.解答题:本大题共6小题,共70分.解答应写出文字说明,证明过程或演算步骤.17.(10分)向量=(3,2),=(﹣1,2),=(4,1):(1)求满足=m+n的实数m,n;(2)若(+k)∥(2﹣),求实数k.18.(12分)△ABC的内角A,B,C所对的边分别为a,b,c.向量=(a,b)与=(cosA,sinB)平行.(Ⅰ)求A;(Ⅱ)若a=,b=2,求△ABC的面积.19.(12分)在平行四边形ABCD中,E,F分别是BC,CD的中点,DE交AF于点H,记、分别为,,则=(用,表示)20.(12分)已知函数f(x)=sin(﹣x)sinx﹣cos2x.(I)求f(x)的最小正周期和最大值;(II)讨论f(x)在[,]上的单调性.21.(12分)在△ABC中,若A=,AB=6,AC=3,点D在BC的边上且AD=BD,则AD=.22.(12分)如图,在等腰直角三角形△OPQ中,∠POQ=90°,OP=2,点M 在线段PQ上.(1)若OM=,求PM的长;(2)若点N在线段MQ上,且∠MON=30°,问:当∠POM取何值时,△OMN 的面积最小?并求出面积的最小值.2016-2017学年江西省南昌十中高一(下)第一次月考数学试卷参考答案与试题解析一、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,恰有一项是符合题目要求的).1.(5分)(2017春•东湖区校级月考)下列函数中,周期是的偶函数是()A.y=sin4x B..y=tan2xC.y=cos22x﹣sin22x D.y=cos2x【分析】根据三角函数的周期公式依次判断各函数的周期即可得到结论.【解答】解:对于A:y=sin4x,其周期T=,根据正弦函数的图象可知,是奇函数,∴A不对.对于B:y=tan2x,其周期T=,根据正切函数的图象可知,是奇函数,∴B不对.对于C:y=cos22x﹣sin22x=cos4x,其周期T=,根据余弦函数的图象可知,是偶函数,∴C对.对于D:y=cos2x,其周期T=,根据余弦函数的图象可知,是偶函数,∴D不对.故选C.【点评】本题主要考查三角函数的图象和性质,比较基础.2.(5分)(2013秋•宿州期末)数列1,,,,的一个通项公式a n是()A.B.C.D.【分析】将原数列中的第一项写成分式的形式:,再观察得出每一项的分子是正整数数列,分母是正奇数数列,从而得出数列1,,,,的一个通项公式a n.【解答】解:将原数列写成:,,,,.每一项的分子是正整数数列,分母是正奇数数列,∴数列1,,,,的一个通项公式a n是.故选B.【点评】本题主要考查了数列的概念及简单表示法、求数列的通项公式.关键推断{a n}中每一项的分式的规律求得数列的通项公式.3.(5分)(2015秋•凯里市校级期末)下列各组向量中,可以作为基底的是()A.,B.,C.,D.,【分析】不共线的向量可作基底,由向量共线的条件逐个选项判断即可.【解答】解:选项A,可得0×(﹣2)﹣0×1=0,故,不可作基底,故错误;选项B,可得2×(﹣)﹣(﹣3)×=0,故,不可作基底,故错误;选项C,可得3×10﹣5×6=0,故,不可作基底,故错误;选项D,可得﹣1×7﹣2×5≠0,故不平行,故可作基底,故正确.故选D【点评】本题考查平面向量基本定理,向量作基底的条件,涉及向量平行的判断,属基础题.4.(5分)(2007•北京)已知O是△ABC所在平面内一点,D为BC边中点,且,那么()A.B. C. D.【分析】先根据所给的式子进行移项,再由题意和向量加法的四边形法则,得到,即有成立.【解答】解:∵,∴,∵D为BC边中点,∴,则,故选:A.【点评】本题考查了向量的加法的四边形法则的应用,即三角形一边上中点的利用,再根据题意建立等量关系,再判断其它向量之间的关系.5.(5分)(2010•湖北模拟)在△ABC中,已知2sinAcosB=sinC,那么△ABC一定是()A.直角三角形B.等腰三角形C.等腰直角三角形 D.正三角形【分析】根据三角形三个内角和为180°,把角C变化为A+B,用两角和的正弦公式展开移项合并,公式逆用,得sin(B﹣A)=0,因为角是三角形的内角,所以两角相等,得到三角形是等腰三角形.【解答】解:由2sinAcosB=sinC知2sinAcosB=sin(A+B),∴2sinAcosB=sinAcosB+cosAsinB.∴cosAsinB﹣sinAcosB=0.∴sin(B﹣A)=0,∵A和B是三角形的内角,∴B=A.故选B【点评】在三角形内会有一大部分题目出现,应用时要抓住三角形内角和是180°,就有一部分题目用诱导公式变形,对于题目中正用、逆用两角和的正弦和余弦公式,必须在复杂的式子中学会辨认公式应用公式.6.(5分)(2017春•东湖区校级月考)已知cos(α﹣)=,则sin2α等于()A.B.﹣C.D.﹣【分析】将已知等式左边利用两角和与差的余弦函数公式及特殊角的三角函数值化简,求出sinα+cosα的值,两边平方并利用同角三角函数间的基本关系及二倍角的正弦函数公式化简,即可求出sin2α的值.【解答】解:∵cos(α﹣)=(cosα+sinα)=,∴cosα+sinα=,两边平方得:(cosα+sinα)2=,即1+2sinαcosα=,则sin2α=2sinαcosα=﹣.故选:D【点评】此题考查了两角和与差的余弦函数公式,同角三角函数间的基本关系,以及二倍角的正弦函数公式,熟练掌握公式是解本题的关键.7.(5分)(2015•重庆)若非零向量,满足||=||,且(﹣)⊥(3+2),则与的夹角为()A.B.C. D.π【分析】根据向量垂直的等价条件以及向量数量积的应用进行求解即可.【解答】解:∵(﹣)⊥(3+2),∴(﹣)•(3+2)=0,即32﹣22﹣•=0,即•=32﹣22=2,∴cos<,>===,即<,>=,故选:A【点评】本题主要考查向量夹角的求解,利用向量数量积的应用以及向量垂直的等价条件是解决本题的关键.8.(5分)(2015•宜宾模拟)如图是函数y=sin(ωx+φ)(ω>0,0<φ<)在区间[﹣,]上的图象,将该图象向右平移m(m>0)个单位后,所得图象关于直线x=对称,则m的最小值为()A.B.C.D.【分析】由周期求出ω,由五点法作图求出φ的值,可得函数的f(x)的解析式.再根据函数y=Asin(ωx+φ)的图象的变换规律,可得结论.【解答】解:由函数f(x)=sin(ωx+φ),(ω>0,|φ|<)的图象可得T==﹣(﹣)=π,∴ω=2.再由五点法作图可得2×(﹣)+φ=0,∴φ=.故函数的f(x)的解析式为f(x)=sin(2x+)=sin2(x+).故把f(x)=sin2(x+)的图象向右平移m(m>0)个单位长度,可得g(x)=sin2(x﹣m+)的图象,∵所得图象关于直线x=对称,∴g(x)=sin2(﹣m+)=±1,∴2(﹣m+)=+kπ,解得:m=﹣kπ,k∈Z,∴当k=0时,φ=.故选:B.【点评】本题主要考查由函数y=Asin(ωx+φ)的部分图象求解析式,函数y=Asin (ωx+φ)的图象的变换规律,属于中档题.9.(5分)(2014•西城区二模)在△ABC中,若a=4,b=3,cosA=,则B=()A.B.C.D.【分析】先利用同角三角函数关系求得sinA的值,进而利用正弦定理求得sinB 的值,最后求得B.【解答】解:∵cosA=,0<∠A<π∴sinA===∵=,即=,∴sinB=,∴∠B=或,∵sinA=>∴∠A>∴∠B=与三角形内角和为180°矛盾.∴∠B=,故选A.【点评】本题主要考查了正弦定理的应用.解题的过程中注意对结果正负号的判断.10.(5分)(2017春•东湖区校级月考)在直角坐标系xOy中,分别是与x轴,y轴同向的单位向量,若直角三角形ABC中,,,则k 的可能值有()A.4个 B.3个 C.2个 D.1个【分析】由向量的运算可得,分三种情况∠A=90°或∠B=90°或∠C=90°利用向量的数量积等于零,建立关系式,再解方程求得所有可能k的值.【解答】解:∵,,∴=﹣=+(k﹣1)∵△ABC为直角三角形,(1)当∠A=90°时,=6+k=0,解得k=﹣6;(2)当∠B=90°时,=2+k﹣1=0,解得k=﹣1;(3)当∠C=90°时,=3+k(k﹣1)=0,方程无实解;综上所述,k=﹣6或﹣1故选:C.【点评】本题考查向量坐标的定义、考查向量的运算法则、考查向量垂直的充要条件,分类讨论是解决问题的关键,属基础题.11.(5分)(2014•四川)如图,从气球A上测得正前方的河流的两岸B,C的俯角分别为75°,30°,此时气球的高是60m,则河流的宽度BC等于()A.m B.m C.m D.m【分析】由题意画出图形,由两角差的正切求出15°的正切值,然后通过求解两个直角三角形得到DC和DB的长度,作差后可得答案.【解答】解:如图,∠DAB=15°,∵tan15°=tan(45°﹣30°)==2﹣.在Rt△ADB中,又AD=60,∴DB=AD•tan15°=60×(2﹣)=120﹣60.在Rt△ADC中,∠DAC=60°,AD=60,∴DC=AD•tan60°=60.∴BC=DC﹣DB=60﹣(120﹣60)=120(﹣1)(m).∴河流的宽度BC等于120(﹣1)m.故选:B.【点评】本题给出实际应用问题,求河流在B、C两地的宽度,着重考查了三角函数的定义、正余弦定理解三角形的知识,属于中档题.12.(5分)(2014秋•内蒙古校级期末)设A1,A2,A3,A4是平面直角坐标系中两两不同的四点,若=λ(λ∈R),=μ(μ∈R),且+=2,则称A3,A4调和分割A1,A2,已知平面上的点C,D调和分割点A,B,则下面说法正确的是()A.C可能是线段AB的中点B.D可能是线段AB的中点C.C、D可能同时在线段AB上D.C、D不可能同时在线段AB的延长线上【分析】由题意可设A(0,0)、B(1,0)、C(c,0)、D(d,0),结合条件+=2,根据题意考查方程+=2的解的情况,用排除法选出正确的答案即可.【解答】解:由已知不妨设A(0,0)、B(1,0)、C(c,0)、D(d,0),则(c,0)=λ(1,0),(d,0)=μ(1,0),∴λ=c,μ=d;代入+=2,得+=2;(*)若C是线段AB的中点,则c=,代入(*)得,d不存在,∴C不可能是线段AB的中点,A错误;同理B错误;若C,D同时在线段AB上,则0≤c≤1,0≤d≤1,代入(*)得,c=d=1,此时C和D点重合,与已知矛盾,∴C错误.若C,D同时在线段AB的延长线上时,则λ>1.μ>1,∴+<2,这与+=2矛盾;∴C、D不可能同时在线段AB的延长线上,D正确.故选:D.【点评】本题考查了新定义应用问题,解题时应正确理解新定义的含义,是易错题目.二、填空题:(本大题共4小题,每小题5分,共20分)13.(5分)(2014•福建)在△ABC中,A=60°,AC=4,BC=2,则△ABC的面积等于2.【分析】利用三角形中的正弦定理求出角B,再利用三角形的面积公式求出△ABC 的面积.【解答】解:∵△ABC中,A=60°,AC=4,BC=2,由正弦定理得:,∴,解得sinB=1,∴B=90°,C=30°,∴△ABC的面积=.故答案为:.【点评】本题着重考查了给出三角形的两边和其中一边的对角,求它的面积.正余弦定理、解直角三角形、三角形的面积公式等知识,属于基础题.14.(5分)(2012•陕西三模)一质点受到平面上的三个力F1,F2,F3(单位:牛顿)的作用而处于平衡状态.已知F1,F2成60°角,且F1,F2的大小分别为2和4,则F3的大小为2.【分析】依题意可画出图象,可知AB=2,BC=4,∠ABC=120°,根据余弦定理求得AC.【解答】解:如图,根据题意可知AB=2,BC=4,∠ABC=120°由余弦定理可知AC2=22+42﹣2×2×4×cos120°=28∴AC=2故答案为2【点评】本题主要考查了解三角形的应用.熟练掌握余弦定理,正弦定理,同角三角函数基本关系,是顺利快速解三角形问题的关键.15.(5分)(2017春•东湖区校级月考)如图,在△ABC中,H为BC上异于B,C的任一点,M为AH的中点,若=λ+μ,则λ+μ=.【分析】根据题意,用、表示出、与,求出λ、μ的值即可.【解答】解:根据题意,==(+)=(+x)=+x(﹣)=(1﹣x)+x∴λ=(1﹣x),μ=x,∴λ+μ=.故答案为:.【点评】本题主要考查了平面向量的线性运算问题,是基础题.16.(5分)(2007•北京)2002年在北京召开的国际数学家大会,会标是以我国古代数学家赵爽的弦图为基础设计的.弦图是由四个全等直角三角形与一个小正方形拼成的一个大正方形(如图).如果小正方形的面积为1,大正方形的面积为25,直角三角形中较小的锐角为θ,那么cos2θ的值等于.【分析】根据两正方形的面积分别求出两正方形的边长,根据小正方形的边长等于直角三角形的长直角边减去短直角边,利用三角函数的定义表示出5cosθ﹣5sinθ=1,两边平方并利用同角三角函数间的基本关系及二倍角的正弦函数公式化简可得sin2θ的值,然后根据θ的范围求出2θ的范围即可判断出cos2θ的正负,利用同角三角函数间的基本关系由sin2θ即可求出cos2θ的值.【解答】解:∵大正方形面积为25,小正方形面积为1,∴大正方形边长为5,小正方形的边长为1.∴5cosθ﹣5sinθ=1,∴cosθ﹣sinθ=.∴两边平方得:1﹣sin2θ=,∴sin2θ=.∵θ是直角三角形中较小的锐角,∴0<θ<.∴cos2θ=.故答案为:【点评】此题考查学生灵活运用同角三角函数间的基本关系及二倍角的正弦函数公式化简求值,是一道中档题.本题的突破点是将已知的两等式两边平方.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.(10分)(2013秋•西山区校级期末)向量=(3,2),=(﹣1,2),=(4,1):(1)求满足=m+n的实数m,n;(2)若(+k)∥(2﹣),求实数k.【分析】(1)由题意和向量的坐标运算求出的坐标,再由向量相等的条件列出方程组,求出m和n的值;(2)由题意和向量的坐标运算求出和的坐标,再由向量共线的条件列出方程.求出k的值.【解答】解:(1)由题意得,=m(﹣1,2)+n(4,1)=(﹣m+4n,2m+n),∵,∴(3,2)=(﹣m+4n,2m+n),即,解得m=,n=,(2)由题意得,=(3,2)+k(4,1)=(3+4k,2+k),=2(﹣1,2)﹣(3,2)=(﹣5,2),∵(),∴2(3+4k)+5(2+k)=0,解得k=.【点评】本题考查了向量的坐标运算,向量相等的条件,以及向量共线的条件,属于基础题.18.(12分)(2015•陕西)△ABC的内角A,B,C所对的边分别为a,b,c.向量=(a,b)与=(cosA,sinB)平行.(Ⅰ)求A;(Ⅱ)若a=,b=2,求△ABC的面积.【分析】(Ⅰ)利用向量的平行,列出方程,通过正弦定理求解A;(Ⅱ)利用A,以及a=,b=2,通过余弦定理求出c,然后求解△ABC的面积.【解答】解:(Ⅰ)因为向量=(a,b)与=(cosA,sinB)平行,所以asinB﹣=0,由正弦定理可知:sinAsinB﹣sinBcosA=0,因为sinB ≠0,所以tanA=,可得A=;(Ⅱ)a=,b=2,由余弦定理可得:a2=b2+c2﹣2bccosA,可得7=4+c2﹣2c,解得c=3,△ABC的面积为:=.【点评】本题考查余弦定理以及正弦定理的应用,三角形的面积的求法,考查计算能力.19.(12分)(2017春•东湖区校级月考)在平行四边形ABCD中,E,F分别是BC,CD的中点,DE交AF于点H,记、分别为,,则=(用,表示)【分析】根据条件,由A,H,F三点共线便可得到,同理,由D,H,E三点共线可以得到,进而得到,这样根据平面向量基本定理便可得到,解出λ,μ即可用表示出.【解答】解:如图,E,F分别为BC,CD的中点;A,H,F三点共线;∴存在实数λ,使===;D,H,E三点共线;∴存在μ,使=;∴=;∴根据平面向量基本定理得:;解得;∴.故答案为:.【点评】考查共线向量基本定理,平面向量基本定理,以及向量加法的几何意义.20.(12分)(2015•重庆)已知函数f(x)=sin(﹣x)sinx﹣cos2x.(I)求f(x)的最小正周期和最大值;(II)讨论f(x)在[,]上的单调性.【分析】(Ⅰ)由条件利用三角恒等变换化简函数的解析式,再利用正弦函数的周期性和最值求得f(x)的最小正周期和最大值.(Ⅱ)根据2x﹣∈[0,π],利用正弦函数的单调性,分类讨论求得f(x)在上的单调性.【解答】解:(Ⅰ)函数f(x)=sin(﹣x)sinx﹣x=cosxsinx﹣(1+cos2x)=sin2x﹣cos2x﹣=sin(2x﹣)﹣,故函数的周期为=π,最大值为1﹣.(Ⅱ)当x∈时,2x﹣∈[0,π],故当0≤2x﹣≤时,即x ∈[,]时,f(x)为增函数;当≤2x﹣≤π时,即x∈[,]时,f(x)为减函数.【点评】本题主要考查三角恒等变换,正弦函数的周期性和最值,正弦函数的单调性,属于中档题.21.(12分)(2016春•蚌埠期末)在△ABC中,若A=,AB=6,AC=3,点D在BC的边上且AD=BD,则AD=.【分析】由已知及余弦定理可解得BC的值,由正弦定理可求得sinB,从而可求cosB,过点D作AB的垂线DE,垂足为E,由AD=BD得:cos∠DAE=cosB,即可求得AD的长.【解答】解:∵A=,AB=6,AC=3,∴在△ABC中,由余弦定理可得:BC2=AB2+AC2﹣2AB•ACcos∠BAC=90.∴BC=3,…4分∵在△ABC中,由正弦定理可得:=,∴sinB=,∴cosB=,…8分∵过点D作AB的垂线DE,垂足为E,由AD=BD得:cos∠DAE=cosB,∴Rt△ADE中,AD===.故答案为:.…12分【点评】本题主要考查了正弦定理,余弦定理在解三角形中的应用,考查了计算能力和转化思想,属于基本知识的考查.22.(12分)(2014•正定县校级三模)如图,在等腰直角三角形△OPQ中,∠POQ=90°,OP=2,点M在线段PQ上.(1)若OM=,求PM的长;(2)若点N在线段MQ上,且∠MON=30°,问:当∠POM取何值时,△OMN 的面积最小?并求出面积的最小值.【分析】(1)在△OPQ中,由余弦定理得,OM2=OP2+MP2﹣2•OP•MPcos45°,解得MP即可.(2)∠POM=α,0°≤α≤60°,在△OMP中,由正弦定理求出OM,同理求出ON,推出三角形的面积,利用两角和与差的三角函数化简面积的表达式,通过α的范围求出面积的最大值.【解答】解:(1)在△OPQ中,∠OPQ=45°,OM=,OP=2,由余弦定理得,OM2=OP2+MP2﹣2•OP•MPcos45°,得MP2﹣4MP+3=0,解得MP=1或MP=3. (6)(2)设∠POM=α,0°≤α≤60°,在△OMP中,由正弦定理,得,所以,同理…8′S△OMN== (10)===== (14)因为0°≤α≤60°,30°≤2α+30°≤150°,所以当α=30°时,sin(2α+30°)的最大值为1,此时△OMN的面积取到最小值.即∠POM=30°时,△OMN的面积的最小值为8﹣4. (16)【点评】本题考查正弦定理以及余弦定理两角和与差的三角函数的应用,考查转化思想以及计算能力.第21页(共21页)。
南昌十中2016-2017学年上学期第二次月考高一地理试题说明:本试卷分第I卷(选择题)和第Ⅱ卷(非选择题)两部分,全卷满分100分。
考试用时100分钟,注意事项:考生在答题前请认真阅读本注意事项及各题答题要求。
1.答题前,请您务必将自己的姓名、IS号用书写黑色字迹的0.5毫米签字笔填写在答题纸上,同时用2B铅笔在规定的位置上认真填涂自己的IS号。
2.作答非选择题必须用书写黑色字迹的0.5毫米签字笔写在答题纸上的指定位置,在其它位置作答一律无效。
作答选择题必须用2B铅笔把答题卡上对应题目的答案标号涂黑。
如需改动,请用橡皮擦干净后,再选涂其它答案,请保持卡面清洁和答题纸清洁,不折叠、不破损。
3.考试结束后,请将答题纸交回监考老师。
一、选择题(本大题共30题,每小题2分,共计60分。
在每小题列出的四个选项中只有一项是最符合题目要求的)下图为冬至日甲、乙、丙、丁四地昼夜长短比例示意图(阴影部分表示夜长),读图完成下列问题。
1. 甲、乙、丙、丁四地地球自转线速度由大到小排序正确的是()A. 甲、丁、丙、乙B. 甲、丙、丁、乙C. 丁、甲、丙、乙D. 乙、丙、甲、丁2. 地球表面做水平运动的物体发生左偏的地点是()A. 甲B. 乙C. 丙D. 丁【答案】1. B 2. D【解析】1. 赤道全年昼夜等长,甲位于赤道,纬度最低。
纬度越高,昼夜长短的变化幅度越大。
各地夜长与赤道夜长的差值比较,差值越大,说明纬度越高。
乙地有极夜现象,夜长差值最大,乙纬度最高。
丙地与赤道昼长差值比丁地小,纬度比丁地低。
自转线速度由赤道向两极递减,四地地球自转线速度由大到小排序正确的是甲、丙、丁、乙,B对。
A、C、D错。
2. 冬至日,北半球昼短夜长,南半球昼长夜短,图中丁是位于南半球,地球表面做水平运动的物体发生左偏的地点是丁,D对。
甲位于赤道,不偏转,A错。
乙、丙位于北半球,B、C错。
点睛:赤道全年昼夜等长,纬度越高,昼夜长短的变化幅度越大。
南昌十中2017--2018年上学期高一第一次月考高一生物科试卷说明:本试卷分第I卷(选择题)和第Ⅱ卷(非选择题)两部分,全卷满分100分。
考试用时60分钟。
一.选择题(本大题共25题,每小题2分,共计50分)1、下列说法不正确的是()A.没有细胞结构的病毒,其生命活动也离不开细胞B.变形虫的细胞能完成各种生命活动C.多细胞生物的生命活动由不同的细胞密切合作完成D.生命系统的各层次层层相依,个体是生命系统最基本的层次2、下列关于细胞学说的建立过程及内容要点,叙述错误的是()A.细胞是一个有机体,一切生物都由细胞发育而来,并由细胞和细胞产物所构成B.细胞学说论证了生物界的统一性C.施莱登和施旺是细胞学说的建立者D.魏尔肖总结提出:细胞通过分裂产生新细胞3. 下列各组合中,能体现生命系统的层次由简单到复杂的正确顺序是()①肝脏②血液③神经元④蓝藻⑤细胞内各种化合物⑥病毒⑦同一片草地上的所有山羊⑧某池塘中的所有鱼⑨一片森林⑩某农田中的所有生物A.⑤⑥③②①④⑦⑩⑨ B.③②①④⑦⑩⑨C.③②①④⑦⑧⑩⑨ D.⑤②①④⑦⑩⑨4. ①②③④⑤是有关显微镜的几个操作步骤。
如图所示是在显微镜下观察到的几何图形,要将图甲转化为图乙,所列A、B、C、D四种操作顺序中,正确的是()①转动粗准焦螺旋②转动细准焦螺旋③调节光圈④转动转换器⑤移动玻片A. ⑤④①② B.④⑤③② C.⑤④③② D.①④⑤③5.下列有关显微镜操作的叙述,不正确的是()A.高倍镜下观察到细胞质流向是逆时针的,则细胞质实际流向应是顺时针的B.将低倍镜换上高倍镜时,一个视野内细胞数目减少,体积变大,视野变暗C.为观察低倍镜视野中位于左下方的细胞,应将装片向左下方移动,再换用高倍镜D.显微镜目镜为10×、物镜为10×时,视野中被相连的256个分生组织细胞所充满,若物镜转换为40×后,则在视野中可检测到的分生组织细胞数为16个6下列生物中具有拟核的一组是()①乳酸菌②酵母菌③草履虫④小球藻⑤发菜⑥念珠藻⑦葡萄球菌⑧霉菌⑨紫菜A.①⑤⑥⑨B. ②⑤⑥⑦C. ①⑤⑥⑦D.②③⑥⑧7. 两种细胞分别拥有如下表所示的特征,下列说法不正确的是()AB.细胞Ⅱ是真核细胞,可能是植物的根尖细胞C.细胞Ⅱ所属的真核生物一定不能进行光合作用D.细胞Ⅱ在化石记录中出现的时间比细胞Ⅰ晚8.下列有关原核细胞与真核细胞的叙述中,错误的是()A.原核细胞具有与真核细胞相似的细胞膜和细胞质B.蓝藻和绿藻细胞中一般都含有核糖体C.蓝藻和绿藻细胞中一般都含有叶绿体D.真核细胞和原核细胞中一般都含有DNA9.科学家在利用无土栽培法培养一些名贵花卉时,培养液中添加了多种必需化学元素。
2016-2017学年江西省南昌十中高一(下)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)What will the two speakers do this morning?A.Do some shopping.B.Go fishing.C.Go skating.2.(1.5分)Why does the woman give a gift to the man?A.It is his birthday.B.He is leavingC.He is going abroad.3.(1.5分)What does the woman tell the man to do?A.Read more books.B.Buy some books.C.Have dinner.4.(1.5分)What will the two speakers do?A.Cry for help.B.Find out what is happening in the room.C.Open the door by force.5.(1.5分)Where are the two speakers now?A.In a hospital.B.In a restaurant.C.At home.6.(3分)听第6段材料,回答第6至7题.6.Who was badly hurt in the traffic accident?A.The bike rider.B.The truck driver.C.The car driver.7.How do the two speakers get the news?A.On TV.B.On the radio.C.In a newspaper.7.(4.5分)听第7段材料,回答第8 至10题.8.Why doesn't Tommy want to get up?A.He is too tired.B.He is too lazy.C.It is too cold.9.What won't Tommy do before breakfast?A.Do morning exercise.B.Wash his face.C.Brush his teeth.10.What time will Tommy have his breakfast?A.At 6:30 am.B.At 6:45 amC.At 7:00 am.8.(4.5分)听第8段材料,回答第11至13题.11.What will the man do for his president?A.Type a report.B.Write a letter.C.Buy some clothes.12.What does the woman want the man to do first?A.Go shopping with her.B.Finish his report.C.See a film.13.Why does the woman want to buy some clothes for the man?A.It is cold now.B.It is his birthday tomorrow.C.The weather turns warm.9.(4.5分)听第9段材料,回答第14至16题.14.What does the woman want to do this afternoon after school?A.Watch a football match.B.Listen to a report.C.Play football on the playground.15.When will the football match be held?A.This afternoon.B.Next Saturday.C.This Saturday.16.What is the weather like when the conversation takes place?A.It is raining heavily.B.It is fine.C.It is cloudy.10.(6分)听第10段材料,回答第17至20题.17.What had Harrison planned to do?A.He had planned to settle down in England.B.He had planned to build a big house in England.C.He had planned to retire in the Mediterranean.18.What did Harrison do the moment he got back to England?A.He bought a nice house.B.He bought a car.C.He called on his old friends.19.What was the weather like in summer in England?A.It was very hot and rainy.B.It was rainy and cold.C.It was cool and sunny.20.Why did Harrison sell his new house?A.It was too small.B.It was too big.C.He didn't like the weather in England.第二部分语言知识运用(共两节,满分10分)第一节单项填空(共10小题;每小题1分,满分10分)11.(1分)The girl burst _________ when she got the news that she failed to be admitted into her ideal university.()A.out laughing B.into tears C.into crying D.out laughter12.(1分)I must warn you that our food is almost __________.We can't survive if there is no one to help us.()A.at an end B.at the end C.by the end D.in the end13.(1分)The English play______ the famous actor acted was a great success.()A.for which B.at which C.in which D.on which14.(1分)During my stay in Paris,my little son _______ some French by playing with the native boys and girls.()A.picked up B.picked out C.took up D.came across15.(1分)﹣Have you moved into your new house?﹣No,it __________ now.()A.is painting B.is paintedC.is being painted D.is being painting16.(1分)The little girl was ________ by the _______film.()A.frightened;frightening B.frightening;frightenedC.frightened;frightened D.frightening;frightening17.(1分)It is the third time that she has won the race,______ has surprised us all.()A.that B.where C.which D.what18.(1分)Some Chinese students find it difficult to understand native speakers when in London Exactly,______ they've learned a lot about grammar and known many words.()A.as if B.even if C.as though D.when19.(1分)You can't imagine ______when they received these nice Christmas presents.()A.how they were excited B.they were how excitedC.how excited were they D.how excited they were20.(1分)The teacher,along with all his students,______ visiting the famous museum now.()A.are B.is C.were D.was完形填空(共1小题;每小题1.5分,满分30分)21.(30分)In August 1999,Yuriko noticed that her daughter,Amy,was looking thin and pale.so she insisted that the 22﹣year﹣old (31) a doctor.As they waited for the rest results,the doctor (32)gave Yuriko a note while her daughter wasn't noticing.In the restroom,Yuriko read the note."It is stomach cancer,"said the doctor."(33)!There is no time."On September 21,Amy had a(n)(34).Three quarters of her stomach was removed.The doctor (35)the situation to Yuriko but the medical terms 术语)sounded like a foreign language.Amy was put on anti﹣cancer drugs,and over the next three months,she(36)from sideeffects,and lost seven kilograms.Yuriko decided to do more to(37)her daughter.She read all kinds of books on cancer.As a single mother,she had no one to (38)her pressure with.Though there were a lot of (39),Yuriko was able to help her daughter.When Amy started experiencing breathing difficulties,Yuriko(40)if it could be a side effect of the anti﹣cancer drug.She told Amy's doctor and he (41)to take her off the drug.Finally,in November 2002,Amy's treatment came to an end.(42)she felt her pain (43),Yuriko couldn't forget how lost and(44)she felt during her daughter's treatment.She wrote a letter to the local newspaper(45)the creation of a support group for cancer patients.Phone calls and letters (46)her idea started pouring in.In December 2002,Yuriko formally (47)Ikkikai,roughly meaning"sharing the joy",(48)provide hope and information for people with cancer,and their families.Ikkikai's message has begun to (49).Yuriko says,"The simple act of talking to other people who understand your problems can make the greatest (50).I hope that more people would join the group."31.A.watch B.see C.find D.ask 32.A.nervously B.kindly C.secretly D.weakly 33.A.Imagine B.Guess C.Hurry D.Look 34.A.preparation B.operation C.decoration D.examination 35.A.explained B.expressed C.informed D.repeated 36.A.suffered B.survived C.learnt D.escaped 37.A.please B.encourage C.help D.comfort 38.A.deal B.share C.take D.fight 39.A.difficulties B.decisions C.jobs D.tasks 40.A.realized B.discussed C.recognized D.wondered 41.A.meant B.regretted C.agreed D.preferred 42.A.Because B.Although C.Even D.until 43.A.raised B.ended C.increased D.reduced 44.A.ashamed B.alone C.lonely D.helpless 45.A.suggesting B.mentioning C.demanding D.commanding 46.A.supporting B.considering C.caring D.opposing 47.A.joined B.discovered C.founded D.assisted48.A.in need of B.so that C.in honor of D.in order to 49.A.spread B.exist C.show D.continue 50.A.progress B.difference C.significance D.effect第三部分阅读理解(共两节,满分40分)22.(8分)The Fourth"21th Century Cup"National English Speaking Competition is tobe held in Shanghai.Organizers:China Daily and Shanghai Broadcasting Network.Coordinator(协调者):China University English Speaking Association (CUESA).Co ﹣sponsors (联办单位):English Speaking Union (ESU),Lotus Software (China)Co.LtD.,Times Publishing Group of Singapore,Hilton Shanghai,Pearson Education,Foreign Language Teaching & Research Press.Date:March 26 (Friday),2017Place:Hilton Shanghai.Competition Format (形式):Each student will present a prepared speech on the given topic,followed by a three﹣minute off﹣hand speech and a three﹣minute question and answer periodwith the judges.Prepared speech period:six minutes.Q & A period:three minutes.Speech topic:People and Nature:In search of harmony (和谐)in a new age+your personal opinion.(Topics for the off﹣hand speech will be given on the day of competition).Prizes:Besides books and certificates (证书),the top two winners will be offered scholarships to travel to the annual international English﹣speaking competition which will beheld by the English Speaking Union in London in May,2017.The third and fourth place winners will be offered a study trip to Singapore,sponsored by the Times Publishing Group.The fifth through 10th place winners will be offered cash prizes.All the competitors will receive certificates from the English Speaking Union and book prizes provided by Pearson Education and Foreign Language Teaching & Research Press.The teachers of the top winners will also receive a one﹣year membership to the International Association of Teachers of English as a Foreign Language (IATEFL).51.The main purpose of this passage is.A.to help to improve your spoken EnglishB.to invite you to take part in the competitionC.to inform you of some information about the competitionD.to show you how to win the competition52.Which of the following is NOT mentioned about the Shanghai English Speaking Competition?A.The number of its competitors.B.It's competition format.C.What each winner will be offered.D.Where and when it will take place.53.Suppose you get the sixth place,you'll.A.travel to London for free.B.become a one﹣year member of IATEFLC.get some money,some books and a certificateD.get a chance to study in Singapore54.According to the passage,an"off﹣hand speech"is.A.a speech not longer than three minutesB.a speech with a piece of paper in handC.a speech which is well preparedD.a speech without preparation.23.(6分)Beauty is only skin deep.This expression means that beauty is only a surface quality.And some beauty products can cause lasting damage that goes far below the surface of the skin.But people,especially women,will risk a lot for beauty.In the United States,many people use sunlight or non﹣natural light to darken their skin.But in the African country of Senegal(塞内加尔),lots of women take health risks trying to do the opposite.The World Health Organization says that 25percent of Senegalese women use skin﹣lightening products on a usual basis.These products can have chemicals which can burn the eyes and skin and possibly cause cancers.A beauty supply shop in Dakar has many kinds of skin﹣lightening products.The shop worker Adama Diagne advises her visitors to avoid the stronger products thatpromise fast results but bring more damage.She says that some women want to lighten their skin every day.But for her,she likes to be a shade of brown.Women in Senegal say they lighten their skin for the same reasons women all over the world make changes to their appearance.They want to look beautiful,to find a husband,to stand out in a crowd or to simply look great for a special event.This thinking troubles Senegalese filmmaker Khardiata Pouye Sall.So she made This Color That Bothers Me,a film about the subject of skin lightening."I used the most shocking images so that women would see the dangers.It is hard to understand why a woman would tell herself that dark skin is not beautiful."Ms.Sall says the government needs to better control the marketing and sale of skin lighteners.But she adds education is the best way to persuade people against using them.55.The underlined part"the opposite"in Para.2refers to the fact that.A.some women pay more attention to skin cancersB.some women think the black skin beautifulC.some women want to get lasting beautyD.some women use products to lighten their skin56、Which of the following does Adama Diagne probably agree with?A.Chemicals should not be used in beauty products.B.The stronger the products are,the more damage they do.C.Beauty products can really improve women's skin quality.D.No effort could be too much for women to become beautiful.57、Ms.Sall made the film This Color That Bothers Me to.A.show the true meaning of beautyB.educate people on the dangers of beauty productsC.tell people how to use beauty products properlyD.explain how beauty products damage the skin.24.(8分)"Our aim is to take our art to the world and make people understand what it is to move,"said David Belle,the founder of park our (also called free running﹣a kind of extreme sport).Do you love running?It is a good exercise,yet many people find it boring.But what if making your morning jog a creative one?Like jumping from walls and over gaps,and ground rolls?Just like the James Bond in the movie Casino Royale?Bond jumps down from a roof to a windowsill and then runs several blocks over obstacles on the way.It is just because of Bond's wonderful performances that the sport has become popular worldwide.Yes,that's park our,an extreme street sport aimed at moving from one point to another as quickly as possible,getting over all the obstacles in the path using only the abilities of thehuman body.Park our is considered an extreme sport.As its participants dash around a city,they may jump over fences,run up walls and even move from rooftop to rooftop.Park our can be just as exciting and charming as it sounds,but its participants see parkour much more than that.Overcoming all the obstacles on the course and in life is part of thephilosophy(理念)behind parkour.This is the same as life.You must determine your destination,go straight,jump over all the barriers as if in parkour and never fall back from them in your life,to reach the destination successfully.A parkour lover said,"I love parkour because its philosophy has become my life,my way to do everything."Another philosophy we've learnt from parkour is freedom.It can be done by anyone,at any time,anywhere in the world.It is a kind of expression of trust in yourself that you earn energy and confidence.58.Parkour has become popular throughout the world because of.A.its risks and tricksB.its founder,David BelleC.the film,Casino RoyaleD.the varieties of participants59.The underlined word"obstacles"in Paragraph 2is closest in meaning to"".A.barriersB.objectsC.roofsD.streets60.Which of the following is true of parkour?A.It is a good but boring sport.B.It challenges human abilities.C.It needs special training.D.It is a team sport.61.Which of the following is the philosophy of parkour?A.Sports and extremes.B.Excitement and popularity.C.Dreams and success.D.Determination and freedom.25.(8分)Watching some children trying to catch butterflies on a hot August afternoon,I was reminded of an incident in my own childhood.When I was a boy of 12in South Carolina,something happened to me that cured me forever of wanting to put any wild creature in a cage.We lived on the edge of a wood,and every evening at dusk the mockingbirds would come and rest in the trees and sing.There isn't a musical instrument made by man that can produce a more beautiful sound than the song of the mockingbird.I decided that I would catch a young bird and keep it in a cage and in that way would have my own private musician.I finally succeeded in catching one and put it in a cage.At first,in its fright at being captured,the bird fluttered about the cage,but eventually it settled down in its new home.I felt very pleased with myself and looked forward to some beautiful singing from my tiny musician.I had left the cage out on our back porch,and on the second day of the bird's captivity my new pet's mother flew to the cage with food in her mouth.The babybird ate everything she brought to it.I was pleased to see this.Certainly the mother knew better than I how to feed her baby.The following morning when I went to see how my captive(被俘虏的)was doing,I discovered it on the floor of the cage,dead.I was shocked!What had happened!I had taken excellent care of my little bird,or so I thought.Arthur Wayne,the famous ornithologist,happened to be visiting my father at the time,hearing me crying over the death of my bird,explained what had occurred."A mother mocking bird,finding her young in a cage,will sometimes bring it poison berries.She thinks it better for her young to die than to live in captivity."Never since then have I caught any living creature and put it in a cage.All living creatures have a right to live free.62.Why did the writer catch a mockingbird when he was a boy of 12?A.He wanted it to sing for him.B.He liked its beautiful feather.C.He had just got a new cage.D.He wanted a pet for a companion.63.The mockingbird died because it.A.was frightened to deathB.ate the poisonous food its mother gave itC.refused to eat anythingD.drank the poisonous water by mistake64.An ornithologist probably means.A.a religious personB.a kind personC.a schoolmasterD.an expert in birds65.What is the most important lesson the writer learned from the incident?A.Be careful about food you give to baby birds.B.All birds put in a cage won't live long.C.You should keep the birds from their mother.D.Freedom is very valuable to all creatures.26.(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.People who are confident really seem to be naturally outstanding and just seem to do everything with more styles than others.(66)It is a habit that everyone can develop in life.Try these simple tips to drill and build up your confidence:1.Admit your shortcomings calmly.Do not try to escape from them or cover them.Face them bravely.(67)Fight against them every day until the day when you could break away and conquer them.2.Dress for self﹣confidence.(68)And therefore pay attention to your dress,display your unique physical advantages and exhibit your best image.In addition,in formal occasions such as a business party or a wedding ceremony,elegant dressing contributes to building your confidence.3.(69)You should get rid of your habit that deals with the work passively(被动地).Concentrate your efforts immediately on overcoming it,because it will make your restless mind at ease and build your self﹣confidence.4.Be positive.Feel pity neither on yourself nor on others.If you are used to hating and blaming yourself,others would tend to do that and believe it.Instead,you should speak positively about yourself,your progress,and your bright future.(70)A.Build your confident vocabulary.B.Don't judge a person by appearance.C.Actually,true self﹣confidence is neither born nor acquired overnight.D.Your appearance could put you into embarrassment or increase your confidence.E.Don't put off what you eventually have to do.F.By doing so,you would encourage your growth in a positive direction.G.Then talk about them to a reliable mate,a friend or a family member.四、第四部分写作(共三节,满分20分)27.(1分)根据首字母、中文或括号里的词,运用恰当形式补全句子.You need(permit)from the World Wide Web before you can access information.28.(1分)Seeing such extraordinary(beautiful),I think every cell in my body woke up.29.(1分)73.She speaks English so well that all her classmates a her very much.30.(1分)74.It is o that he failed the exam;he looks so sad.31.(1分)75.He a to the teacher for not arriving on time just now.32.(1分)76.Jack agreed to enter the"ghost room"with his brother just out of c.33.(1分)77.The teacher advised us not to c the dictionary for every new word.34.(1分)78.Mr.Wang is good at c education with pleasure.35.(1分)79.As we all know,eating more vegetables b our health a lot.36.(1分)80.Lily is going on a diet these days.Don't you see that she is much s than before?37.(10分)根据课文内容以及所给提示,在空白处填入适当的内容(1个单词),或者括号内单词的正确形式填空.Festivals and celebrations of all kinds (81)(held)everywhere since ancient times.Different festivals are celebrated for different reasons.Some festivals are held to honor the dead or (82)(satisfy)the ancestors,who might return either to help (83)to do harm.Festivals can also be held to honor the famouspeople.In the USA,Columbus Day is in memory of the (84)(arrive)of Christopher Columbus in the New World.India had a (85)(nation)festival to honor Gandhi,who helped gain India's(86)(independent)from Britain.Harvest and Thanksgiving festivals can be very happy events.People are very grateful because their food (87)(gather)and the agricultural work is over.In China,(88)most energetic and important festival isthe Spring Festival,which looks forward to the end of the winter and the coming of spring.In Japan,there is Cherry Blossom Festival.The country,(89)(cover)with cherry tree flowers,looks as though it is covered with pink snow.Festivals let us enjoy life,be (90)(pride)of our customs and forget our work for a little while.书面表达38.(20分)假定你是李华,你原本和Peter约好一起去观看一场足球比赛,但是昨天下午你突然接到母亲的电话说她要来看你,结果你没能按时赴约.为此你感到十分抱歉.请根据以下要点给Peter写一封信,以表歉意.主要内容包括:1.表达歉意;2.解释没能去的原因;3.希望下周三邀请Peter吃饭作为补偿.注意:1.词数100左右;2.可以适当增加细节,以使行文连贯;3.信的开头和结尾已为你写好,但不计人总词数.Dear Peter,Yours'LiHua2016-2017学年江西省南昌十中高一(下)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)What will the two speakers do this morning?A.Do some shopping.B.Go fishing.C.Go skating.【解答】A2.(1.5分)Why does the woman give a gift to the man?A.It is his birthday.B.He is leavingC.He is going abroad.【解答】A3.(1.5分)What does the woman tell the man to do?A.Read more books.B.Buy some books.C.Have dinner.【解答】C4.(1.5分)What will the two speakers do?A.Cry for help.B.Find out what is happening in the room.C.Open the door by force.【解答】B5.(1.5分)Where are the two speakers now?A.In a hospital.B.In a restaurant.C.At home.【解答】B6.(3分)听第6段材料,回答第6至7题.6.Who was badly hurt in the traffic accident?A.The bike rider.B.The truck driver.C.The car driver.7.How do the two speakers get the news?A.On TV.B.On the radio.C.In a newspaper.【解答】CC7.(4.5分)听第7段材料,回答第8 至10题.8.Why doesn't Tommy want to get up?A.He is too tired.B.He is too lazy.C.It is too cold.9.What won't Tommy do before breakfast?A.Do morning exercise.B.Wash his face.C.Brush his teeth.10.What time will Tommy have his breakfast?A.At 6:30 am.B.At 6:45 amC.At 7:00 am.【解答】CAC11.What will the man do for his president?A.Type a report.B.Write a letter.C.Buy some clothes.12.What does the woman want the man to do first?A.Go shopping with her.B.Finish his report.C.See a film.13.Why does the woman want to buy some clothes for the man?A.It is cold now.B.It is his birthday tomorrow.C.The weather turns warm.【解答】AAB9.(4.5分)听第9段材料,回答第14至16题.14.What does the woman want to do this afternoon after school?A.Watch a football match.B.Listen to a report.C.Play football on the playground.15.When will the football match be held?A.This afternoon.B.Next Saturday.C.This Saturday.16.What is the weather like when the conversation takes place?A.It is raining heavily.B.It is fine.C.It is cloudy.【解答】ABB17.What had Harrison planned to do?A.He had planned to settle down in England.B.He had planned to build a big house in England.C.He had planned to retire in the Mediterranean.18.What did Harrison do the moment he got back to England?A.He bought a nice house.B.He bought a car.C.He called on his old friends.19.What was the weather like in summer in England?A.It was very hot and rainy.B.It was rainy and cold.C.It was cool and sunny.20.Why did Harrison sell his new house?A.It was too small.B.It was too big.C.He didn't like the weather in England.【解答】AABC第二部分语言知识运用(共两节,满分10分)第一节单项填空(共10小题;每小题1分,满分10分)11.(1分)The girl burst _________ when she got the news that she failed to be admitted into her ideal university.()A.out laughing B.into tears C.into crying D.out laughter【解答】答案:BA:突然大笑起来;B:突然大哭起来;C:无此表达,正确应为burst into tears;D:无此表达,正确应为burst into laughter.根据句意"听到她没能被意向的大学录取的消息时,这个女孩突然大哭起来"及所给选项分析可知本题答案为B选项.拓展:burst into sth/burst out doing sth"突然爆发出…".12.(1分)I must warn you that our food is almost __________.We can't survive if there is no one to help us.()A.at an end B.at the end C.by the end D.in the end【解答】答案:A at an end耗尽;at the end在…的末端;by the end到…为止;in the end最后;句意表达的是食物几乎要耗尽了.故选A.13.(1分)The English play______ the famous actor acted was a great success.()A.for which B.at which C.in which D.on which【解答】答案是C.考查定语从句,被修饰的先行词是play,在定语从句the famous actor acted 中做状语,在戏剧中"in the play"which指代play,填空内容应该是in which,所以答案选C.14.(1分)During my stay in Paris,my little son _______ some French by playing with the native boys and girls.()A.picked up B.picked out C.took up D.came across【解答】答案:A考查动词短语:A.picked up 学会、捡起;B.picked out 挑选出;C.took up 占据、从事;D.came across 偶然遇到.本句句意:我们在法国巴黎逗留期间,我的小儿子在和当地的小朋友玩耍的过程中学会了一些法语.结合选项,将答案带入句中,选A.15.(1分)﹣Have you moved into your new house?﹣No,it __________ now.()A.is painting B.is paintedC.is being painted D.is being painting【解答】答案:C.根据now可知本句为现在进行时态,而new house和paint 是被动关系,故使用现在进行时的被动语态,即is being painted,故选C.16.(1分)The little girl was ________ by the _______film.()A.frightened;frightening B.frightening;frightenedC.frightened;frightened D.frightening;frightening【解答】答案是A.本题考查分词的用法.题干中主要考查frightened 和frightening的区别:说明"(人)感到…害怕的状态"用过去分词,说明"(物)令人…害怕的特性"用现在分词;所以第一空中用frightened表示"小女孩感到害怕",第二空用frightening表示film令人害怕的特性,所以答案选择A.17.(1分)It is the third time that she has won the race,______ has surprised us all.()A.that B.where C.which D.what【解答】答案:C分析句子结构可知,"_____has surprised us all"为非限制性定语从句,修饰前面的整个句子,因为what不能引导定语从句,所以排除;另外本句关系词在从句中作主语,where引导定语从句时只充当地点状语,所以排除B;that不能引导非限制性定语从句,排除A.故答案为C.18.(1分)Some Chinese students find it difficult to understand native speakers when in London Exactly,______ they've learned a lot about grammar and known many words.()A.as if B.even if C.as though D.when【解答】答案:Bas if好像;even if尽管,即使;as though好像;when当…时候;句意表达的是让步,要用even if来连接让步状语从句.故选B.19.(1分)You can't imagine ______when they received these nice Christmas presents.()A.how they were excited B.they were how excitedC.how excited were they D.how excited they were【解答】D 本题考查由感叹句变来的宾语从句.此时,位于从句句首的应为连接词how,形容词或副词紧随其后,然后是正常的陈述句语序.根据这一要求,本题应该选择D.20.(1分)The teacher,along with all his students,______ visiting the famous museum now.()A.are B.is C.were D.was【解答】答案:B.考查主谓一致.在英语中,主语后有along with,as well as,except,in addition to,instead of,rather than,together with,with 等引起的介词短语或其他从属结构时,谓语动词的单复数同主语一致,不受其后面名词单复数的影响,再根据句中的单词now可知答案选B.完形填空(共1小题;每小题1.5分,满分30分)21.(30分)In August 1999,Yuriko noticed that her daughter,Amy,was looking thin and pale.so she insisted that the 22﹣year﹣old (31)B a doctor.As they waited for the rest results,the doctor (32)C gave Yuriko a note while her daughter wasn't noticing.In the restroom,Yuriko read the note."It is stomach cancer,"said the doctor."(33)C!There is no time."On September 21,Amy had a(n)(34)B.Three quarters of her stomach was removed.The doctor (35)A the situation to Yuriko but the medical terms 术语)sounded like a foreign language.Amy was put on anti﹣cancer drugs,and over the next three months,she(36)A from sideeffects,and lost seven kilograms.Yuriko decided to do more to(37)C her daughter.She read all kinds of books on cancer.As a single mother,she had no one to (38)B her pressure with.Though there were a lot of (39)A,Yuriko was able to help her daughter.When Amy started experiencing breathing difficulties,Yuriko(40)D if it could be a side effect of the anti﹣cancer drug.She told Amy's doctor and he (41)C totake her off the drug.Finally,in November 2002,Amy's treatment came to an end.(42)B she felt her pain (43)D,Yuriko couldn't forget how lost and(44)C she felt during her daughter's treatment.She wrote a letter to the local newspaper(45)A the creation of a support group for cancer patients.Phone calls and letters (46)A her idea started pouring in.In December 2002,Yuriko formally (47)C Ikkikai,roughly meaning"sharing the joy",(48)D provide hope and information for people with cancer,and their families.Ikkikai's message has begun to (49)A.Yuriko says,"The simple act of talking to other people who understand your problems can make the greatest (50)B.I hope that more people would join the group."31.A.watch B.see C.find D.ask 32.A.nervously B.kindly C.secretly D.weakly 33.A.Imagine B.Guess C.Hurry D.Look 34.A.preparation B.operation C.decoration D.examination 35.A.explained B.expressed C.informed D.repeated 36.A.suffered B.survived C.learnt D.escaped 37.A.please B.encourage C.help D.comfort 38.A.deal B.share C.take D.fight 39.A.difficulties B.decisions C.jobs D.tasks 40.A.realized B.discussed C.recognized D.wondered 41.A.meant B.regretted C.agreed D.preferred 42.A.Because B.Although C.Even D.until 43.A.raised B.ended C.increased D.reduced 44.A.ashamed B.alone C.lonely D.helpless 45.A.suggesting B.mentioning C.demanding D.commanding 46.A.supporting B.considering C.caring D.opposing 47.A.joined B.discovered C.founded D.assisted48.A.in need of B.so that C.in honor of D.in order to。
2017-2018学年江西省南昌市第十中学高一上学期第二次月考数学试题一、单选题1.下列各角中与60︒角终边相同的角是( )A. -300°B. -60°C. 600°D. 1 380° 【答案】A【解析】与60︒角终边相同的角为: 60360k,k Z ︒+︒∈. 当k 1=-时,即为-300°. 故选A.2.代数式sin120cos210 的值为( ) A. 34-B. 4C. 32-D. 14 【答案】A【解析】试题分析:由诱导公式以可得,sin120°cos210°=sin60°×(-cos30°)=-2×2=34-,选A.【考点】诱导公式的应用.3.已知扇形的面积为2cm 2,扇形圆心角θ的弧度数是4,则扇形的周长为( ) (A)2cm (B)4cm (C)6cm (D)8cm 【答案】C【解析】设扇形的半径为R,则R 2θ=2,∴R 2=1⇒R=1,∴扇形的周长为2R+θ·R=2+4=6(cm). 4.函数y =)A. [)1,+∞ B. 2,3⎛⎫+∞ ⎪⎝⎭ C. 2,13⎡⎤⎢⎥⎣⎦ D. 2,13⎛⎤ ⎥⎝⎦【答案】D【解析】试题分析: 因为要使函数y =有意义,则满足()12320320{ { l o g 320321x x x x ->->∴-≥-≤,解得x 的取值范围是2,13⎛⎤⎥⎝⎦,选D. 【考点】本题主要考查了函数定义域的求解问题的运用。
点评:解决该试题的关键是理解对数真数大于零,同时偶此根式下被开方数为非负数,并且从内向外依次保证表达式有意义即可。
易错点就是忽略对数真数大于零这个前提条件。
5.sin 4y x π⎛⎫=-⎪⎝⎭的图象的一个对称中心是( ) A. (),0π- B. 3,04π⎛⎫- ⎪⎝⎭ C. 3,02π⎛⎫ ⎪⎝⎭ D. ,02π⎛⎫⎪⎝⎭【答案】B【解析】函数sin 4y x π⎛⎫=-⎪⎝⎭,令k π,k Z 4x π-=∈.即得对称中心k π,0,k Z 4π⎛⎫+∈⎪⎝⎭. 当k 1=-时,即为3,04π⎛⎫- ⎪⎝⎭.故选B.6.已知()cos ,2n f n n Z π=∈,则()()()()1232017f f f f ++++= ( ) A. 1- B. 0 C. 1 D. 2【答案】B【解析】由()cos,2n f n n Z π=∈知 当4k 3n =-时, ()3π43cos 02f k ⎛⎫-=-= ⎪⎝⎭; 当4k 2n =-时, ()()42cos π1f k -=-=-; 当4k 1n =-时, ()π41cos 02f k ⎛⎫-=-= ⎪⎝⎭; 当4k n =时, ()4cos01f k ==;有: ()()()()4?41?42?430f k f k f k f k ++++++=所以()()()()()()123201750402017010f f f f f f ++++=⨯+=+= . 故选B.7.已知()62log f x x =,那么()8f 等于( )A.43 B. 8 C. 18 D. 12【答案】D【解析】由()62log f x x =得()618log 2f f ⎡⎤===⎢⎥⎣⎦.故选D.8.将函数()sin 2y x ϕ=+的图象沿x 轴向左平移8π个单位后,得到一个偶函数的图象,则ϕ的一个可能取值为( )A.34π B. 4π C. 0 D. 4π- 【答案】B【解析】试题分析:由题意得sin 2sin 284y x x ππϕϕ⎛⎫⎛⎫⎛⎫=++=++ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭关于y 轴对称,所以()(),,424k k Z k k Z πππϕπϕπ+=+∈=+∈ ϕ的一个可能取值为4π,选B.【考点】三角函数图像变换【思路点睛】三角函数的图象变换,提倡“先平移,后伸缩”,但“先伸缩,后平移”也常出现在题目中,所以也必须熟练掌握.无论是哪种变形,切记每一个变换总是对字母x 而言. 函数y =Asin(ωx +φ),x ∈R 是奇函数⇔φ=kπ(k ∈Z);函数y =Asin(ωx +φ),x ∈R 是偶函数⇔φ=kπ+(k ∈Z);函数y =Acos(ωx +φ),x ∈R 是奇函数⇔φ=kπ+(k ∈Z);函数y =Acos(ωx +φ),x ∈R 是偶函数⇔φ=kπ(k ∈Z); 9.已知函数()()2sin f x x ωϕ=+对任意x 都有,66f x f x ππ⎛⎫⎛⎫+=- ⎪ ⎪⎝⎭⎝⎭则6f π⎛⎫⎪⎝⎭等于( )A. 2B. 0C. 2-或2D. 2- 【答案】C【解析】因为函数()()2sin f x x ωϕ=+对任意x 都有,66f x f x ππ⎛⎫⎛⎫+=- ⎪ ⎪⎝⎭⎝⎭所以()f x 关于直线x 6π=对称.则6f π⎛⎫⎪⎝⎭为()()2sin f x x ωϕ=+的最大值或最小值,即26f π⎛⎫=- ⎪⎝⎭或2. 故选C.10.设()f x 是(),-∞+∞上的奇函数, ()()2f x f x +=-,当01x ≤≤时有()2f x x =,则()2015f =( )A. 1-B. 2-C. 1D. 2 【答案】B【解析】∵f (x +2)=−f (x ),得f (x +4)=f (x ), ∴周期为T =4,又∵函数为奇函数, ()2015f =f (504×4−1)=f (−1)=−f (1)=−2, 故选B11.已知如图是函数()2sin y x ωϕ=+其中||<2πφ的图象,那么( )A. ω=2, φ=6πB. ω=1011 , φ=-6πC. ω=1011, φ= 6πD. ω=2, φ=-6π【答案】A【解析】由图可知, 12sin 1sin 2ϕϕ==,. 又||<2πφ,所以π6ϕ=.又由图知: 11π2k π,k Z 126πω+=∈. 解得: 224,1111k k Z ω=-+∈. 又知最小正周期21112ππω>,所以2411ω<.所以12k ω==,. 故选A.点睛:由图象确定函数()()sin f x A x ωϕ=+解析式的方法 (1)A 由图象上的最高(低)点的纵坐标确定。
2017-2018学年江西省南昌十中高一(上)第二次月考数学试卷一、选择题(本大题共12题,每小题5分,共计60分.在每小题列出的四个选项中只有一项是最符合题目要求的)1.(5分)下列各角中,与60°角终边相同的角是()A.﹣300°B.﹣60°C.600° D.1380°2.(5分)代数式sin120°cos210°的值为()A.﹣ B.C.﹣ D.3.(5分)已知扇形的面积为2 cm2,扇形圆心角的弧度数是4,则扇形的周长为()A.2 B.4 C.6 D.84.(5分)函数的定义域是:()A.[1,+∞)B.C. D.5.(5分)函数的图象的一个对称中心是()A.(﹣π,0)B.C.D.6.(5分)已知,则f(1)+f(2)+f(3)+…+f(2017)=()A.﹣1 B.0 C.1 D.27.(5分)已知f(x6)=log2x,那么f(8)等于()A.B.8 C.18 D.8.(5分)函数y=sin(2x+φ)的图象沿x轴向左平移个单位后,得到一个偶函数的图象,则φ的一个可能的值为()A. B.C.0 D.9.(5分)已知函数f(x)=2sin(ωx+φ)对任意x都有,则等于()A.2或0 B.﹣2或2 C.0 D.﹣2或010.(5分)设f(x)是(﹣∞,+∞)上的奇函数,f(x+2)=﹣f(x),当0≤x ≤1时有f(x)=2x,则f(2015)=()A.﹣1 B.﹣2 C.1 D.211.(5分)已知如图示是函数y=2sin(ωx+φ)(|φ|<)的图象,那么()A. B.C.D.12.(5分)已知函数f(x)=sin(2x+φ),其中φ为实数,若f(x)≤|f()|对x∈R恒成立,且f()>f(π),则f(x)的单调递增区间是()A.[kπ﹣,kπ+](k∈Z)B.[kπ,kπ+](k∈Z)C.[kπ+,kπ+](k∈Z)D.[kπ﹣,kπ](k∈Z)二、选择题(本大题共4题,每小题5分,共计20分.把答案填在答题纸的横线上)13.(5分)已知角α的终边经过点,则cosα=.14.(5分)函数y=﹣cos(﹣)的单调递增区间是.15.(5分)若函数f(x)=|4x﹣x2|﹣a的零点个数为3,则a=.16.(5分)已知函数f(x)=﹣3x﹣x3,x∈R,若时,不等式f(cos2θ﹣2t)+f(4sinθ﹣3)≥0恒成立,则实数t的取值范围是.三、解答题(本大题共6题,共计70分.解答应写出文字说明、证明过程或演算步骤)17.(10分)已知函数的最小正周期为π.(1)求f(π);(2)先列表,再利用“五点法”在给定的坐标中作出函数f(x)的简图.18.(12分)已知tanα=2.(1)求的值;(2)求的值;(3)若α是第三象限角,求cosα的值.19.(12分)已知0<θ<π,且,求(1)sinθ﹣cosθ的值;(2)tanθ的值.20.(12分)已知函数f(x)=Asin(ωx+φ),x∈R,(其中A>0,ω>0,0<φ<)的图象与x轴的交点中,相邻两个交点之间的距离为,且图象上一个最低点为M(,﹣2).(1)求f(x)的解析式;(2)将函数f(x)的图象向右平移个单位后,再将所得图象上各点的横坐标缩小到原来的,纵坐标不变,得到y=g(x)的图象,求函数y=g(x)的解析式.21.(12分)已知.(1)求f(x)的单调递增区间;(2)当时,f(x)的最大值为4,求a的值;(3)在(2)的条件下,求满足f(x)=1且x∈[﹣π,π]的x的取值集合.22.(12分)已知函数f(x)=ln2x﹣2aln(ex)+3,x∈[e﹣1,e2](1)当a=1时,求函数f(x)的值域;(2)若f(x)≤﹣alnx+4恒成立,求实数a的取值范围.2017-2018学年江西省南昌十中高一(上)第二次月考数学试卷参考答案与试题解析一、选择题(本大题共12题,每小题5分,共计60分.在每小题列出的四个选项中只有一项是最符合题目要求的)1.(5分)下列各角中,与60°角终边相同的角是()A.﹣300°B.﹣60°C.600° D.1380°【分析】与60°终边相同的角一定可以写成k×360°+60°的形式,k∈z,检验各个选项中的角是否满足此条件.【解答】解:与60°终边相同的角一定可以写成k×360°+60°的形式,k∈z,令k=﹣1 可得,﹣300°与60°终边相同,故选:A.【点评】本题考查终边相同的角的特征,凡是与α 终边相同的角,一定能写成k ×360°+α,k∈z的形式.2.(5分)代数式sin120°cos210°的值为()A.﹣ B.C.﹣ D.【分析】原式中的角度变形后,利用诱导公式及特殊角的三角函数值计算即可得到结果.【解答】解:原式=sin(180°﹣60°)cos(180°+30°)=﹣sin60°cos30°=﹣×=﹣.故选:A.【点评】此题考查了运用诱导公式化简求值,熟练掌握诱导公式是解本题的关键.3.(5分)已知扇形的面积为2 cm2,扇形圆心角的弧度数是4,则扇形的周长为()A.2 B.4 C.6 D.8【分析】根据扇形的面积公式建立等式关系,求出半径,以及弧长公式求出弧长,再根据扇形的周长等于2个半径加弧长即可求出周长.【解答】解:设扇形的半径为R,则R2α=2,∴R2=1,∴R=1,∴扇形的周长为2R+α•R=2+4=6故选:C.【点评】本题主要考查了扇形的面积公式,以及扇形的周长和弧长等有关基础知识,属于基础题.4.(5分)函数的定义域是:()A.[1,+∞)B.C. D.【分析】无理式被开方数大于等于0,对数的真数大于0,解答即可.【解答】解:要使函数有意义:≥0,即:可得0<3x﹣2≤1解得x∈故选:D.【点评】本题考查对数函数的定义域,考查学生发现问题解决问题的能力,是基础题.5.(5分)函数的图象的一个对称中心是()A.(﹣π,0)B.C.D.【分析】利用正弦函数的对称性质可知,x﹣=kπ⇒x=kπ+,从而可得其对称中心为(kπ+,0),k∈Z.,再赋值即可得答案.【解答】解:由x﹣=kπ,得:x=kπ+,k∈Z.所以函数的图象的对称中心为(kπ+,0),k∈Z.当k=﹣1时,就是函数的图象的一个对称中心,故选:B.【点评】本题考查正弦函数的对称性,求得其对称中心为(kπ+,0)是关键,考查赋值法的应用,属于中档题.6.(5分)已知,则f(1)+f(2)+f(3)+…+f(2017)=()A.﹣1 B.0 C.1 D.2【分析】根据余弦函数的周期性,得出f(n)是以4为周期的函数,求值即可.【解答】解:,则f(1)+f(2)+f(3)+…+f(2017)=cos+cosπ+cos+cos2π+…+cos=0﹣1+0+1+…+0=0×504+0=0.故选:B.【点评】本题考查了余弦函数的周期性与函数值的计算问题,是基础题.7.(5分)已知f(x6)=log2x,那么f(8)等于()A.B.8 C.18 D.【分析】考查f(x6)=log2x的形式,把f(8)化为f(x6)的形式,即可.【解答】解:∵f(x6)=log2x,∴f(8)=故选:D.【点评】本题考查函数的含义,是基础题;本题也可以先求函数f(x)的解析式,代入求值即可.8.(5分)函数y=sin(2x+φ)的图象沿x轴向左平移个单位后,得到一个偶函数的图象,则φ的一个可能的值为()A. B.C.0 D.【分析】利用函数y=Asin(ωx+φ)的图象变换可得函数y=sin(2x+φ)的图象沿x轴向左平移个单位后的解析式,利用其为偶函数即可求得答案.【解答】解:令y=f(x)=sin(2x+φ),则f(x+)=sin[2(x+)+φ]=sin(2x++φ),∵f(x+)为偶函数,∴+φ=kπ+,∴φ=kπ+,k∈Z,∴当k=0时,φ=.故φ的一个可能的值为.故选:B.【点评】本题考查函数y=Asin(ωx+φ)的图象变换,考查三角函数的奇偶性,属于中档题.9.(5分)已知函数f(x)=2sin(ωx+φ)对任意x都有,则等于()A.2或0 B.﹣2或2 C.0 D.﹣2或0【分析】函数f(x)=2sin(ωx+φ)对任意x都有,说明故取最大值或者是最小值,由解析式得出即可其值【解答】解:∵函数f(x)=2sin(ωx+φ)对任意x都有∴函数图象的对称轴是,∴取最大值或者是最小值∵函数的最大值是2,最小值是﹣2∴等于﹣2或2故选:B.【点评】本题考查由y=Asin(ωx+φ)的部分图象确定其解析式,解题的关键是根据函数图象的对称性判断出函数的最值.10.(5分)设f(x)是(﹣∞,+∞)上的奇函数,f(x+2)=﹣f(x),当0≤x ≤1时有f(x)=2x,则f(2015)=()A.﹣1 B.﹣2 C.1 D.2【分析】利用抽象函数求出函数的周期,结合已知函数的解析式求解即可.【解答】解:∵f (x+2)=﹣f (x),得f (x+4)=f (x),∴周期为T=4,又∵函数为奇函数,f (2015)=f (504×4﹣1)=f (﹣1)=﹣f (1)=﹣2,故选:B.【点评】本题考查函数的周期性的应用,函数的值的求法,考查计算能力.11.(5分)已知如图示是函数y=2sin(ωx+φ)(|φ|<)的图象,那么()A. B.C.D.【分析】利用x=0,y=1,结合φ的范围,求出φ的值,结合选项ω的值,确定函数的周期,利用图象判断正确选项.【解答】解:f(0)=1,即.得,又当对应的周期T为,又由图可知,且,故,于是有T=π,则ω=2,故选:D.【点评】本题考查选择题的解法,图象的应用能力,若非选择题,条件是不够的,不能由图得到周期的值,当然也不能得到ω的值.12.(5分)已知函数f(x)=sin(2x+φ),其中φ为实数,若f(x)≤|f()|对x∈R恒成立,且f()>f(π),则f(x)的单调递增区间是()A.[kπ﹣,kπ+](k∈Z)B.[kπ,kπ+](k∈Z)C.[kπ+,kπ+](k∈Z)D.[kπ﹣,kπ](k∈Z)【分析】由题意求得φ的值,利用正弦函数的性质,求得f(x)的单调递增区间.【解答】解:若f(x)≤|f()|对x∈R恒成立,则f()为函数的函数的最大值或最小值,即2×+φ=kπ+,k∈Z,则φ=kπ+,k∈Z,又f()>f(π),sin(π+φ)=﹣sinφ>sin(2π+φ)=sinφ,sinφ<0.令k=﹣1,此时φ=﹣,满足条件sinφ<0,令2x﹣∈[2kπ﹣,2kπ+],k∈Z,解得:x∈[kπ+,kπ+](k∈Z).则f(x)的单调递增区间是[kπ+,kπ+](k∈Z).故选:C.【点评】本题考查的知识点是函数y=Asin(ωx+φ)的图象变换、三角函数的单调性,属于基础题.二、选择题(本大题共4题,每小题5分,共计20分.把答案填在答题纸的横线上)13.(5分)已知角α的终边经过点,则cosα=﹣.【分析】由题意可得x=﹣1,y=,r==2,由此求得cosα=的值.【解答】解:∵角α的终边经过点,∴x=﹣1,y=,r==2,故cosα==﹣.【点评】本题主要考查任意角的三角函数的定义,属于基础题.14.(5分)函数y=﹣cos(﹣)的单调递增区间是[4kπ+,4kπ+],k∈Z.【分析】所求区间即为函数的单调递减区间,由2kπ≤≤2kπ+π解之即可.【解答】解:由复合函数的单调性可知:函数的单调递增区间即为函数的单调递减区间,由2kπ≤≤2kπ+π,解得4kπ+≤x≤4kπ+,k∈Z故所求函数的单调递增区间为:[4kπ+,4kπ+],k∈Z故答案为:[4kπ+,4kπ+],k∈Z【点评】本题考查三角函数的单调性,整体代入是解决问题的关键,属基础题.15.(5分)若函数f(x)=|4x﹣x2|﹣a的零点个数为3,则a=4.【分析】令f(x)=0,判断得到a>0,利用绝对值的代数意义化简,得到两个一元二次方程,由f(x)的零点个数为3,得到两方程共有3个解,即一个方程△>0,一个方程△=0,即可求出a的值.【解答】解:令f(x)=0,得到|4x﹣x2|﹣a=0,即|4x﹣x2|=a,可得4x﹣x2=a或4x﹣x2=﹣a,即x2﹣4x+a=0或x2﹣4x﹣a=0,若a=0,解得:x=0或x=4,只有两个解,舍去,∴a>0,由f(x)的零点个数为3,得到两方程共有3个解,即一个方程△>0,一个方程△=0,若x2﹣4x+a=0中的△=16﹣4a>0,即a<4;x2﹣4x﹣a=0的△=16+4a=0,即a=﹣4,不合题意,舍去;若x2﹣4x+a=0中的△=16﹣4a=0,即a=4;x2﹣4x﹣a=0的△=16+4a>0,即a>﹣4,满足题意,则a=4,故答案为:4【点评】此题考查了函数零点的判定定理,以及一元二次方程根的判别式,函数的零点的研究就可转化为相应方程根的问题,函数与方程的思想得到了很好的体现.16.(5分)已知函数f(x)=﹣3x﹣x3,x∈R,若时,不等式f(cos2θ﹣2t)+f(4sinθ﹣3)≥0恒成立,则实数t的取值范围是.【分析】先研究函数f(x)=﹣3x﹣x3,x∈R的单调性,求导既得,由不等式恒成立进行转化,再研究时cos2θ﹣2t与4sinθ﹣3取值范围,分离出参数t,利用三角函数的性质求其范围即得实数t的取值范围.【解答】解:由于f′(x)=﹣3﹣3x2<0恒成立,故函数函数f(x)=﹣3x﹣x3,x ∈R是一个减函数,由解析式可知,函数也是一个奇函数,又不等式f(cos2θ﹣2t)+f(4sinθ﹣3)≥0恒成立,故f(cos2θ﹣2t)≥﹣f(4sinθ﹣3)=f(﹣4sinθ+3)在时恒成立即cos2θ﹣2t≤﹣4sinθ+3在时恒成立即cos2θ﹣3+4sinθ≤2t在时恒成立即2t≥﹣sin2θ+4sinθ﹣2=﹣(sinθ﹣2)2+2在时恒成立∵时sinθ∈[0,1],∴=﹣(sinθ﹣2)2+2≤1∴2t≥1,t故答案为【点评】本题考查函数单调性的性质,本题是一个恒成立的问题,通过函数的单调性将其转化为三角不等式恒成立的问题,再分离常数,通过求三角函数的最值得到参数t的取值范围.本题考查了转化化归的思想,解题的关键是将恒等式进行正确转化,且能根据所得的形式判断应该求出三角形函数的最值以得到参数满足的不等式,求参数,本题思维量较大,难度不小.易因为转化时不等价出错.三、解答题(本大题共6题,共计70分.解答应写出文字说明、证明过程或演算步骤)17.(10分)已知函数的最小正周期为π.(1)求f(π);(2)先列表,再利用“五点法”在给定的坐标中作出函数f(x)的简图.【分析】(1)根据函数的周期公式求解ω,即可(2)利用五点法列表然后作图即可.【解答】解:(1)∵函数的周期为π,∴,则f(x)=sin(2x﹣)+1,则f(π)=sin(2π﹣)+1=sin(﹣)+1=1﹣(2)列表如下【点评】本题主要考查三角函数的图象和性质,根据周期公式求出ω是解决本题的关键.18.(12分)已知tanα=2.(1)求的值;(2)求的值;(3)若α是第三象限角,求cosα的值.【分析】(1)原式分子分母除以cosα,利用同角三角函数间的基本关系变形,将tanα的值代入计算即可求出值;(2)原式利用诱导公式化简后,再利用同角三角函数间的基本关系弦化切后,将tanα的值代入计算即可求出值;(3)利用同角三角函数间的基本关系列出关系式,tanα的值代入计算即可求出cosα的值.【解答】解:(1)∵tanα=2,∴原式===8;(2)∵tanα=2,∴原式==﹣=﹣=﹣;(3)∵tanα=2,∴cos2α===,∵α为第三象限角,∴cosα<0,∴cosα=﹣.【点评】此题考查了运用诱导公式化简求值,以及同角三角函数基本关系的运用,熟练掌握诱导公式是解本题的关键.19.(12分)已知0<θ<π,且,求(1)sinθ﹣cosθ的值;(2)tanθ的值.【分析】(1)由题意利用同角三角函数的基本关系、二倍角公式,求得sinθ﹣cosθ的值.(2)由(1)根据sinθ+cosθ和sinθ﹣cosθ的值,可得tanθ的值.【解答】解:(1)∵sinθ+cosθ=,①∴(sinθ+cosθ)2=,解得sinθcosθ=﹣.∵0<θ<π,且sinθcosθ<0,∴sinθ>0,cosθ<0,∴sinθ﹣cosθ>0.又∵(sinθ﹣cosθ)2=1﹣2sinθcosθ=,∴sinθ﹣cosθ=②,(2)由①②si nθ+cosθ=,sinθ﹣cosθ=.解得sinθ=,cosθ=﹣,∴tanθ==﹣.【点评】本题主要考查同角三角函数的基本关系、二倍角公式,属于基础题.20.(12分)已知函数f(x)=Asin(ωx+φ),x∈R,(其中A>0,ω>0,0<φ<)的图象与x轴的交点中,相邻两个交点之间的距离为,且图象上一个最低点为M(,﹣2).(1)求f(x)的解析式;(2)将函数f(x)的图象向右平移个单位后,再将所得图象上各点的横坐标缩小到原来的,纵坐标不变,得到y=g(x)的图象,求函数y=g(x)的解析式.【分析】(1)根据条件求出A,ω和φ的值即可看求出函数的解析式.(2)根据函数平移变换关系进行求解即可.【解答】解:(1)由函数图象的最低点为M(,﹣2),得A=2由x轴上相邻两个交点间的距离为,得=,即T=π…∴ω==2.又点M(,﹣2)在图象上,得2sin(2×+φ)=﹣2,即sin(+φ)=﹣1,故+φ=2kπ﹣,k∈Z,∴φ=2kπ﹣,又φ∈(0,),∴φ=.综上可得f(x)=2sin(2x+);(2)将f(x)=2sin(2x+)的图象向右平移个单位,得到f1(x)=2sin[2(x﹣)+],即f1(x)=2sin2x的图象,然后将f1(x)=2sin2x的图象上各点的横坐标缩小到原来的,纵坐标不变,得到g(x)=2sin(2•2x),即g(x)=2sin4x.【点评】本题主要考查三角函数解析式的求解,以及三角函数图象变换关系,根据条件求出A,ω和φ的值是解决本题的关键.21.(12分)已知.(1)求f(x)的单调递增区间;(2)当时,f(x)的最大值为4,求a的值;(3)在(2)的条件下,求满足f(x)=1且x∈[﹣π,π]的x的取值集合.【分析】(1)由题意利用正弦函数的单调性,求得f(x)的单调递增区间.(2)由题意利用正弦函数的定义域和值域,求得a的值.(3)(3)由(2)可得,再结合x的范围,求得x取值的集合.【解答】解:(1)对于函数,由2kπ﹣≤2x+≤2kπ+,k∈Z,可得kπ﹣≤x≤kπ+,k∈Z,所以f(x)的单调递增区间为,k∈Z.(2)x∈[0,],2x+∈[,],当2x=,即x=时,f(x)取得最大值4,故有,∴a=1.(3)由(2)可得,可得,则2x+=+2kπ,k∈Z或2x+=π+2kπ,k∈Z,即x=+kπ,k∈Z或x=+kπ,k∈Z,又x∈[﹣π,π],可解得x=﹣,﹣,,,所以x的取值集合为.【点评】本题主要考查正弦函数的单调性、最值,根据三角函数的值求角,属于基础题.22.(12分)已知函数f(x)=ln2x﹣2aln(ex)+3,x∈[e﹣1,e2](1)当a=1时,求函数f(x)的值域;(2)若f(x)≤﹣alnx+4恒成立,求实数a的取值范围.【分析】(1)求得y=f(x)=ln2x﹣2lnx+1,令t=lnx∈[﹣1,2],y=t2﹣2t+1=(t ﹣1)2,运用二次函数的值域求法,即可得到所求值域;(2)由题意可得ln2x﹣alnx﹣2a﹣1≤0恒成立,令t=lnx∈[﹣1,2],t2﹣at﹣2a ﹣1≤0恒成立,设y=t2﹣at﹣2a﹣1,讨论对称轴和区间的关系,可得最大值,解不等式即可得到所求范围.【解答】解:(1)当a=1时,y=f(x)=ln2x﹣2lnx+1,令t=lnx∈[﹣1,2],∴y=t2﹣2t+1=(t﹣1)2,当t=1时,取得最小值0;t=﹣1时,取得最大值4.∴f(x)的值域为[0,4];(2)∵f(x)≤﹣alnx+4,∴ln2x﹣alnx﹣2a﹣1≤0恒成立,令t=lnx∈[﹣1,2],∴t2﹣at﹣2a﹣1≤0恒成立,设y=t2﹣at﹣2a﹣1,∴当时,y max=﹣4a+3≤0,∴,当时,y max=﹣a≤0,∴a>1,综上所述,.【点评】本题考查函数的值域的求法,注意运用换元法,及二次函数的值域求法,考查不等式恒成立问题的解法,注意运用分类讨论的思想方法,讨论对称轴和区间的关系,考查运算能力,属于中档题.。
【题文】作文请以“秋天的怀念”为题写一篇记叙文。
要求:不少于700字;不要套作,不得抄袭;写出真情实感;卷面整洁工整;不得透露考生真实校名,班名,人名。
【答案】秋天的怀念黄叶打着旋儿从树上飘落,鲜花敛起笑容装饰变作春泥。
我站在窗边,望着天际振翅远飞的雁,忽地想起了半句诗:“自古逢秋悲寂寥”。
秋,总能用它特有的惆怅勾起我们对既往事物的怀念。
然后,可得愁思一发不可收拾……这恼人的季节!秋天是冷的,阵阵寒风吹得人自觉地想起三月暖春,那个阳光和煦,风光动人的季节,那时候,雨水多么滋润啊!风儿多么醉人啊!万事万物都从冰雪中苏醒了,绽放出缤纷的生机。
那时候的我,搀着奶奶的手,行走在春光里,看草长莺飞,燕裁柳叶。
春天里的奶奶一点也不像年老又犯病的人,她看着树木抽新,精神饱满地对我说:囡囡,你看啊!那树枝又发新芽了,这人跟树叶都是一个理吧!生出来,长大,过了最漂亮的时光,又落叶归根,再重新长出来,真好!”我为奶奶难得的诗情和活力感动高兴,撒着娇问她:“奶奶,现在是春天,那您就是一片新生的绿叶吧!”她缓缓地笑了,轻摸着我的头发,把目光投向远方,轻轻地说了声:“你这孩子!”奶奶没能熬过那个春天,她走的时候用爱怜的眼神抚摸着泣不成声的我,然后,望着春天,留下了一个永恒的微笑,我知道,她是在告诉我:“落叶归根,明年还会长出来,人死了,还会有下不念旧恶轮回,”我本是不相信来世今生的,但我愿意,相信奶奶,相信让刀物复苏的春。
中可怜这个秋,昔人已乘黄宿主去,没人嘱咐我着上大衣。
秋天是个单调的季节,一眼望不到头金黄让人情不自禁地想起绚烂的夏季,那个热烈的,缤纷的季节,那时候,百花争艳,紫嫣红,那时候,似火的骄阳总能点燃人心中蜇伏已久的激情,那时候的我,正面临着人生的第一次高考,以及一次意义重大的别离,有人说七月是黑色的,但我知道,七月是一个只为胜利者敞开的舞台,它是黑色,只是为了祭尊一段逝去的青春,以及某些破碎的梦想,永远忘不了,送朝夕相处的同学们登上火车的那一天,我们都哭了,彼此的心情就像一首老歌所唱的一样:那片笑声让我想起我的那些花儿,在我生命某个角落静静为我开着,我曾以为我会永远守在它身旁,可惜我们已被吹落散落在天涯,夏天,实在是一个让梦想腾飞的季节,它的激情与拼搏注定了它的无法被取代,只可惜,秋天来到之后,世界都变得沉寂,只有如水般的淡定,没有似火的冲动。
南昌十中2016-2017学年上学期第二次月考高一地理试题说明:本试卷分第I卷(选择题)和第Ⅱ卷(非选择题)两部分,全卷满分100分。
考试用时100分钟,注意事项:考生在答题前请认真阅读本注意事项及各题答题要求。
1.答题前,请您务必将自己的姓名、IS号用书写黑色字迹的0.5毫米签字笔填写在答题纸上,同时用2B铅笔在规定的位置上认真填涂自己的IS号。
2.作答非选择题必须用书写黑色字迹的0.5毫米签字笔写在答题纸上的指定位置,在其它位置作答一律无效。
作答选择题必须用2B铅笔把答题卡上对应题目的答案标号涂黑。
如需改动,请用橡皮擦干净后,再选涂其它答案,请保持卡面清洁和答题纸清洁,不折叠、不破损。
3.考试结束后,请将答题纸交回监考老师。
一、选择题(本大题共30题,每小题2分,共计60分。
在每小题列出的四个选项中只有一项是最符合题目要求的)下图为冬至日甲、乙、丙、丁四地昼夜长短比例示意图(阴影部分表示夜长),读图完成下列问题。
1. 甲、乙、丙、丁四地地球自转线速度由大到小排序正确的是()A. 甲、丁、丙、乙B. 甲、丙、丁、乙C. 丁、甲、丙、乙D. 乙、丙、甲、丁2. 地球表面做水平运动的物体发生左偏的地点是()A. 甲B. 乙C. 丙D. 丁【答案】1. B 2. D【解析】1. 赤道全年昼夜等长,甲位于赤道,纬度最低。
纬度越高,昼夜长短的变化幅度越大。
各地夜长与赤道夜长的差值比较,差值越大,说明纬度越高。
乙地有极夜现象,夜长差值最大,乙纬度最高。
丙地与赤道昼长差值比丁地小,纬度比丁地低。
自转线速度由赤道向两极递减,四地地球自转线速度由大到小排序正确的是甲、丙、丁、乙,B对。
A、C、D错。
2. 冬至日,北半球昼短夜长,南半球昼长夜短,图中丁是位于南半球,地球表面做水平运动的物体发生左偏的地点是丁,D对。
甲位于赤道,不偏转,A错。
乙、丙位于北半球,B、C错。
点睛:赤道全年昼夜等长,纬度越高,昼夜长短的变化幅度越大。
南昌十中2017-2018学年度上学期第二次月考高一英语试题本试卷分第I卷(选择题)和第Ⅱ卷(非选择题)两部分,全卷满分150分。
考试用时120分钟注意事项:考生在答题前请认真阅读本注意事项及各题答题要求1.答题前,请您务必将自己的姓名、考试证号用书写黑色字迹的0.5毫米签字笔填写在答题卡和答题纸上。
2.作答非选择题必须用书写黑色字迹的0.5毫米签字笔写在答题纸上的指定位置,在其它位置作答一律无效。
作答选择题必须用2B铅笔把答题卡上对应题目的答案标号涂黑。
如需改动,请用橡皮擦干净后,再选涂其它答案,请保持卡面清洁和答题纸清洁,不折叠、不破损。
3.考试结束后,答题纸交回。
第I卷(选择题)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,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.Who is the woman?A. Mary’s sister.B. Mary.C. Mary’s mother.2.When did the man live in London?A. Last year.B. Last month.C. When he was a child.3.What happened to the man just now?A. He met an old friend on the street.B. He mistook the woman for his friend.C. Lydia paid an unexpected visit to him.4.Why did the man change his mind probably?A. He forgot his wallet.B. He didn’t bring enough money.C. He didn’t need that much fruit.5.What are the speakers mainly talking about?A. The role of shopping in people’s lives.B. How to promote sales.C. The importance of mass media.第二节(每小题1.5分,满分22.5分)听下面5段对话或独白。
南昌十中2017-2018学年上学期第一次月考
高一数学试题
命题人:刘丽华、章小云 审题人:刘丽华、章小云
说明:本试卷分第I 卷(选择题)和第Ⅱ卷(非选择题)两部分,全卷满分150分。
考试用时120分钟。
注 意 事 项:
考生在答题前请认真阅读本注意事项及各题答题要求。
1.答题前,请您务必将自己的姓名、考试证号用书写黑色字迹的0.5毫米签字笔填写在答题卡和答题纸上。
2.作答非选择题必须用书写黑色字迹的0.5毫米签字笔写在答题纸上的指定位置,在其它位置作答一律无效。
作答选择题必须用2B 铅笔把答题卡上对应题目的答案标号涂黑。
如需改动,请用橡皮擦干净后,再选涂其它答案,请保持卡面清洁和答题纸清洁,不折叠、不破损。
3.考试结束后,请将答题纸交回。
第I 卷
一、选择题(本大题共12题,每小题5分,共计60分。
在每小题列出的四个选项中只有一项是最符合题目要求的)
1. 已知集合{1,2,3,4}A =,那么A 的真子集的个数是( )
A 、15
B 、16
C 、3
D 、4 2. 设集合{|101},{|5}A x Z x B x Z x =∈--=∈≤≤≤,则A
B 中元素的个数是
( )
A 、11
B 、10
C 、16
D 、15
3.
已知集合{
}40≤≤=x x P ,}20{≤≤=y y Q 下列不能表示从P 到Q
的映射的是( )
A B C
D 4. 设⎪⎩
⎪⎨⎧<=>+=)0(,0)0(,)
0(,1)(x x x x x f π,则=-)]}2([{f f f ( )
A 、1+π
B 、0
C 、π
D 、1- 5. 已知函数2
3212---=x x x y 的定义域为( )
A 、]1,(-∞
B 、]2,(-∞
C 、]1,2
1
()21
,(-
⋂--∞ D 、 ]1,2
1()21,(-
⋃--∞
6. 已知}5,53,2{2+-=a a M ,}3,106,1{2
+-=a a N ,且}3,2{=⋂N M ,则a 的值( )
A 、1或2
B 、2或4
C 、2
D 、1
7.已知集合{(,)|2},{(,)|4}M x y x y N x y x y =+==-=,那么集合M N 为( )
A 、3,1x y ==-
B 、(3,1)-
C 、{3,1}-
D 、{(3,1)}- 8. 下列命题之中,U 为全集时,不正确的是 ( )
A 、若
B A ⋂= φ,则U B
C A C U U =⋃)()(
B 、若B A ⋂= φ,则A = φ或B = φ
C 、若B A ⋃= U ,则=
⋂)()(B C A C U U φ
D 、若B A ⋃= φ,则==B A φ
9. 设函数x x x
f =+-)11(,则)(x f 的表达式为( )
A 、x x -+11
B 、 11-+x x
C 、x
x +-11
D 、
1
2+x x
10. 将二次函数c bx ax y ++=2
的图像向左平移1个单位,再向下平移2个单位,便得到
函数1322
++=x x y 的图像,则( ) A 、2,1,2===c b a B 、2,1,2=-==c b a
C 、2,1,2-==-=c b a
D 、2,1,2-=-=-=c b a
11. 下列图中,画在同一坐标系中,函数bx ax y +=2
与)0,0(≠≠+=b a b ax y 函数的图
象只可能是( )
12. 已知定义在]1,1[-上的函数满足对任意的,有
,且,则x 的取值范围是( )
()f x 1212,[1,1]()x x x x ∈-≠2121()(()())0x x f x f x -->(2)(1)f x f x -<-
A 、]2
3,1[
B 、]2
3,1(
C 、)2
3,1[
D 、)2
3,(-∞
第Ⅱ卷
二、填空题(本大题共4题,每小题5分,共计20分。
把答案填在答案的横线上。
) 13.已知()f x 是定义在()0,+∞上的函数,满足()()()(),31f xy f x f y f =+=, 则()9f = ;
14.已知()f x 满足()()324f x f x x +-=,则()f x = ;
15.点
)
2在幂函数()f x 的图像上,点12,4⎛
⎫- ⎪⎝
⎭在幂函数()g x 的图像上,则满足
()()f x g x >的解集为 ;
16.已知定义域为R 的奇函数()f x ,当0x >时,()2
2f x x x =-+,若函数()f x 在区间
[]1,2a --上单调递增,则实数a 的取值范围为 .
三、解答题(本大题共6题,共计70分。
解答应写出文字说明、证明过程或演算步骤。
) 17.(本小题满分10分)
已知函数⎩⎨⎧<+≥+=.
0,4,
0,4)(2x x x x x f
(1)求()()2f f -;
(2)画出函数的图象,并写出函数()f x 在区间()2,2-上的值域.
18.(本小题满分12分)
设全集{}{}{}|2,|210,|28U x x A x x B x x =≥-=<<=≤≤,求,,
U C A A B ⋂()(),U U C A B C A B ⋂⋂.
19.(本小题满分12分)
奇函数()f x 的定义域为[]1,1-,若()f x 在[]0,1上单调递减,且
0)()1(<++m f m f ,求实数m 的取值范围.
20.(本小题满分12分)
已知集合{}{}
|,|15A x a b x a b B x x x =-<<+=<->或. (1)若1,b A B =⊆,求实数a 的取值范围; (2)若1a A B =⋂=∅,求实数b 的取值范围.
22.(本小题满分12分)
南昌十中2017-2018学年上学期第一次月考
高一数学参考答案
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,
13. 2 ; 14. x 4 ;
15.; }1-1{<>x x x 或 16. (1,3]
三、解答题:(本大题共6小题,共70分. 解答应写出文字说明.证明过程或演算步骤. ) 17. 【解析】 解:(Ⅰ)∵f (-2)=2,f (2)=8,
∴f (f (-2))=f (2)=8. ……………4分。