2018-2019年上海市七宝中学高二上期中数学试卷(有答案)
2018-2019学年上海中学高二(上)期中数学试卷(解析版)
2018-2019学年上海中学高二(上)期中数学试卷一、填空题1.直线2x﹣y+3=0的倾斜角为.2.行列式中元素0所对应的代数余子式的值为.3.若直线ax+2y+1=0与直线x+y﹣2=0互相垂直,则a=.4.若=(1,﹣2),=(x,1),=(1,2),且()⊥,则x=.5.以=(﹣3,2)为方向向量的直线平分圆x2+y2+2y=0,直线l的方程为.6.经过两条直线2x+3y+1=0和3x﹣y+4=0的交点,并且平行于直线3x+4y﹣7=0的直线方程是.7.若直线ax+by﹣3=0与圆x2+y2+4x﹣1=0相切于点P(﹣1,2),则a+b=.8.如图,△ABC中D在边BC上,且=2,E为AD的中点,记=,=,则=(用、的线性组合表示)9.二阶方阵A=称矩阵为A的转置矩阵记作A T,设M、N是两个二阶矩阵,对于下列四个结论:(1)(M T)T=M;(2)(M+N)T=M T+N T;(3)(MN)T=M T N T;(4)“M=”是“M T=M”的充分不必要条件;其中真命题的序号为.10.在平面直角坐标系xOy中,A为直线l:y=2x上在第一象限内的点,B(5,0),以AB为直径的圆C与直线l交于另一点D.若=0,则点A的横坐标为.11.设动点M在x轴正半轴上,过动点M与定点P(2,1)的直线l交y=x(x>0)于点Q,那么的最大值为.12.如图,已知向量的夹角为,|﹣|=6,向量,的夹角为,|﹣|=2,则与的夹角为,的最大值为.二.选择题13.“D2=4F且E≠0”是“圆x2+y2+Dx+Ey+F=0与x轴相切”的()条件A.充分不必要B.必要不充分C.充要D.既不充分也不必要14.在下列向量组中,可以把向量=(3,2)表示出来的是()A.=(0,0),=(1,2)B.=(﹣1,2),=(5,﹣2)C.=(3,5),=(6,10)D.=(2,﹣3),=(﹣2,3)15.实数x,y满足,若z=2x+y的最大值为9,则实数m的值为()A.1B.2C.3D.416.如图,△ABC的AB边长为2,P,Q分别是AC,BC中点,记•+•=m,•+•=n,则()A.m=2,n=4B.m=3,n=1C.m=2,n=6D.m=3n,但m,n的值不确定三、解答题(共4小题,满分0分)17.已知二元一次方程组的增广矩阵为,请利用行列式求解此方程组.18.已知||=4,||=3,(2﹣3)•(2+)=61.(1)求与的夹角θ;(2)若,且=0,求t及||19.在平面直角坐标系xOy中,平行于x轴且过点A的入射光线l1被直线l:反射,反射光线l2交y轴于B点.圆C过点A且与l1、l2相切.(1)求l2所在的直线的方程和圆C的方程;(2)设P、Q分别是直线l和圆C上的动点,求PB+PQ的最小值及此时点P的坐标.20.已知a、b、c为△ABC的三边长,直线l的方程ax+by+c=0,圆M:(x+a)2+(y+b)2=c2.(1)若△ABC为直角三角形,c为斜边长,且直线l与圆M相切,求c的值;(2)若△ABC为正三角形,对于直线l上任意一点P,在圆M上总存在一点Q,使得线段|PQ|的长度为整数,求c的取值范围;(3)点E(﹣1,﹣1)、F(﹣1,1)、G(1,1)、H(1,﹣1),设E、F、G、H四点到直线l的距离之和为S,求S的取值范围.2018-2019学年上海中学高二(上)期中数学试卷参考答案与试题解析一、填空题1.直线2x﹣y+3=0的倾斜角为arctan2.【解答】解:因为直线2x﹣y+3=0的斜率为k=﹣=2,设直线的倾斜角为θ,则tanθ=2,所以θ=arctan2,故填:arctan2.2.行列式中元素0所对应的代数余子式的值为﹣6.【解答】解:行列式中元素0所对应的代数余子式的值为:(﹣1)5•=﹣6.故答案为:﹣6.3.若直线ax+2y+1=0与直线x+y﹣2=0互相垂直,则a=﹣2.【解答】解:直线ax+2y+1=0与直线x+y﹣2=0互相垂直,由于直线的斜率存在,所以斜率乘积为﹣1,即﹣1•()=﹣1,所以a=﹣2.故答案为:﹣2.4.若=(1,﹣2),=(x,1),=(1,2),且()⊥,则x=1.【解答】解:;∵;∴;∴x=1.故答案为:1.5.以=(﹣3,2)为方向向量的直线平分圆x2+y2+2y=0,直线l的方程为2x+3y+3=0.【解答】解:根据题意,要求直线的方向向量为=(﹣3,2),设其方程为2x+3y+m =0,圆x2+y2+2y=0,即x2+(y+1)2=1,其圆心为(0,﹣1),若要求直线平分圆,则圆心在要求直线上,则有2×0+3×(﹣1)+3=0,解可得m=3,则要求直线的方程为2x+3y+3=0;故答案为:2x+3y+3=0.6.经过两条直线2x+3y+1=0和3x﹣y+4=0的交点,并且平行于直线3x+4y﹣7=0的直线方程是3x+4y+=0.【解答】解:联立直线的方程,得到两直线的交点坐标为(﹣,),平行于直线3x+4y﹣7=0的直线方程是3x+4y+c=0,则3(﹣)+4×+c=0,解得c=,所以直线方程为3x+4y+=0.故填:3x+4y+=0.7.若直线ax+by﹣3=0与圆x2+y2+4x﹣1=0相切于点P(﹣1,2),则a+b=3.【解答】解:根据题意,圆x2+y2+4x﹣1=0的圆心为(﹣2,0),若直线ax+by﹣3=0与圆x2+y2+4x﹣1=0相切于点P(﹣1,2),则有,解可得a=1,b=2;则a+b=3;故答案为:3.8.如图,△ABC中D在边BC上,且=2,E为AD的中点,记=,=,则=(用、的线性组合表示)【解答】解:∵E为AD的中点,,∴=====,故答案为:.9.二阶方阵A=称矩阵为A的转置矩阵记作A T,设M、N是两个二阶矩阵,对于下列四个结论:(1)(M T)T=M;(2)(M+N)T=M T+N T;(3)(MN)T=M T N T;(4)“M=”是“M T=M”的充分不必要条件;其中真命题的序号为(1)(2)(4).【解答】解:对于(1),设M=,则M T=,(M T)T=,所以(M T)T=M,(1)正确;对于(2),设M=,N=,则M+N=,∴(M+N)T=;M T=,N T=,则M T+N T=,∴(M+N)T=M T+N T,(2)正确;对于(3),设M=,N=,则MN=,∴(MN)T=;M T=,N T=,则M T N T=,∴(MN)T≠M T N T,(3)错误;对于(4),M=时,M T=,充分性成立,M T=M时,M不一定为,如M=,即必要性不成立,是充分不必要条件,(4)正确.综上,其中真命题的序号是(1)、(2)、(4).故答案为:(1)、(2)、(4).10.在平面直角坐标系xOy中,A为直线l:y=2x上在第一象限内的点,B(5,0),以AB为直径的圆C与直线l交于另一点D.若=0,则点A的横坐标为3.【解答】解:设A(a,2a),a>0,∵B(5,0),∴C(,a),则圆C的方程为(x﹣5)(x﹣a)+y(y﹣2a)=0.联立,解得D(1,2).∴=.解得:a=3或a=﹣1.又a>0,∴a=3.即A的横坐标为3.故答案为:3.11.设动点M在x轴正半轴上,过动点M与定点P(2,1)的直线l交y=x(x>0)于点Q,那么的最大值为.【解答】解:设l:y=k(x﹣2)+1,要它与y=x(x>0)相交,则k>1或k<0.令y=0,可得:M(2﹣,0),令y=x,得Q.∴|MP|=,|PQ|=.∴u==.于是u2==g(k),k>1或k<0.g′(k)=,可得:k=﹣2,函数g(k)取得极大值,g(﹣2)=5.∴u max=.此时M(﹣,0).故答案为:.12.如图,已知向量的夹角为,|﹣|=6,向量,的夹角为,|﹣|=2,则与的夹角为,的最大值为.【解答】解:如图,设,则,,,∴AB=6,,AC=,又,∴A,O,B,C四点共圆,在△ABC中,由正弦定理得,即,∴sin∠ABC=,则.由同弧所对圆周角相等,可得,即与的夹角为;设∠OAC=θ,则,在△AOC中,由正弦定理得:,∴OC=,,∴===64×=64×()=64×()=64×()=64[].∴当,即时,有最大值为.故答案为:,.二.选择题13.“D2=4F且E≠0”是“圆x2+y2+Dx+Ey+F=0与x轴相切”的()条件A.充分不必要B.必要不充分C.充要D.既不充分也不必要【解答】解:圆的方程可化为:,故若D2=4F且E≠0,则圆x2+y2+Dx+Ey+F=0与x轴相切,若圆x2+y2+Dx+Ey+F=0与x轴相切,则D2=4F且E≠0,综上“D2=4F且E≠0”是“圆x2+y2+Dx+Ey+F=0与x轴相切”的充要条件.故选:C.14.在下列向量组中,可以把向量=(3,2)表示出来的是()A.=(0,0),=(1,2)B.=(﹣1,2),=(5,﹣2)C.=(3,5),=(6,10)D.=(2,﹣3),=(﹣2,3)【解答】解:根据,选项A:(3,2)=λ(0,0)+μ(1,2),则3=μ,2=2μ,无解,故选项A不能;选项B:(3,2)=λ(﹣1,2)+μ(5,﹣2),则3=﹣λ+5μ,2=2λ﹣2μ,解得,λ=2,μ=1,故选项B能.选项C:(3,2)=λ(3,5)+μ(6,10),则3=3λ+6μ,2=5λ+10μ,无解,故选项C 不能.选项D:(3,2)=λ(2,﹣3)+μ(﹣2,3),则3=2λ﹣2μ,2=﹣3λ+3μ,无解,故选项D不能.故选:B.15.实数x,y满足,若z=2x+y的最大值为9,则实数m的值为()A.1B.2C.3D.4【解答】解:作出不等式组对应的平面区域如图:(阴影部分).由z=2x+y得y=﹣2x+z,平移直线y=﹣2x+z,由图象可知当直线y=﹣2x+z经过点B时,直线y=﹣2x+z的截距最大,此时z最大,此时2x+y=9.由,解得,即B(4,1),∵B在直线y=m上,∴m=1,故选:A.16.如图,△ABC的AB边长为2,P,Q分别是AC,BC中点,记•+•=m,•+•=n,则()A.m=2,n=4B.m=3,n=1C.m=2,n=6D.m=3n,但m,n的值不确定【解答】解:∵P,Q分别是AC,BC中点,∴m=•+•=====2;∵P,Q分别是AC,BC中点,∴,,∴n=•+•=+===6.故选:C.三、解答题(共4小题,满分0分)17.已知二元一次方程组的增广矩阵为,请利用行列式求解此方程组.【解答】解:对于增广矩阵,当m=2时,矩阵化为,此时方程组有无数个解;当m=﹣2时,矩阵化为,此时方程组无解;当m≠±2,矩阵第二行有,(2+m)(2﹣m)•y=(m+1)(2﹣m),得进第一行得,综上所述,当m=2时,方程有无数个解;当m=﹣2时,方程组无解;当m≠±2时,,.18.已知||=4,||=3,(2﹣3)•(2+)=61.(1)求与的夹角θ;(2)若,且=0,求t及||【解答】解(1)∵||=4,||=3,(2﹣3)•(2+)=61,∴•=﹣6.﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣∴cos θ===﹣,﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣又0≤θ≤π,∴θ=.﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(2)=()=t+(1﹣t)=﹣15t+9=0∴t=﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣∴||2=(+)2=,∴||=﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣19.在平面直角坐标系xOy中,平行于x轴且过点A的入射光线l1被直线l:反射,反射光线l2交y轴于B点.圆C过点A且与l1、l2相切.(1)求l2所在的直线的方程和圆C的方程;(2)设P、Q分别是直线l和圆C上的动点,求PB+PQ的最小值及此时点P的坐标.【解答】解:(Ⅰ)直线l1:y=2,设l1交l于D,则D(2,2).∵l的倾斜角为30°,∴l2的倾斜角为60°,…∴,∴反射光线l2所在的直线方程为y﹣2=(x﹣2).即.…已知圆C与l1切于点A,设C(a,b),∵圆心C在过点D且与l垂直的直线上,∴①…又圆心C在过点A且与l1垂直的直线上,∴②,由①②得,圆C的半径r=3.故所求圆C的方程为.…(Ⅱ)设点B(0,﹣4)关于l的对称点B'(x0,y0),则,…得.固定点Q可发现,当B'、P、Q共线时,PB+PQ最小,故PB+PQ的最小值为为B'C﹣3.…,得,最小值.…(16分)20.已知a、b、c为△ABC的三边长,直线l的方程ax+by+c=0,圆M:(x+a)2+(y+b)2=c2.(1)若△ABC为直角三角形,c为斜边长,且直线l与圆M相切,求c的值;(2)若△ABC为正三角形,对于直线l上任意一点P,在圆M上总存在一点Q,使得线段|PQ|的长度为整数,求c的取值范围;(3)点E(﹣1,﹣1)、F(﹣1,1)、G(1,1)、H(1,﹣1),设E、F、G、H四点到直线l的距离之和为S,求S的取值范围.【解答】解(1)因为若△ABC为直角三角形,c为斜边长,所以a2+b2=c2,直线l与圆M相切,所以圆心(a,b)到直线ax+by+c=0的距离为c,即,所以,即c2﹣c=±c2,得c=,或者c=0(舍).(2)若△ABC为正三角形,若△ABC为正三角形,则此时圆是以{c,c}为圆心,c为半径的圆,直线方程为x+y+1=0,设圆心(c,c)到直线的距离为d,则d=,要使直线l上任意一点P,在圆M上总存在一点Q,使得线段|PQ|的长度为整数,需满足同时成立,即,解得c≥.(3)依题意S=+++,因为三角形的两边之和大于第三边,所以S可化为:S=,∵c<a+b,,∴S≤=4,下面求S的最小值,从几何意义上看,S代表(1,1)到直线l的距离的二倍,而直线l在x轴上的截距为﹣,在y轴上的截距为﹣,三边中若c为最大值,则直线l在两坐标轴上的截距均小于﹣1,此时(1,1)到直线l 的最小距离大于2,即S>4.若c不是最大值,不妨设a为最大值,则S=>==2.综上2<S<.。
上海市七宝中学2018-2019学年高二上学期期末考试数学试卷(解析版)
2018-2019学年上海市闵行区七宝中学高二(上)期末数学试卷一、选择题(本大题共4小题)1.与圆相切,且横截距与纵截距相等的直线条数是( )A. 2B. 4C. 6D. 以上说法都不对【答案】B【解析】【分析】根据题意,分析圆的圆心与半径,分2种情况讨论,①直线过原点,设直线的方程为,②直线不过原点,设其方程为,由直线与圆的位置关系分析直线的条数,综合2种情况即可得答案。
【详解】解:根据题意,圆的圆心为,半径,分2种情况讨论,①直线过原点,设直线的方程为,即,则有,解可得,此时直线的方程为:,②直线不过原点,由于直线横截距与纵截距相等,设其方程为,即,则有,解可得,此时直线的方程为,故一共有4条符合条件的直线;故选:B.【点睛】本题考查直线与圆的位置关系,涉及直线在坐标轴上的截距,注意直线过原点的情况,属于基础题。
2.直线:与直线:的夹角为( )A. B.C. D. 以上说法都不对【答案】B【解析】【分析】先求出两条直线的倾斜角和斜率,可得两条直线的夹角。
【详解】解:直线:的斜率为,倾斜角为,直线:的斜率不存在,倾斜角为,故直线:与直线:的夹角为,故选:B.【点睛】本题主要考查直线的倾斜角和斜率,两条直线的夹角,属于基础题。
3.下列说法正确的是()A. 平面中两个定点A,B,k为非零常数,若,则动点P的轨迹是双曲线B. 定圆C上有一定点A和一动点不与A重合,O为坐标原点,若,则动点P的轨迹是椭圆C. 斜率为定值的动直线与抛物线相交于A,B两点,O为坐标原点,,则动点P 的轨迹是直线D. 以上说法都不对【答案】C【解析】【分析】由双曲线的定义可判断A错误;由P为AB的中点,,可得P的轨迹为圆,可判断B错误;由抛物线的方程,可设,,运用直线的斜率公式和中点坐标公式,即可判断C正确,进而可得到答案。
【详解】解:设A,B是两个定点,k为非零常数,若,则轨迹为两条射线;若,则轨迹不存在,若,则轨迹为双曲线,故A错误;过定圆C上一定点A作圆的动弦AB,O为坐标原点若,则P为AB的中点,,即恒为直角,则动点P的轨迹为以AC为直径的圆,故B错误;斜率为定值t的动直线与抛物线相交于A,B两点,设,,,可得P为AB的中点,,即有,则动点P的轨迹是直线,故C正确.故选:C.【点睛】本题考查双曲线的定义,考查动点的轨迹,考查中点坐标公式和直线的斜率公式,以及运算能力和推理能力,属于中档题。
2018-2019学年上海市闵行区七宝中学高三(上)期中数学试卷(精编含解析)
2018-2019学年上海市闵行区七宝中学高三(上)期中数学试卷一、选择题(本大题共4小题,共20分)1.“函数存在反函数”是“函数在R上为单调函数”的A. 充分而不必要条件B. 必要而不充分条件C. 充分必要条件D. 既不充分也不必要条件【答案】B【解析】【分析】函数f(x)(x∈R)存在反函数,至少还有可能函数f(x)在R上为减函数,充分条件不成立;而必要条件显然成立【详解】“函数f(x)在R上为增函数”⇒“函数f(x)(x∈R)存在反函数”;反之取f(x)=﹣x(x∈R),则函数f(x)(x∈R)存在反函数,但是f(x)在R上为减函数.故选:B.【点睛】本题考查充要条件的判断及函数存在反函数的条件,属基本题.2.若函数的反函数为,则函数与的图象可能是A. B. C. D.【答案】A【解析】【分析】f(x)和f﹣1(x)关于y=x对称是反函数的重要性质;而将f(x)的图象向右平移a个单位后,得到的图象的解析式为f(x﹣a)而原函数和反函数的图象同时平移时,他们的对称轴也相应平移.【详解】函数f(x﹣1)是由f(x)向右平移一个单位得到,f﹣1(x﹣1)由f﹣1(x)向右平移一个单位得到,而f(x)和f﹣1(x)关于y=x对称,从而f(x﹣1)与f﹣1(x﹣1)的对称轴也是由原对称轴向右平移一个单位得到即y=x﹣1,排除B,D;A,C选项中各有一个函数图象过点(2,0),则平移前的点坐标为(1,0),则反函数必过点(0,1),平移后的反函数必过点(1,1),由此得A选项有可能,C选项排除;故答案为:A【点睛】本题主要考查函数与其反函数的关系,考查函数的图像的变换,意在考查学生对这些知识的掌握水平和分析推理能力. 用整体平移的思想看问题,是解决本题的关键.3.在△中,角、、所对的边分别为、、,给出四个命题:(1)若,则△为等腰三角形;(2)若,则△为直角三角形;(3)若,则△为等腰直角三角形;(4)若,则△为正三角形;以上正确命题的个数是()A. 1B. 2C. 3D. 4【答案】B【解析】【分析】对每一个命题逐一分析得解.【详解】(1)若,则2A=2B或2A+2B=π,所以A=B或A+B=,所以△ABC是等腰三角形或直角三角形,所以该命题是错误的.(2) 若,所以sinA=sin(,所以则△不一定为直角三角形,所以该命题是错误的.(3) 若,所以A=C=,则△为等腰直角三角形,所以该命题是真命题.(4)若,所以所以A=B=C,所以△ABC是正三角形.所以该命题是真命题.故答案为:B【点睛】本题主要考查正弦定理和三角恒等变换,考查三角函数的图像和性质,意在考查学生对这些知识的掌握水平和分析推理能力.4.是定义在上的函数,且,若的图像绕原点逆时针旋转后与原图像重合,则在以下各项中,的可能取值只能是()A. 0B. 1C. 2D. 3【答案】C【解析】【分析】直接利用定义函数的应用求出结果.【详解】由题意得到:问题相当于圆上由12个点为一组,每次绕原点逆时针旋转个单位后与下一个点会重合.我们可以通过代入和赋值的方法当f()=,,3时,此时得到的圆心角为,,,然而此时x=0或者x=时,都有2个y与之对应,而我们知道函数的定义就是要求一个x只能对应一个y,因此只有当=,此时旋转,此时满足一个x只会对应一个y,故答案为:C【点睛】本题考查函数的定义的运用,意在考查学生对这些知识的掌握水平和分析推理能力.二、填空题(本大题共12小题,共36.0分)5.集合的真子集有________个【答案】【解析】【分析】直接写出集合A的真子集即得解.【详解】集合A的真子集有,{0},{1},{2018},{0,1},{0,2018},{1,2018},所以集合A的真子集个数为7,故答案为:7【点睛】本题主要考查集合的真子集及其个数,意在考查学生对这些知识的掌握水平和分析推理能力.6.设全集,,,则图中阴影部分所表示的集合是________(用区间表示)【答案】【解析】【分析】先化简集合M和N,再求M∩N,再求即得阴影部分所表示的集合.【详解】由题得M={x|x>2或x<-2},N={x|x≥0},所以M∩N={x|x>2},所以.所以阴影部分所表示的集合为[0,2].故答案为:【点睛】本题主要考查韦恩图和集合的运算,意在考查学生对这些知识的掌握水平和分析推理能力.7.命题“若实数、满足,则或”是________命题(填“真”或“假”)【答案】真【解析】【分析】先考虑其逆否命题“a>2且b>3则a+b>5”的真假,即得原命题的真假.【详解】由题得原命题的逆否命题为“a>2且b>3则a+b>5”,由不等式同向可加的性质得其逆否命题为真命题,所以原命题是真命题.故答案为:真【点睛】(1)本题主要考查原命题及其逆否命题,考查命题真假性的判断,意在考查学生对这些知识的掌握水平和分析推理能力.(2)互为逆否关系的命题同真同假,即原命题与逆否命题的真假性相同,原命题的逆命题和否命题的真假性相同.所以,如果某些命题(特别是含有否定概念的命题)的真假性难以判断,一般可以判断它的逆否命题的真假性.8.某个时钟时针长6cm,则在本场考试时间120分钟内,该时针扫过的面积是______【答案】.【解析】时针所扫过的面积是以时针的长度为半径,圆心角为×2=的扇形的面积,据此解答即可.【详解】∵设扇形的弧长为l,圆心角大小为α(rad),半径为r,∴则时针所扫过的面积是以时针的长度为半径,圆心角为×2=的扇形的面积,即:r=6,α=,∴扇形的面积为S=r2α==6π.故答案为:6π.【点睛】本题弄清楚分针时针的运动轨迹,是解答本题的关键,属于基础题.9.设为奇函数,则______.【答案】【解析】【分析】根据对数的基本运算以及函数奇偶性的性质建立条件关系即可求a的值;【详解】∵f(x)是奇函数,∴f(﹣x)=﹣f(x),∴,∴,即(1+ax)(1﹣ax)=﹣(x+1)(x﹣1),即1﹣a2x2=1﹣x2,即a2=1,∴a=﹣1或a=1,若a=1,则=不满足条件,舍去,故答案为:a=﹣1.【点睛】本题主要考查利用函数奇偶性的应用求参数的值,注意取舍,属于基础题.10.函数在上单调递增,则实数的取值范围为________【答案】【解析】先对函数求导得在(1,2)上恒成立,再分离参数求出a的范围.【详解】由题得在(1,2)上恒成立,所以.故答案为:【点睛】(1)本题主要考查利用导数研究不等式的单调性和恒成立问题,意在考查学生对这些知识的掌握水平和分析推理能力.(2)一般地,函数在某个区间可导,在某个区间是增函数≥0 .11.在△中,角、、所对的边分别为、、,若,,,则△的面积为________ 【答案】【解析】【分析】利用余弦定理可得b,再利用三角形面积计算公式即可得出.【详解】∵a=,∴a2=b2+c2﹣2bccosA,∴3=4+b2﹣4b×,化为b2﹣2b+1=0,解得b=1.∴S△ABC===.故答案为:.【点睛】本题主要考查了余弦定理、三角形面积计算公式,意在考查学生对这些知识的掌握水平和分析推理能力与计算能力.12.已知函数,则的解集是________【答案】【解析】【分析】由于函数是定义域在上的增函数,所以,解不等式即得解.【详解】由于函数是定义域在上的增函数,所以故答案为:【点睛】(1)本题主要考查幂函数的图像和性质,意在考查学生对这些知识的掌握水平和分析推理能力.(2)处理函数的问题,一定要注意“定义域优先的原则”,本题不要漏了3x-1≥0.13.若关于的不等式在上恒成立,则正实数的取值范围为________【答案】【解析】【分析】由题得|2x-a|>-x+1,再分1<x≤2和0≤x≤1两种情况讨论恒成立问题,即得解.【详解】由题得|2x-a|>-x+1,当1<x≤2时,-x+1<0,所以不等式恒成立.当0≤x≤1时,-x+1≥0,所以2x-a>-x+1或2x-a<x-1,所以a<3x-1或a>x+1在[0,1]上恒成立,所以a<-1或a>2,因为a>0,综合得a>2.故答案为:a>2【点睛】本题主要考查绝对值不等式的恒成立问题,意在考查学生对这些知识的掌握水平和分析推理能力.14.已知常数,函数的图像经过点、,若,则________【答案】【解析】【分析】直接利用函数的关系式,利用恒等变换求出相应的a值.【详解】函数f(x)=的图象经过点P(p,),Q(q,).则:,整理得:=1,解得:2p+q=a2pq,由于:2p+q=16pq,所以:a2=16,由于a>0,故:a=4.故答案为:4【点睛】本题主要考查函数的性质和指数幂的运算,意在考查学生对这些知识的掌握水平和分析推理计算能力.15.已知函数,若,则的最大值是________【答案】【解析】【分析】设g(x)=f(x)-3,再判断函数g(x)的奇偶性和单调性,再由得,再利用三角换元求的最大值.【详解】设g(x)=f(x)-3,所以g(x)=,所以所以g(-x)=-g(x),所以函数g(x)是奇函数,由题得,所以函数g(x)是减函数,因为,所以,所以g=0,所以g=g(1-,所以不妨设,所以==,所以的最大值为.故答案为:【点睛】(1)本题主要考查函数的奇偶性和单调性,考查函数的图像和性质,考查三角函数的图像和性质,意在考查学生对这些知识的掌握水平和分析推理计算能力.(2)本题的解题关键有三点,其一是构造函数g(x)得到函数g(x)的奇偶性和单调性,其二是由得,其三是利用三角换元求的最大值.16.已知函数,如果函数恰有三个不同的零点,那么实数的取值范围是________【答案】【解析】【分析】先求出函数的解析式,作出函数的图像,由题得有三个不同的实根,数形结合分析得到实数k 的取值范围.【详解】当1<x≤2时,f(x)=-x+2,当时,1<2x≤2,所以f(x)=,当时,<2x≤1,所以f(x)=,当时,<2x≤,所以f(x)=,当时,<2x≤,所以f(x)=,所以函数的图像为:其图像为线段PA,EB,GC,HD,,(不包括上端点A,B,C,D,)直线y=k(x-1)表示过定点P(1,0)的直线系,由题得C(),D(),当直线在PD(可以取到)和直线PC(不能取到)之间时,直线和函数f(x)的图像有三个不同的交点,由题得.所以k的取值范围为.故答案为:【点睛】(1)本题主要考查函数的图像和性质,考查求函数的解析式,考查函数的零点问题,意在考查学生读这些知识的掌握水平和数形结合分析推理能力.(2)解答本题的关键是求出函数f(x)的解析式作出函数的图像.(3)函数的零点问题常用的方法有:方程法、图像法、方程+图像法.三、解答题(本大题共5小题,共60.0分)17.已知锐角和钝角的终边分别与单位圆交于、两点,其中点坐标.(1)求的值;(2)若,求点坐标.【答案】(1);(2).【解析】【分析】(1)先求出,再求的值.(2)由题得,解方程组即得点B的坐标. 【详解】由题得,,所以=-7.由题设B(x,y),因为是钝角,所以,所以点B的坐标为.【点睛】本题主要考查三角函数的坐标定义,考查三角恒等变换求值,意在考查学生对这些知识的掌握水平和分析推理能力.18.如图,某公园有三个警卫室、、有直道相连,千米,千米,千米.(1)保安甲沿从警卫室出发行至点处,此时,求的直线距离;(2)保安甲沿从警卫室出发前往警卫室,同时保安乙沿从警卫室出发前往警卫室,甲的速度为1千米/小时,乙的速度为2千米/小时,若甲乙两人通过对讲机联系,对讲机在公园内的最大通话距离不超过3千米,试问有多长时间两人不能通话?(精确到0.01小时)【答案】(1);(2).【解析】【分析】(1)由解直角三角形可得∠C=30°,在△BPC中由余弦定理可得BP的值;(2)设甲出发后的时间为t小时,则由题意可知0≤t≤4,设甲在线段CA上的位置为点M,则AM=4﹣t,讨论0≤t≤1时,当1≤t≤4时,分别在△AMQ和△AMB中,运用余弦定理和二次不等式的解法,即可得到所求结论.【详解】(1)在Rt△ABC中,AB=2,BC=2,所以∠C=30°,在△PBC中PC=1,BC=2,由余弦定理可得BP2=BC2+PC2﹣2BC•PCcos30°=(2)2+1﹣2×2×1×=7,即BP=;(2)在Rt△ABC中,BA=2,BC=2,AC=4,设甲出发后的时间为t小时,则由题意可知0≤t≤4,设甲在线段CA上的位置为点M,则AM=4﹣t,①当0≤t≤1时,设乙在线段AB上的位置为点Q,则AQ=2t,如图所示,在△AMQ中,由余弦定理得MQ2=(4﹣t)2+(2t)2﹣2•2t•(4﹣t)cos60°=7t2﹣16t+16>9,解得t<或t>,所以0≤t≤;②当1≤t≤4时,乙在警卫室B处,在△ABM中,由余弦定理得MB2=(4﹣t)2+4﹣2•2t•(4﹣t)cos60°=t2﹣6t+12>9,解得t<3﹣或t>3+,又1≤t≤4,不合题意舍去.综上所述0≤t ≤时,甲乙间的距离大于3千米,所以两人不能通话的时间为小时.【点睛】本题考查解三角形的实际问题的解法,注意运用余弦定理,考查化简整理的运算能力,属于 中档题.19.问题:正数、满足,求的最小值.其中一种解法是:,当且仅当且时,即且时取等号.学习上述解法并解决下列问题:(1)若实数、、、满足,试比较和的大小,并指明等号成立的条件;(2)利用(1)的结论,求函数的值域.【答案】(1),且等号成立;(2).【解析】 【分析】(1)先化简=( ,再利用基本不等式求最值即得解.(2) 令再利用结论求函数的值域.【详解】=(当时取等.令由(1)得,因为f(t)>0,所以.所以函数的值域为.【点睛】(1)本题主要考查常量代换和基本不等式求最值,意在考查学生对这些知识的掌握水平和分析推理能力.(2) 本题的解题关键是对“1”的常量代换,再利用基本不等式求函数的最小值. 利用基本不等式求最值时,要注意“一正二定三相等”,三个条件缺一不可.20.定义区间、、、的长度均为,已知不等式的解集为.(1)求的长度;(2)函数(,)的定义域与值域都是(),求区间的最大长度;(3)关于的不等式的解集为,若的长度为6,求实数的取值范围.【答案】(1);(2);(3).【解析】【分析】解不等式得其解集即得区间长度.(2) 由题意求出f(x)的定义域并化简解析式,判断出区间的范围和f(x)的单调性,由题意列出方程组,转化为m,n是方程f(x)的同号的相异实数根,利用韦达定理表示出mn和m+n,由判别式大于零求出a 的范围,表示出n﹣m 利用配方法化简后,由二次函数的性质求出最大值和a的值.(3)先求出A∩B⊆(0,6),再转化为不等式组,当x∈(0,6)时恒成立. 分析两个恒成立问题即得t的取值范围.【详解】解不等式得其解为-1≤x<6,所以解集A区间长度为6-(-1)=7.(2) 由题意得,函数f(x)的定义域是{x|x≠0},∵[m,n]是其定义域的子集,∴[m,n]⊆(﹣∞,0)或(0,+∞).∵f(x)=在[m,n]上是增函数,∴由条件得,则m,n是方程f(x)=x的同号相异的实数根,即m,n是方程(ax)2﹣(a2+a)x+1=0同号相异的实数根.∴mn=,m+n==,则△=(a2+a)2﹣4a2>0,解得a>1或a<﹣3.∴n﹣m====,∴n﹣m的最大值为,此时,解得a=3.即在区间[m,n]的最大长度为.(3) 因为x>0,A=[-1,6),的长度为6,所以A∩B⊆(0,6).不等式log2x+log2(tx+3t)<2等价于又A∩B⊆(0,6),不等式组的解集的各区间长度和为6,所以不等式组,当x∈(0,6)时恒成立.当x∈(0,6)时,不等式tx+3t>0恒成立,得t>0当x∈(0,6)时,不等式tx2+3tx﹣4<0恒成立,即恒成立当x∈(0,6)时,的取值范围为,所以实数综上所述,t的取值范围为【点睛】本题考查一个新定义问题,即区间的长度,本题解题的关键是对于条件中所给的三种不同的题目进行整理变化,灵活解答函数的最值问题和恒成立问题.21.已知定义在上的函数满足:对任意的实数都成立,当且仅当时取等号,则称函数是上的函数,已知函数具有性质:(,)对任意的实数()都成立,当且仅当时取等号.(1)试判断函数(且)是否是上的函数,说明理由;(2)求证:是上的函数,并求的最大值(其中、、是△三个内角);(3)若定义域为,①是奇函数,证明:不是上的函数;②最小正周期为,证明:不是上的函数.【答案】(1),是S函数;,不是S函数;(2)见解析,最大值;(3)见解析.【解析】【分析】(1)利用S函数的定义证明当0<a<1时,不是上的函数.当a大于1时,不是上的函数.(2)利用S函数的定义证明是上的函数,并利用S函数的性质求的最大值.(3)利用举反例证明.【详解】任取,当同理可证,当0<a<1时,不是上的函数.(2),,,所以是上的函数.由S函数的性质有所以(3)用举反例证明,令f(x)=sinx,所以f(x)=sinx是R上的周期为π的奇函数,取所以而即在R上,f(x)=sinx不是S函数,故原命题得证.【点睛】本题主要考查新定义解题,考查学生对新定义的理解和掌握水平和利用新定义处理数学问题的能力.解题的关键是对新定义理解透彻.。
上海市七宝中学2018-2019学年高二上学期期中考试数学试题 Word版含答案
七宝中学高二期中数学试卷2018.11一. 填空题1. 若线性方程组的增广矩阵是122301c c ⎛⎫⎪⎝⎭,其解为11x y =⎧⎨=⎩,则12c c += 2.已知行列式4513732xx 中元素4的代数余子式是1,则实数x 的值是3. 求26100lim 3110045nn n n n n →∞⎧≤⎪⎪⎨+⎪>⎪+⎩(*n ∈N )=4.在△ABC 中,(0,0)A ,(3,5)B ,(4,4)C ,则△ABC 面积为5. 已知(1,2)a =r,(2,3)b =-r,(2)a b +r r∥()a kb +rr,则实数k 的值是6. 秦九韶是我国南宋时期的数学家,他在所著的《数书九章》中提出的多项式求值的秦九韶算法,至今仍是比较先进的算法,如图所示的程序框图给出了利用秦九韶算法求某多项式值的一个实例,若输入的n 、x 的值分别是3、2,则输出v的值为7.设111()122f n n n n=++⋅⋅⋅+++,*n ∈N ,若*k ∈N ,则(1)()f k f k +=+ 8.已知||1a =r ,||1b =r ,a r、b r 的夹角是60°,若向量c r 满足||1c a b --=r r r ,则||c r 的最小 值为9. 设函数2y nx n =-+和1122y x n =-+(*n ∈N ,2n ≥)的图像与两坐标轴围成的封闭图形的面积为n S ,则lim n n S →∞=10.已知圆O 中,弦3AB =,5AC =,则AO BC ⋅u u u r u u u r的值是11.定义平面向量之间的一种运算“*”如下:对任意的(,)a m n =r ,(,)b p q =r,m na b p q*=r r ,有下列说法:①若a r 与b r 垂直,则0a b *=r r ;②a b b a *=*r r r r ;③对任意的λ∈R ,有()()a b a b λλ*=*r r r r ;④2222()()a b a b a b *+⋅=r r r r r r ;正确的是(写出所有正确的序号)12. 已知无穷等比数列{}n a 的公比为q ,前n 项和为n S ,且lim n n S S →∞=,若对于任意*n ∈N ,2n S S <恒成立,则公比q 的取值范围是二. 选择题13. 用数学归纳法证明11112321nn +++⋅⋅⋅+<-(*n ∈N ,2n ≥)时,第一步应验证()A.1122+<B. 111223++<C.111323++< D. 11113234+++<14. 已知A 、B 、C 是平面不同三点,则“0AB BC CA ⋅+=u u u r u u u r u u u r r”是“A 、B 、C 三点能构成三角形”的()A.充分非必要条件B. 必要非充分条件C.充要条件D. 既非充分又非必要条件15. 若等比数列{}n a 的公比为q ,则关于x 、y 的二元一次方程组132423a x a y a x a y +=⎧⎨+=⎩的解的情况,下列说法正确的是()A.对任意q ∈R ,0q ≠,方程组都有唯一解B. 对任意q ∈R ,0q ≠,方程组都无解C.当且仅当32q =时,方程组无解 D. 当且仅当32q =时,方程组无穷多解16. 正六边形ABCDEF 中,令AB a =u u u r r ,AF b =u u u r r,P是△CDE 内含边界的动点(如图),AP xa yb =+u u u r r r,则x y +的最大值是()A. 1B. 3C. 4D. 5三. 解答题17. 求证:对任意的*n ∈N ,22389n n +--能被64整除.18. 上海市旅游节刚落下帷幕,在旅游节期间,甲、乙、丙三位市民顾客分别获得一些景区门票的折扣消费券,数量如表1,已知这些景区原价和折扣价如表2(单位:元). 表1:表2:(1)按照上述表格的行列次序分别写出这三位市民获得的折扣消费券数量矩阵A 和三个景区的门票折扣后价格矩阵B ;(2)利用你所学的矩阵知识,计算三位市民各获得多少元折扣?19. 已知平面直角坐标系内三点A 、B 、C 在一条直线上,满足(3,1)OA m =-+u u u r ,(,3)OB n =u u u r ,(7,4)OC =u u u r,且OA OB ⊥u u u r u u u r ,其中O 为坐标原点.(1)求实数m 、n 的值;(2)设△AOC 的重心为G ,且23OG OB =u u u ru u ur ,且1P 、2P 为线段AB 的三等分点, 求12OA AB OP AB OP AB OB AB ⋅+⋅+⋅+⋅u u u r u u u r u u u r u u u r u u u r u u u r u u u r u u u r的值.20. 已知一列非零向量n a u u r 满足:1a =u u r ,11111(,)(,)2n n n n n n n a x y x y x y ----==-+u u r .(1)写出数列{||}n a u u r的通项公式;(2)求出向量n a u u r 与1n a -u u u r 的夹角θ,并将12,,,n a a a ⋅⋅⋅u u r u u r u u r中所有与1a u u r 平行的向量取出来,按原来的顺序排成一列,组成新的数列{}n b u u r ,12n n OB b b b =++⋅⋅⋅+u u u u r u r u u r u u r,O 为坐标原点,求点列{}n B 的坐标;(3)令8128n n S a a a =++⋅⋅⋅+u u u r u u r u u r u u u r (*n ∈N ),求8{}n S u u u r 的极限点位置.21. 几位大学生响应国家的创业号召,开发了A 、B 、C 三款应用软件,为激发大家学习数学的兴趣,他们推出了“解数学题获取软件激活码”的活动,这三款软件的激活码分别为下面数学问题的三个答案:已知数列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,⋅⋅⋅,其中第一项是02,接下来的两项是02,12,再接下来的三项是02,12,22,以此类推,试根据下列条件分别求三款软件的激活码.(1)A 款应用软件的激活码是该数列中第四个三位数的项数的平方; (2)B 款应用软件的激活码是该数列中第一个四位数及其前所有项的和; (3)C 款应用软件的激活码是满足如下条件的最小整数0N :①01000N >;②该数列的前0N 项和为2的整数幂.参考答案一. 填空题1.62. 53.04. 45.16. 187.1(21)(22)k k ++19. 1410. 8 11.(1)(3)(4)12. 1((0,)22-U二. 选择题13. B 14. B 15. D 16. C三. 解答题17. 设22389n n S n +=--,∴8(91)n n a =-,即证91n -能被8整除,数学归纳法证.18.(1)022301410A ⎛⎫⎪= ⎪⎪⎝⎭,()406080B =; (2)034201210C ⎛⎫⎪= ⎪⎪⎝⎭,()203040D =,()140100110D C ⋅=,即三位市民各获得140、100和110元折扣. 19.(1)1m =,2n =;或8m =,9n =;(2)26.20.(1)(2n ;(2)1()]4(,0)5n--;(3).21.(1)2809;(2)4083;(3)(1)1000452m m m +>⇒≥,21m m b =-, 11222m m b b b m +++⋅⋅⋅+=--,∴12124221k k m -+=+++⋅⋅⋅+=-,2log (3)6k m =+≥,此时61m =,∴06162618972N ⨯=+=。
2018-2019学年七宝中学高二上期中试卷 答案
2018 Mid-term exam Senior Two Semester One [The Key]I. Listening Comprehension (30’)Section A (10’)1-5 DCCAB 6-10 DDBCDSection B (20’) [2 Points for Each]11-13 AAB 14-16 CDB 17-18 CC 19-20 ADII. Grammar (10’)21. haven’t been married / haven’t married22. themselves 23. what 24. requested 25. less patient 26. However 27. to be reasoned 28. regardless of 29. as long as / so long as 30. is callingIII. Vocabulary (20’)(A)31-35 BC/ABC/A/C/AC 36-40 AD/B/CD/D/BD(B)41-45 AC/B/CD/AB/C 46-50 A/D/AD/BC/BDIV. Cloze (15’)51-55 DAACD 56-60 BBCDA 61-65 CABDCV. Reading Comprehension (32’) [2 Points for Each](A)66-69 ADBA(B)70-73 CADB(C)74-77 DCAD(D)78-81 B/C/AB/DVI. Translation (18’) [3’+3’+3’+4’+5’]1.When will / would it be convenient for you to come to get your physical examination report?What time will / would it be convenient for you to come to get your health examination report?2. A qualified college graduate should have a good command of English.3.He often complains to me about not being appreciated at work.4.Busy as my father was at that time, he still kept his promise and took me and my mother to Shanghai DisneyResort.5.The writer illustrated / has illustrated his opinion about love that only when a man loses the one he loves willhe start to truly love her with all the exhibits displayed in his museum.VII. Writing (25’)听力文字:Section A1.M: Do you know what time the train goes into the city?W: Normally it’s every 20 minutes. But, this’s weekend, so I’m no t sure.Q: What does the woman imply the man should do?2.W: There you are, finally. We’d better hurry up if we expect to get a seat. The lectures start in fifteen minutes.M: These things never start on time. And anyway, I don’t think it would be crowded.Q: What does the man imply?3.W: I studied French in high school, but I never really learned it until I spent the summer in Paris.M: Really using the language makes all the differences, doesn’t it?Q: What does the man mean?4.M: Can you believe that Doctor Foster actually gives us an extra week to hand in the papers?W: That’s time I can certainly use. Believe me, I am not even half way through.Q: What is the most probable relationship between the two speakers?5.W: My parents will come next weekend. I’ve chec ked every hotel in the area, but all seem to be full.M: Why not call the Cliffside Inn? It’s not so near the campus but it always gets a few vacant rooms.Q: What does the man suggest the woman do?6.M: I heard you’re looking for someone to pick up your mai l when you are away at the conference next week.W: Oh, could you?Q: What does the woman want the man to do?7.M: I’m sorry. I shouldn’t have volunteered to help you memorize your lines to the play. I still haven’tfinished my essay.W: That’s OK. Sue said she would help me if you couldn’t do it.Q: What will the woman probably do next?8.M: I am really hoping to get that job as a journalist. But I still haven’t heard from the news agency.W: Don’t worry. There must be plenty of other places that would be happ y to have you.Q: What does the woman imply the man should do?9.M: You know, Tim is hard to talk to now, ever since he won that tennis game last week and got thephotograph in the local paper.W: Yeah, I know what you mean. I guess it’s all going to his hea d.Q: What does the woman mean about Tim?10.W: If I can keep up the pace, I will graduate in just 3 years.M: That may be true, but I never want to give up my summer breaks.Q: What does the man imply?Section BQuestions 11 through 13 are based on the following passage.Good evening. I’m Peter Crane with the six o’clock news.At least 17 people have been injured in the road accident that took place on the M1 near Leeds last night. The police said that the truck which caused the accident was traveling about 85 miles an hour, well over the 60-mile-an-hour speed limit for heavy goods vehicles.2,600 workers have walked out of the Peugeot car factory in Coventry in protest against the company’s pay offer. The unions have asked for a rise of 8.5%. There’ll be a me eting between their leaders and management later today.The latest unemployment figures have been released for this year. They show an increase of 150,263 on last year’s figures. This brings the total number of unemployed to approximately 1,490,000. The Em ployment Minister says this increase has been caused by the relocation of several factories from Britain to the Far East.Estate agents are predicting that house prices will continue to rise this year, making it extremely difficult for first-time buyers to get onto the property ladder. It’s estimated that house prices have increased by a third in the last five years. The average price of a three-bedroom house in southeast England is now £ 255,900.And, the weather for the weekend…Questions:11. Which of the following is a reason for the car accident on the Ml?12. What have workers in Coventry protested against?13. What do estate agents say about house prices?Questions 14 through 16 are based on the following passage.This semester we’ve been looking at f ear from biological perspective. And someone asked whether the tendency to be fearful is genetic. Some study done with mice indicates that mammals do inherit fearfulness to some degree. In one study for instance, a group of mice were placed in the brightly-lit open boxes with no hiding places. Some of the mice wandered around the box and didn’t appear to be bothered about being so exposed. But other mice didn’t move. They stayed up against one wall, which indicates that they were afraid. When mice like this were bred with one another repeatedly, after about 12 or so generations, then all of the mice show similar signs of fearfulness. And even when the new born mice from this generation were raised by a mother and with other mice who were not fearful, those m ice still tend to be fearful. Now, why is this? Well, it’s thought that the specific gene in animal bodies have influenced the anxious behavior. These genes are associated with particular nerve cells in the brain. And the degree of overall fearfulness in the mammal seems to depend in large part on the presence or absence of these nerve cells. And this appears to apply to humans as well by the way.Questions:14. What is the passage mainly about?15. How did some of the mice in the box show that they were afraid?16. According to the passage, what leads to a mammal’s tendency to be fearful?Questions 17 through 18 are based on the following conversation.M: You say you worked as a manager at Computer Country. Could you tell me a little more about your responsibilities there?W: Sure. I was in charge of the after-sales service there.M: What was the most difficult part of your job?W: Probably handling angry customers. I had to take care of them properly.M: How long did you work there?W: I was there for three and a half years.M: And why did you leave?W My husband has been transferred to your county and I understand your company has an opening there, too. M: But the position won’t start until the middle of next month. Would that be a problem for you?W: No, not at all.M: Tell me, why are you interested in this position?W: I know that your company has a great reputation and a wonderful product. I would like to be a part of it.M: Well, I’m glad you did.Questions:17. What’s the most probable relationship be tween the two speakers?18. Why did the woman leave her previous job?Questions 19 through 20 are based on the following conversation.M: Hi, Jane, mind if I eat lunch with you?W: No, Mr. Evens, not at all.M: Thanks. I’m wondering if you will be interest ed in a small project we are doing this term.W: What is the project all about?M: More and more students choose not to have meals here and we want to attract them back.W: Well, actually, a lot of students have complained to me about the food. We really need to make some changes. M: I’d like to make some changes in the way we prepare our food. For example, just look at what we had to choose from today. You got a fried hamburger and I got fried chicken. They both contain too much fat.W: But you’d better not get rid of them. They are everybody’s favorites.M: Well, we can certainly keep them but we need to give the people who are health-conscious some choices. For example we could also prepare chicken without the fatty skin, and served on some rice with light sauce. Do you think that would appeal to the students?W: Well, I’d like fat. But you are right, you’d better find out what others think. Oh, sorry, I’ve got to get back to work. I’d like to hear more though. I’ll drop by your office later.M: OK, see you then.Questions:19. What is the purpose of the project that the man is going to do?20. What measure is the man going to take?Keys:1-10 DCCAB DDBCD 11-20 AABCD BCCAD。
上海市2018-2019学年七宝中学高二上期末数学期末试卷
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2018-2019 学年上海市闵行区七宝中学高二(上)期末
数学试卷
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上海市七宝中学2019_2020学年高二数学上学期期中试题(含解析)
上海市七宝中学2019-2020学年高二数学上学期期中试题(含解析)一、填空题1.已知向量()()1,1,2,2m n λλ=+=+,若()()m n m n +⊥-,则=λ_________ . 【答案】3- 【解析】试题分析:因为()()m n m n +⊥-,所以()()220m n m n m n +⋅-=-=,22m n =,即222(1)1(2)2λλ++=++,解得3λ=-.考点:向量垂直的性质,考查学生的基本运算能力. 2.把22111133332223x y x y x y x y x y x y ++表示成一个三阶行列式是________. 【答案】112233213x y x y x y - 【解析】 【分析】根据行列式第一列进行展开,由其逆运算即可得结果. 【详解】根据行列式按第一列展开式,可得22111133332223x y x y x y x y x y x y ++ ()()()()112131221111333322=211131x y x y x y x y x y x y +++⋅-⋅+-⋅-⋅+⋅-⋅112233213x y x y x y =- 故答案为: 112233213x y x y x y - 【点睛】本题考查了行列式按列展开的概念和运算,注意运算的格式,属于基础题. 3.已知向量(1,2)a =,(3,4)b =-,则向量a 在向量b 上的投影为________.【答案】1- 【解析】 【分析】根据向量在向量上的投影的定义,结合向量数量积和模长公式计算可得. 【详解】由定义可得向量a 在向量b 上的投影为||cos ,||a ba ab b⋅<>==1=-.故答案为:1-.【点睛】本题考查了向量在向量上的投影,平面向量数量积和模长公式,属于基础题., 4.若11342x y -=,22342x y -=,则过()11,A x y 、()22,B x y 两点的直线l 的方程为________.【答案】3420x y --= 【解析】 【分析】根据()11,A x y 、()22,B x y 都在同一直线上,结合两点确定一条直线可知直线的唯一性,即得直线方程.【详解】若11342x y -=,22342x y -= 则点()11,A x y 在直线3420x y --=上, 点()22,B x y 在直线3420x y --=上 即()11,A x y 、()22,B x y 都在同一直线3420x y --=上因为两点确定一条直线,所以由()11,A x y 、()22,B x y 确定的直线即为3420x y --=故答案为: 3420x y --=【点睛】本题考查了直线方程的意义,两点确定一条直线,属于基础题.5.已知点(3,1)和(- 4,6)在直线3x-2y+a=0的同侧,则a 的取值范围是【答案】【解析】试题分析:因为点(3,1)和(- 4,6)在直线3x-2y+a=0的同侧,所以,解得a<-7或a>24考点:二元一次不等式表示的平面区域6.直线l 过点()5,3A --且在两坐标轴上的截距相等,则直线l 方程是__________. 【答案】x +y +8=0或3x ﹣5y =0. 【解析】 【分析】当直线经过原点时,直线方程为y =35x ;当直线不经过原点时,设直线方程为x +y =a ,把点A 的坐标代入即可得出.【详解】当直线经过原点时,直线方程为y =35x ,即3x ﹣5y =0; 当直线不经过原点时,设直线方程为x +y =a ,∵直线l 过点A (﹣5,﹣3), ∴﹣3﹣5=a ,∴a =﹣8,∴直线方程为x +y ﹣8=0. 综上,直线方程为x +y +8=0或3x ﹣5y =0. 故答案为:x +y +8=0或3x ﹣5y =0.【点睛】本题考查了直线方程的求法,考查了分类讨论思想,属于基础题. 7.点()1,5A -关于直线90x y -+=的对称点坐标为________. 【答案】()4,8- 【解析】 【分析】设出对称点坐标,根据两个对称点的中点位于直线上,及两直线垂直时的斜率关系,联立方程组即可得对称点的坐标.【详解】设对称点的坐标为(),B a b则AB 中点坐标为15,22a b M -+⎛⎫⎪⎝⎭则M 在直线90x y -+=上,即159022a b -+-+=根据AB 与直线垂直,斜率的关系可得()5111b a -⋅=--- 即()1590225111a b b a -+⎧-+=⎪⎪⎨-⎪⋅=---⎪⎩,解方程组可得48a b =-⎧⎨=⎩即对称点的坐标为()4,8- 故答案为: ()4,8-【点睛】本题考查了点关于直线对称点的坐标求法,两直线垂直的斜率关系,属于基础题. 8.已知P 是ABC △内部一点23PA PB PC ++=0,记PBC 、PAC 、PAB △的面积分别为1S 、2S 、3S ,则::123S S S =________. 【答案】1:2:3 【解析】 【分析】延长PB 到'B ,使得'2PB PB =;延长PC 到'C,使得'3PC PC =,构造出''AB C∆,根据线段关系及三角形面积公式即可求得面积比. 【详解】延长PB 到'B ,使得'2PB PB=;延长PC 到'C ,使得'3PC PC =,如下图所示:则230PA PB PC ++=可化为''0PA PB PC ++=所以P 为''AB C ∆的重心设''''PAB PAC PB C S S S k ∆∆∆=== 则3'1122PAB PAB S S S k ∆∆=== 3'1122PAB PAB S S S k ∆∆=== 2'1133PAC PAC S S S k ∆∆=== ''11111sin sin 2223PBC S S PB PC BPC PB PC BPC ∆⎛⎫⎛⎫==⨯⨯∠=⨯⨯∠ ⎪ ⎪⎝⎭⎝⎭''''1111sin 6266PB C PB PC BPC S k ∆⎛⎫=⨯⨯⨯∠== ⎪⎝⎭ 所以123111::::1:2:3632S S S k k k ⎛⎫⎛⎫⎛⎫== ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭故答案为: 1:2:3【点睛】本题考查了向量加法法则的应用,三角形面积的表示方法,需要构造三角形解决问题,属于中档题.9.在ABC ∆中,5AB =,7AC =,D 是BC 边的中点,则AD BC ⋅=________. 【答案】12 【解析】 【分析】根据三角形中线可知()12AD AB AC =+,结合向量减法运算即可表示出AD BC ⋅,转化为AC 与AB 的等式,即可求得AD BC ⋅的值.【详解】在ABC ∆中,D 是BC 边的中点 所以()12AD AB AC =+ 因为BC AC AB =- 所以()()12AD BC AB AC AC AB ⋅=+- ()()22221122AC AB AC AB =-=- 因为5AB =,7AC =所以()()222211751222AC AB -=-=即12AD BC ⋅= 故答案为: 12【点睛】本题考查了向量的加法及减法运算,平面向量数量积的应用,属于基础题.10.在平面直角坐标系中,O 是坐标原点,两定点,A B 满足·2OA OB OAOB===,由点集{|,1,,}P OP OA OB R λμλμλμ=++≤∈所表示的区域的面积是__________.【答案】43 【解析】【详解】由|OA |=|OB |=OA ·OB =2,知cos∠AOB =12,又0≤∠AOB ≤π,则∠AOB =3π,又A ,B 是两定点,可设A (3,1),B (0,2),P (x ,y ), 由OP =λOA +μOB ,可得3{2x y λλμ=,=+⇒3{32x y x λμ==-.因为|λ|+|μ|≤1,所以33x +32y x -≤1, 等价于由可行域可得S 0=1233,所以由对称性可知点P 所表示的区域面积S =4S 0=4311.在平面上,12AB AB ⊥,12||||1OB OB ==,12AP AB AB =+,若1||2OP <,则||OA 的取值范围是________.【答案】7,2⎛⎤⎥ ⎝【解析】 【分析】根据题意,建立平面直角坐标系,设出A 、1B 、2B 、P 的坐标,由12||||1OB OB ==及1||2OP <可得关于O 点坐标的不等式组,结合两点间距离公式即可表示出||OA 的取值范围. 【详解】因为12AB AB ⊥,12AP AB AB =+则12AB PB 为矩形,以1AB 所在直线为x 轴,以2AB 为y 轴建立平面直角坐标系.如下图所示:设12,AB m AB n ==,(),O x y则()0,0A ,()1,0B m ,()20,B n ,()P m n ,因为12||||1OB OB ==所以()()222211x m y x y n ⎧-+=⎪⎨+-=⎪⎩ 变形可得()()222211x m yy n x⎧-=-⎪⎨-=-⎪⎩ 因为1||2OP <,即()()2214x m y n -+-<由以上两式可得221114x y -+-<即2274x y +>因为()()220,0x m y n -≥-≥,即2210,10y x -≥-≥ 所以221,1y x ≤≤ 则222x y +≤ 综上可知22724x y <+≤ 因为22||x y OA =+所以72OA <≤,即7,22OA ⎛⎤∈ ⎥ ⎝ 故答案为: 7,22⎛⎤⎥⎝ 【点睛】本题考查了平面向量在坐标系中的综合应用,向量的加法运算与向量的模长,通过建立平面直角坐标系,用坐标研究向量关系是常见方法,属于中档题. 12.已知集合(){},,0A x y x y a a =+=>,(){},1B x y xy x y =+=+.若A B ∩是平面上正八边形的顶点所构成的集合,则a 的值为________. 【答案】2或22+ 【解析】【详解】如图所示,正八边形在第一象限的两个顶点坐标应满足,1.x y a xy a +=⎧⎨=-⎩解得111,1.x y a =⎧⎨=-⎩或221,1.x a y =-⎧⎨=⎩据题意218OP k tgπ==-,于11121,2a a -=或1111,2a a-===+二、选择题13.关于x y、的二元一次方程组341310x yx y+=⎧⎨-=⎩的增广矩阵为()A.3411310-⎛⎫⎪-⎝⎭B.3411310-⎛⎫⎪--⎝⎭C.3411310⎛⎫⎪-⎝⎭D.3411310⎛⎫⎪⎝⎭【答案】C【解析】【分析】根据二元一次方程方程组与增广矩阵的关系,即可求得结果.【详解】关于x y、的二元一次方程组341310x yx y+=⎧⎨-=⎩的增广矩阵为31⎛⎝43-110⎫⎪⎭,故选C. 【点睛】本题考查二元一次方程组与系数矩阵及增广矩阵的关系,意在考查对基础知识的掌握情况,属于基础题.14.若变量x,y满足约束条件222xyx y≤⎧⎪≤⎨⎪+≥⎩则目标函数2z x y=+的取值范围是A. [2,6] B. [2,5] C. [3,6] D. [3,5] 【答案】A 【解析】【分析】画出不等式组对应的可行域,将目标函数变形,画出目标函数对应的直线,由图得到当直线过A点时纵截距最大,z最大,当直线过(2,0)时纵截距最小,z最小.【详解】画出可行域,如图所示:将2z x y =+变形为122zy x =-+,平移此直线, 由图知当直线过A (2,2)时,z 最大为6,当直线过(2,0)时,z 最小为2, ∴目标函数Z =x +2y 的取值范围是[2,6] 故选:A .【点睛】本题考查画不等式组表示的平面区域:直线定边界,特殊点定区域结合图形求函数的最值,属于基础题.15.点()5,0A ,(1,43B -到直线的距离都是4,满足条件的直线有( ) A. 一条 B. 两条C. 三条D. 四条【答案】C 【解析】 【分析】先求得两点间距离,根据距离判断中垂线的情况;再由两条平行线可满足条件即可得解. 【详解】因为点()5,0A ,(1,43B - 所以由两点间距离公式可得()()22514316488AB =-+=+=则点()5,0A ,(1,43B -到线段AB 中垂线的距离都等于4位于直线AB 两侧且与AB 直线平行的直线,有两条满足点()5,0A ,(1,43B -到直线的距离都是4综上可知,共有3条直线满足点()5,0A ,(1,43B -到直线的距离都是4 故选:C【点睛】本题考查了两点间距离公式,两平行线距离,属于基础题.16.已知O 是ABC ∆所在平面上的一点,若aPA bPB cPC PO a b c++=++(其中P 是ABC ∆所在平面内任意一点),则O 点是ABC ∆的( )A. 外心B. 内心C. 重心D. 垂心 【答案】B【解析】【分析】将所给向量表达式进行变形,表示成AB 与AC 方向上的单位向量的形式,由向量加法运算的性质即可知O 在角平分线上,即可得解. 【详解】因为aPA bPB cPC PO a b c++=++ 则()a b c PO aPA bPB cPC ++=++,即aPO bPO cPO aPA bPB cPC ++=++ 移项可得0aPA aPO bPB bPO cPC cPO -+-+-=即()()()0a PA PO b PB PO c PC PO -+-+-=则0aOA bOB cOC ++=因为,,OB OA AB OC OA AC =+=+所以()()0OA AB aOA b c OA AC ++++=化简可得0OA AB OA aOA b b c c AC ++++=,即()A a B b OA b cAC c --++= 设i 为AB 方向上的单位向量,j 为AC 方向上的单位向量所以AB ci =,AC b j =则()a b c OA bci bc j ++=--()()i j a b c OA bc -+++= 所以()bc OA a b c i j =++-+ 则O 在BAC ∠的角平分线上同理可知 O 在CBA ∠的角平分线上因而O 为ABC ∆的内心故选:B【点睛】本题考查了向量线性运算的化简及应用,三角形内心的向量表示形式,化简过程较为复杂,属于中档题.三、解答题17.讨论关于x 、y 的方程组()()()()3114541123k x k y k k x k y k ⎧++-=+⎪⎨++-=⎪⎩解的情况. 【答案】当0k ≠且2k ≠时,有唯一解;当0k =时,无解;当2k =时,有无数解.【解析】【分析】先根据方程组中x ,y 的系数及常数项计算出D ,D x ,D y ,下面对k 的值进行分类讨论:(1)当k ≠0且k ≠2时,(2)当k =0时,(3)当k =2时,分别求解方程组的解即可. 【详解】依题意,31142(2)112k kD k k k k +-==--+-,54142(1)(2)312x k kD k k k k +-==---,31542(2)(21)13y k k D k k k k ++==-++,(1)当k ≠0且k ≠2时,D ≠0,方程组有唯一解121x y D k x D k D k y D k -⎧==⎪⎪⎨+⎪==⎪-⎩; (2)当k =0时,D =0,D x =4≠0,方程组无解;(3)当k =2时,D =D x =D y =0方程组有无数解.【点睛】本题主要考查二元一次方程组的矩阵形式、线性方程组解的存在性,唯一性、二元方程的解法等基础知识,考查运算求解能力与转化思想.属中档题.18.ABC ∆的顶点()3,1A -,AB 的中线方程为610590x y +-=,B 的平分线方程为4100x y -+=,求:(1)点B 的坐标;(2)BC 边所在的直线方程.【答案】(1)()10,5B ;(2)29650x y +-=.【解析】【分析】(1)设点B 的坐标,根据点B 在直线4100x y -+=上,及A 与B 的中点在610590x y +-=上,可得关于B 点坐标的方程组,解方程组即可得B 的坐标.(2)根据两点间斜率公式先求得,AB k 由角平分线及两直线的夹角公式即可求得BC k ,再根据点斜式即可求得BC 边所在的直线方程.【详解】(1)设(),B x y ,由题意可知B 在直线4100x y -+=上,及A 与B 的中点在610590x y +-=上,即31610590224100x y x y +-⎧⨯+⨯-=⎪⎨⎪-+=⎩ 解方程组可得105x y =⎧⎨=⎩即()10,5B(2)设B 的平分线与直线AC 交于点E. 则1563107AB k --==-,14BE k = ABE EBC ∠=∠ 由两直线夹角公式可得11BC BE BE AB BE AB BE BC k k k k k k k k --=++,即16147416147411BC BC k k -=+⨯-+ 解得29BC k =- 由点斜式方程可得()25109y x -=-- 化简后可得29650x y +-=【点睛】本题考查了直线方程的求法,两直线夹角公式的用法,属于基础题.19.已知ABC ∆的三边长8AB =,7BC =,3AC =.(1)求AB AC ⋅;(2)A 的半径为3,设PQ 是A 的一条直径,求BP CQ ⋅的最大值和最小值.【答案】(1)12;(2)最大值24,最小值为18-. 【解析】【分析】(1)先根据余弦定理,求得cos BAC ∠,再根据平面向量数量积的定义即可求得AB AC ⋅.(2)根据向量加法与减法的线性运算,将BP CQ ⋅化简为2BA CA AP AP CB ⋅-+⋅,设向量AP 与CB 夹角为θ,进而转化为余弦函数的表达式,根据余弦函数的值域即可求得最大值与最小值.【详解】(1)设BAC θ∠= 则2222228371cos 22832AB AC BC AB AC θ+-+-===⨯⨯⨯⨯, ||||cos AB AC AB AC θ⋅=183122=⨯⨯= (2)()()BP CQ BA AP CA AQ ⋅=+⋅+ ()()BA AP CA AP =+⋅-()2BA CA AP AP CA BA =⋅-+⋅-2BA CA AP AP CB =⋅-+⋅设向量AP ,CB 夹角为()0θθπ≤≤,则上式12937cos 321cos θθ=-+⨯=+,最大值为24,最小值为18-【点睛】本题考查了平面向量数量积的应用,向量加法与减法的应用,余弦定理在求角中的用法,属于中档题.20.在ABC ∆中,2AC =,6BC =,60ACB ∠=︒,点O 为ABC ∆所在平面上一点,满足OC mOA nOB =+(,m n ∈R 且1m n +≠).(1)证明:11m n CO CA CB m n m n =++-+-; (2)若点O 为ABC ∆的重心,求m 、n 的值;(3)若点O 为ABC ∆的外心,求m 、n 的值.【答案】(1)证明见解析;(2)1m =-,1n =-;(2)3757m n ⎧=⎪⎪⎨⎪=-⎪⎩. 【解析】【分析】(1)根据条件OC mOA nOB =+,结合向量的加法运算,化简即可证明.(2)根据重心的向量表示为0OA OB OC ++=,即可求得m 、n 的值.(3)根据点O 为ABC ∆的外心,求得21||2CO CB CB ⋅=,21||2CO CA CA ⋅=,CA CB ⋅,再根据已知分别求得CO CB ⋅,CO CA ⋅,结合平面向量基本定理即可求得m 、n 的值.【详解】(1)CO mAO nBO =+()()m AC CO n BC CO =+++ mAC mCO nBC nCO =+++即CO mAC mCO nBC nCO =+++所以CO mCO nCO mAC nBC --=+则()1m n CO mAC nBC --=+ 所以11m n CO CA CB m n m n =++-+-; (2)若点O 为ABC ∆的重心则0OA OB OC ++=因为OC mOA nOB =+所以0mOA nOB OC --+=则1m =-,1n =-(3)由O 是ABC △的外心 得21||182CO CB CB ⋅==,21||22CO CA CA ⋅==,6CA CB ⋅=, 所以,1111m n CO CB CA CB CB CB m n m n m n CO CA CA CA CB CA m n m n ⎧⋅=⋅+⋅⎪⎪+-+-⎨⎪⋅=⋅+⋅⎪+-+-⎩即23321m n m n -=⎧⎨+=-⎩, 解得3757m n ⎧=⎪⎪⎨⎪=-⎪⎩. 【点睛】本题考查了平面向量加法和减法的运算,三角形重心和外心的向量表示,对向量线性运算的化简要熟练掌握,属于中档题.21.(1)已知直线l 过点53,22P ⎛⎫- ⎪⎝⎭,它的一个方向向量为()3,3m =. ①求直线l 的方程;②一组直线1l ,2l ,,n l ,,2n l ()*n ∈N 都与直线l 平行,它们到直线l 的距离依次为d ,2d ,,nd ,,2nd (0d >),且直线n l 恰好经过原点,试用n 表示d 的关系式,并求出直线(1,2,,2)i l i n =的方程(用n 、i 表示);(2)在坐标平面上,是否存在一个含有无穷多条直线1L ,2L ,,n L ,的直线簇,使它同时满足以下三个条件:①点()1,1n L ∈;②1n n n k a b +=-,其中1n k +是直线1n L +的斜率,n a 和n b 分别为直线n L 在x 轴和y 轴上的截距;③10n n k k +>()*n ∈N .【答案】(1)①40x y -+=;②)*d n =∈N ,410i x y n ⎛⎫-+-= ⎪⎝⎭;(2)不存在. 【解析】【分析】 (1)根据直线的方向向量可得直线的斜率,结合点斜式即可求得直线方程;根据直线平行且过原点,可得直线n l 的方程,由平行线间距离公式可得n 与d 的关系式,设出直线i l 的方程,根据点到直线距离公式可求得直线方程.(2)假设存在这样的直线簇.先求得n a ,n b 的表达式,进而表示出1n k +.通过迭加法求得1n n k k +-,即可证明当21n k >时,10n k +<与10n k +>不能成立.【详解】(1)①直线l 方向向量为()3,3m = 所以直线的斜率为313k == 直线l 过点53,22P ⎛⎫- ⎪⎝⎭,由点斜式方程可得 35122y x ⎛⎫-=⨯+ ⎪⎝⎭ 即直线l 的方程为:40x y -+=;②直线//n l l 且经过原点,∴直线n l 的方程为:0x y -=由题意知直线n l 到l 的距离为nd ,nd =则)*d n n=∈N 设直线(1,2,,2)i l i n =方程为:()04i i x y C C -+=< 由题意知:直线(1,2,,2)i l i n=到直线l id =, 41i i C n ⎛⎫∴=- ⎪⎝⎭所以直线(1,2,,2)i l i n =的方程为:410i x y n ⎛⎫-+-= ⎪⎝⎭; (2)假设存在满足题意的直线簇.由①知n L 的方程为:()11n y k x -=-,1,2,3,n =, 分别令0y =,0x =得11n n a k =-,1nn b k =-, 由11n n n n n k a b k k +=-=-,即11n n nk k k +-=-,1,2,3,n =, 迭加得1112111n n k k k k k +⎛⎫=-+++ ⎪⎝⎭. 由③知所有的()1,2,3,,,i k i n =同号,仅讨论0n k >的情形, 由111110n n n n nk k k k k ++-=-<⇒>, 所以111121111n n n k k k k k k k +⎛⎫=-+++<- ⎪⎝⎭ 显然,当21n k >时,10n k +<与10n k +>矛盾!故满足题意的直线簇不存在.【点睛】本题考查了直线的方向向量与点斜式方程,点到直线距离公式的应用,直线方程的新定义应用,正确理解题目所给条件是关键,属于难题.。
2018-2019学年上海市闵行区七宝中学高二(上)期末数学试卷解析版
2018-2019学年上海市闵行区七宝中学高二(上)期末数学试卷一、选择题(本大题共4小题,共12.0分)1. 与圆x 2+(y +5)2=3相切,且横截距与纵截距相等的直线条数是( )A. 2B. 4C. 6D. 以上说法都不对 【答案】B【解析】解:根据题意,圆x 2+(y +5)2=3的圆心为(0,−5),半径r =√3, 分2种情况讨论,①,直线过原点,设直线的方程为y =kx ,即kx −y =0, 则有|5|√1+k 2=√3,解可得k =±√663,此时直线的方程为:y =±√663x ,②,直线不过原点,由于直线横截距与纵截距相等,设其方程为x +y =a ,即x +y −a =0, 则有|5+a|√2=√3,解可得a =±√6−5,此时直线的方程为x +y ±√6+5=0, 故一共有4条符合条件的直线; 故选:B .根据题意,分析圆的圆心与半径,分2种情况讨论,①,直线过原点,设直线的方程为y =kx ,②,直线不过原点,设其方程为x +y =a ,由直线与圆的位置关系分析直线的条数,综合2种情况即可得答案.本题考查直线与圆的位置关系,涉及直线在坐标轴上的截距,注意直线过原点的情况,属于基础题,2. 直线l 1:2x −3y +1=0与直线l 2:x =3的夹角为( )A. π−arccos 2√1313B. arccos 2√1313C. arcsin 2√1313D. 以上说法都不对【答案】B【解析】解:直线l 1:2x −3y +1=0的斜率为23,倾斜角为arctan 23,直线l 2:x =3的斜率不存在,倾斜角为90∘,故直线l 1:2x −3y +1=0与直线l 2:x =3的夹角为,故选:B .先求出两条直线的倾斜角和斜率,可得两条直线的夹角.本题主要考查直线的倾斜角和斜率,两条直线的夹角,属于基础题.3. 下列说法正确的是( )A. 平面中两个定点A ,B ,k 为非零常数,若||PA|−|PB||=k ,则动点P 的轨迹是双曲线B. 定圆C 上有一定点A 和一动点B(不与A 重合),O 为坐标原点,若OP ⃗⃗⃗⃗⃗ =12(OA ⃗⃗⃗⃗⃗ +OB⃗⃗⃗⃗⃗⃗ ),则动点P 的轨迹是椭圆 C. 斜率为定值的动直线与抛物线y 2=2px(p >0)相交于A ,B 两点,O 为坐标原点,OP ⃗⃗⃗⃗⃗ =12(OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗⃗ ),则动点P 的轨迹是直线 D. 以上说法都不对【答案】C【解析】解:设A ,B 是两个定点,k 为非零常数,若|k|>|AB|,可得P 的轨迹为双曲线的一支;若|k|=|AB|,即为射线;若|k|>|AB|,则轨迹不存在,故A 错误;过定圆C 上一定点A 作圆的动弦AB ,O 为坐标原点.若OP ⃗⃗⃗⃗⃗ =12(OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗⃗ ), 则P 为AB 的中点,CP ⊥AB ,则动点P 的轨迹为以AC 为直径的圆,故B 错误;斜率为定值t 的动直线与抛物线y 2=2px(p >0)相交于A ,B 两点,设A(y 122p,y 1),B(y 222p ,y 2),OP ⃗⃗⃗⃗⃗ =12(OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗⃗ ),可得P 为AB 的中点, t =y 1−y 2y 122p −y 222p=2p y1+y 2=py P ,即有y P =p t,则动点P 的轨迹是直线,故C 正确. 故选:C .由双曲线的定义可判断A ;由P 为AB 的中点,CP ⊥AB ,可得P 的轨迹为圆,可判断B ;由抛物线的方程,可设A(y 122p,y 1),B(y 222p ,y 2),运用直线的斜率公式和中点坐标公式,即可判断C ,进而判断D .本题考查椭圆和双曲线的定义和方程、性质,考查中点坐标公式和直线的斜率公式,以及运算能力和推理能力,属于中档题.4. 点A 为椭圆C :x 2a 2+y 2b 2=1(a >b >0)的右顶点,P 为椭圆C 上一点(不与A 重合),若PO ⃗⃗⃗⃗⃗ ⋅PA ⃗⃗⃗⃗⃗ =0(O 是坐标原点),则ca(c 为半焦距)的取值范围是(( ) A. (12,1)B. (√22,1) C. (√32,1) D. 以上说法都不对【答案】B【解析】解:∵设P(x,y),∵PO ⃗⃗⃗⃗⃗ ⋅PA ⃗⃗⃗⃗⃗ =0(O 是坐标原点), ∴{(x −a2)2+y 2=a 24b 2x 2+a 2y 2=a 2b2⇒c 2x 2−a 3x +a 2b 2=0,⇒(c 2x −ab 2)(x −a)=0. ⇒x =a ,x =ab 2c 2,∴0<ab 2c 2<a .∴b 2<c 2. ∴c a>√22, ∴则ca 的取值范围是(√22,1)故选:B .设P(x,y),由PO ⃗⃗⃗⃗⃗ ⋅PA ⃗⃗⃗⃗⃗ =0,可得{(x −a2)2+y 2=a 24b 2x 2+a 2y 2=a 2b2⇒c 2x 2−a 3x +a 2b 2=0,⇒x =a ,x =ab 2c2,0<ab 2c 2<a.即可求解.本题考查椭圆的离心率的取值范围的求法,考查椭圆性质等基础知识,考查推理论证能力、运算求解能力,考查化归与转化思想、函数与方程思想,是中档题.二、填空题(本大题共12小题,共36.0分) 5. 抛物线y =4x 2的准线方程为______. 【答案】y =−116【解析】解:整理抛物线方程得x 2=14y ,∴p =18 ∵抛物线方程开口向上, ∴准线方程是y =−116 故答案为:y =−116.先把抛物线方程整理成标准方程,进而求得p ,再根据抛物线性质得出准线方程. 本题主要考查抛物线的标准方程和简单性质.属基础题.6. 直线的倾斜角范围是______. 【答案】[0,π)【解析】解:直线的倾斜角的范围是[0,π), 故答案为:[0,π).根据直线倾斜角的定义判断即可.本题考查了直线的倾斜角的范围,考查基础知识的掌握.7. 直线l 1:2x −y −1=0,ax +y +2=0,若11⊥12,则a =______. 【答案】12【解析】解:∵直线l 1:2x −y −1=0,ax +y +2=0,11⊥12, ∴2a −1=0, 解得a =12. 故答案为:12.利用直线与直线垂直的性质直接求解.本题考查实数值的求法,考查直线与直线垂直的性质等基础知识,考查运算求解能力,是基础题.8.直线x−y+5=0被圆x2+y2−2x−4y−4=0所截得的弦长等于______.【答案】2【解析】解:x2+y2−2x−4y−4=0可变为(x−1)2+(y−2)2=9,故圆心坐标为(1,2),半径为3圆心到直线x−y+5=0的距离是√2=2√2故弦长的一半是√9−8=1所以弦长为2故答案为:2.先求出圆心到直线的距离既得弦心距,求出圆的半径,利用勾股定理求出弦长的一半,即可求得弦长本题考查直线与圆相交的性质,解题的关键是了解直线与圆相交的性质,半径,弦心距,弦长的一半构成一个直角三角形,掌握点到直线的公式,会用它求点直线的距离.9.P是双曲线x29−y216=1上的一点,F1,F2为焦点,若|PF1|=7,则|PF2|=______.【答案】13【解析】解:双曲线x29−y216=1,其中a=3,b=4,c=5又由P是双曲线上一点,则有||PF1|−|PF2||=2a=6,又由|PF1|=7,则|PF2|=1<c−a=2(舍去)或13,故答案为:13.由双曲线的标准方程分析可得a、c的值,结合双曲线的定义可得||PF1|−|PF2||=2a= 6,计算可得|PF2|分析可得答案.本题考查双曲线的定义,注意由双曲线的标准方程求出a的值,属于基础题.10.过点(2,3)且与原点距离为2的直线方程是______.【答案】5x−12y+26=0或x=2【解析】解:当直线的斜率不存在时,直线x=2时满足条件;当直线的斜率存在时,设直线的方程为:y−3=k(x−2),化为kx−y+3−2k=0,∴√k2+1=2,解得k=512.∴直线的方程为:512x−y+3−56=0,化为5x−12y+26=0.综上可得:直线的方程为:5x−12y+26=0;x=2.故答案为:5x−12y+26=0或x=2.分直线的斜率存在与不存在讨论,利用点到直线的距离公式即可得出.本题考查了点到直线的距离公式、点斜式、分类讨论方法,考查了计算能力,属于基础题.11.已知p:曲线C上的点的坐标都是方程F(x,y)=0的解,q:曲线C是方程F(x,y)=0的曲线,则p成立是q成立的______条件.【答案】必要不充分【解析】解:若曲线C是方程F(x,y)=0的曲线,则曲线C上的点的坐标都是方程F(x,y)=0的解,即充分性成立,若曲线C上的点的坐标都是方程F(x,y)=0的解,则曲线不一定是方程的曲线,即充分性不成立,比如:曲线y =x(x ≥0)上的点的坐标都满足方程x −y =0, 而方程x −y =0对应的曲线为直线y =x , 则p 成立是q 成立的必要不充分条件, 故答案为:必要不充分条件根据充分条件和必要条件的定义进行判断即可. 本题主要考查充分条件和必要条件的判断,结合曲线方程和方程曲线的关系是解决本题的关键.12. 已知P 是椭圆x 2a 2+y 2b 2=1(a >b >0)上的一点,F 1,F 2为焦点,若PF 1⃗⃗⃗⃗⃗⃗⃗ ⋅PF 2⃗⃗⃗⃗⃗⃗⃗ =0,tan∠PF 1F 2=12,则椭圆的焦距与长轴的比值为______. 【答案】【解析】解:设PF 1=2m ,∵PF 1⃗⃗⃗⃗⃗⃗⃗ ⋅PF 2⃗⃗⃗⃗⃗⃗⃗ =0,tan∠PF 1F 2=12,则PF 2=m . ∴2m +m =2a ,可得PF 1=4a3,PF 2=2a 3,∴(4a3)2+(2a3)2=(2c)2. ⇒c 2a 2=59,则椭圆的焦距与长轴的比值为ca=√53.故答案为:√53.可得PF 1=4a3,PF 2=2a3,由勾股定理可得c 2a 2=59,即可.本题考查了椭圆的性质,转化思想,属于基础题.13. 直线y =kx −2与双曲线x 2−y 2=1有且仅有一个公共点,则k =______. 【答案】±1或±√5【解析】解:联立{x 2−y 2=1y=kx−2,化为(1−k 2)x 2+4kx −5=0.①当1−k 2=0时,可得k =±1,此时直线l 的方程为y =±x +1, 分别与等轴双曲线的渐近线y =±x 平行,此时直线l 与双曲线有且只有一个交点,满足题意;②当1−k 2≠0时,由直线与双曲线有且只有一个公共点, 可得△=16k 2+20(1−k 2)=0, 解得k =±√5.此时满足条件. 综上可得:k =±1,±√5. 故答案为:±1,±√5.联立直线与双曲线方程,化为(1−k 2)x 2+4kx −5=0.分类讨论:当1−k 2=0时,可得k =±1,此时直线l 与等轴双曲线的渐近线,满足题意;当1−k 2≠0时,由直线与双曲线有且只有一个公共点,可得△=0,解出即可.本题考查了直线与双曲线的位置关系及其性质、一元二次方程与△的关系、分类讨论等基础知识与基本方法,属于中档题和易错题.14. 若x ,y 满足约束条件{x ≥0x +y −3≥0x −2y ≤0,则z =x +2y 的取值范围是______.【答案】[4,+∞)【解析】解:x ,y 满足约束条件{x ≥0x +y −3≥0x −2y ≤0, 表示的可行域如图:目标函数z =x +2y 经过C 点时,函数取得最小值,由{x −2y =0x+y−3=0解得C(2,1),目标函数的最小值为:4目标函数的范围是[4,+∞). 故答案为:[4,+∞).画出约束条件的可行域,利用目标函数的最优解求解即可.本题考查线性规划的简单应用,画出可行域判断目标函数的最优解是解题的关键.15. 已知曲线Γ的参数方程为{x =−t 31+t2y =t21+t 2(t 为参数),则以下曲线Γ的说法中: ①关于原点对称;②在直线y =1下方;③关于y 轴对称;④是封闭图形,正确的有______. 【答案】②【解析】解:∵曲线Γ的参数方程为{x =−t 31+t 2y =t 21+t2(t 为参数), ∴y =−1t x ,且0<y <1,在①中,曲线Γ不能关于原点对称,故①错误;在②中,曲线Γ在直线y =1下方,故②正确; 在③中,曲线Γ关于y 轴不对称,故③错误; 在④中,曲线Γ不是封闭图形,故④错误. 故答案为:②.由曲线Γ的参数方程推导出y =−1t x ,且0<y <1,由此能求出结果.本题考查命题真假的判断,考查参数方程、直角坐标方程的互化等基础知识,考查运算求解能力,是中档题.16. 已知F 1,F 2分别为双曲线C :x 2a 2−y 2b 2=1(a,b >0)的左、右焦点,过F 2的直线l 与双曲线的右支分别交于A ,B 两点,△AF 1F 2的内切圆半径为r 1,△BF 1F 2的内切圆半径为r 2,若r 1=2r 2,则直线l 的斜率为______. 【答案】±2√2【解析】解:记△AF1F2的内切圆圆心为C,边AF1、AF2、F1F2上的切点分别为M、N、E,易见C、E横坐标相等,则|AM|=|AN|,|F1M|=|F1E|,|F2N|=|F2E|,由|AF1|−|AF2|=2a,即|AM|+|MF1|−(|AN|+|NF2|)=2a,得|MF1|−|NF2|=2a,即|F1E|−|F2E|=2a,记C的横坐标为x0,则E(x0,0),于是x0+c−(c−x0)=2a,得x0=a,同样内心D的横坐标也为a,则有CD⊥x轴,设直线的倾斜角为θ,则∠OF2D=θ2,∠CF2O=90∘−θ2,在△CEF2中,tan∠CF2O=tan(90∘−θ2)=r1|EF|,在△DEF2中,tan∠DF2O=tanθ2=r2|EF|,由r1=2r2,可得2tanθ2=tan(90∘−θ2)=cotθ2,解得tanθ2=√22,则直线的斜率为tanθ=2tanθ21−tan2θ2=√21−12=2√2,由对称性可得直线l的斜率为±2√2.故答案为:±2√2.充分利用平面几何图形的性质解题.因从同一点出发的切线长相等,得|AM|=|AN|,|F1M|=|F1E|,|F2N|=|F2E|,再结合双曲线的定义得|F1E|−|F2E|=2a,从而即可求得△AF1F2的内心的横坐标a,即有CD⊥x轴,在△CEF2,△DEF2中,运用解直角三角形知识,运用正切函数的定义和二倍角公式化简即可得到直线的斜率.本题考查双曲线的定义、方程和性质,考查三角形的内心的概念,考查三角函数的化简和求值,考查直线斜率的求法,属于中档题三、解答题(本大题共5小题,共60.0分)17.已知两点A(1,2),B(5,−1).(1)求直线AB的方程;(2)若A,B的到直线l的距离都是2,求直线l的方程.【答案】解:(1)∵两点A(1,2),B(5,−1) ∴直线AB 的方程为:y+1x−5=2+11−5,整理,得3x +4y −11=0.(2)∵两点A(1,2),B(5,−1).A ,B 的到直线l 的距离都是2, ∴当直线l 的斜率不存在时,直线l 的方程为x =3,成立;当直线l 的斜率k 存在时,设直线l 为y =kx +b ,即kx −y +b =0, ∵A ,B 的到直线l 的距离都是2,∴√k 2+1=2√k 2+1=2,解得k =−34,b =14或k =−34,b =214,或k =724,b =−38, ∴直线l 的方程为y =−34x +14或y =−34x +214或y =724x −38, 整理,得:3x +4y −1=0,或3x +4y −21=0,或7x −24y −9=0.综上,直线l 的方程为3x +4y −1=0,或3x +4y −21=0,或7x −24y −9=0,或x =3.【解析】(1)利用两点式方程能求出直线AB 的方程.(2)由两点A(1,2),B(5,−1).A ,B 的到直线l 的距离都是2,当直线l 的斜率不存在时,直线l 的方程为x =3;当直线l 的斜率k 存在时,设直线l 为kx −y +b =0,由A ,B 的到直线l 的距离都是2,能求出直线l 的方程.本题考查直线方程的求法,考查直线的两点式方程、点到直线的距离公式等基础知识,考查运算求解能力,是基础题.18. 双曲线M :x 2a 2−y 2b 2=1过点P(4,√62),且它的渐近线方程是x ±2y =0.(1)求双曲线M 的方程; (2)设椭圆N 的中心在原点,它的短轴是双曲线M 的实轴,且椭圆N 中斜率为−3的弦的中点轨迹恰好是M 的一条渐近线截在椭圆N 内的部分,试求椭圆N 的方程. 【答案】解:(1)∵双曲线M :x 2a 2−y 2b 2=1过点P(4,√62),且它的渐近线方程是x ±2y =0, ∴16a 2−64b 2=1,b a =12,解得a 2=10,b 2=52, ∴双曲线M 的方程为x 210−2y 25=1,(2)椭圆N 的中心在原点,它的短轴是双曲线M 的实轴,则设椭圆的方程为:y 2a2+x 2b 2=1,∴b 2=10,设椭圆N 中斜率为−3的弦所在直线方程为y =−3x +m , 两端点分别为A(x 1,y 1),B(x 2,y 2),AB 的中点为P(x 0,y 0),联立方程组{y 2a 2+x 210=1y =−3x +m ,消y 可得(90+a 2)x 2−60mx +10m 2−10a 2=0,∴x 1+x 2=60m90+a 2, 则x 0=30m90+a 2,∴y 0=−3x 0+m =a 2m 90+a 2,∴P(30m90+a 2,a 2m90+a 2),∵点P 在双曲线M 渐近线x −2y =0上, ∴30m90+a 2=2×a 2m90+a 2, 解得a 2=15, ∴椭圆N 的方程为y 215+x 210=1.【解析】(1)根据双曲线的简单性质即可求出,(2)设椭圆N 中斜率为−3的弦所在直线方程为y =−3x +m ,两端点分别为A(x 1,y 1),B(x 2,y 2),AB 的中点为P(x 0,y 0),根据韦达定理求出点P 的坐标,再将点P 的坐标代入双曲线M 渐近线x −2y =0上,即可求出a 2=15,问题得以解决.本题考查了双曲线和椭圆的方程,以及直线和椭圆的位置关系,考查了运算求解能力和转化与化归能力,属于中档题19. 已知椭圆的两个焦点为F 1(−1,0),F 2(1,0),且椭圆过点(1,√22).(1)求椭圆的方程.(2)已知斜率为k(k ≠0)的直线11过F 2,与椭圆分别交于P ,Q ;直线l 2过F 2,与直线11垂直,与椭圆分别交于M ,N ,求四边形PMQN 面积的函数解析式f(k). 【答案】解:(1)设椭圆的方程为x 2a 2+y 2b 2=1,a >b >0由题意可得{c =11a 2+12b 2=1a 2=b 2+c 2,解得a 2=2,b 2=1(2)设直线l 1的方程为y =k(x −1),则直线l 2的方程为y =−1k (x −1) 设P(x 1,y 1),Q(x 2,y 2),联立方程{x 22+y 2=1y =k(x +1),化简得(2k 2+1)x 2−4k 2x +2k 2−2=0.则x 1+x 2=4k 21+2k 2,x 1x 2=2k 2−21+2k 2,∴|PQ|=√1+k 2⋅|x 1−x 2|=√1+k 2⋅√(x 1+x 2)2−4x 1x 2=√1+k 2⋅√16k 4(1+2k 2)2−8k 2−81+2k 2=2√2⋅1+k 21+2k 2, 同理,得|MN|=2√2⋅1+k 22+k 2,∴S 四边形PMNQ ═12|PQ||MN|=4(1+k 2)2(1+2k 2)(2+k 2), ∴f(k)=4(1+k 2)2(1+2k 2)(2+k 2),k ≠0.【解析】(1)设椭圆的方程为x 2a 2+y 2b 2=1,a >b >0,由题意可得{c =11a 2+12b 2=1a 2=b 2+c 2,解得即可,(2)设直线l 1的方程为y =k(x −1),则直线l 2的方程为y =−1k (x −1),设P(x 1,y 1),Q(x 2,y 2),根据弦长公式,分别求出|PQ|,|MN|,即可表示四边形的面积.本题考查直线和椭圆的位置关系,考查四边形面积的求法,解题时要认真审题,注意椭圆弦长公式的合理运用.20. 设直线l :x =ky +b 与抛物线y 2=4x 相交于不同的两点A ,B ,M 为线段AB 中点, (1)若M(3,2),且b =1,求线段AB 的长;(2)若直线l 与圆C :(x −5)2+y 2=16相切于点M ,求直线l 的方程;(3)若直线l 与圆C :(x −5)2+y 2=r 2(0<r <5)相切于点M ,写出符合条件的直线l 的条数.(直接写出结论即可)【答案】解:(1)当b =1时,直线l 的方程为x =ky +1,设点A(x 1,y 1)、B(x 2,y 2), 将直线l 的方程与抛物线的方程联立得{y 2=4x x=ky+1,消去x 得y 2−4ky −4=0, 由韦达定理可得y 1+y 2=4k ,y 1y 2=−4, 由于点M(3,2)是线段AB 的中点,则y 1+y 22=2k =2,则k =1,所以,y 1+y 2=4,由弦长公式可得|AB|=√1+12⋅|y 1−y 2|=√2⋅√(y 1+y 2)2−4y 1y 2=√2×√42−4×(−4)=8;(2)设点A(x 1,y 1)、B(x 2,y 2),将直线l 的方程与抛物线的方程联立{y 2=4x x=ky+b,消去x 得y 2−4ky −4b =0, 由韦达定理可得y 1+y 2=4k ,y 1y 2=−4b , 设点M 的坐标为(x 0,y 0),则y 0=y 1+y 22=2k ,x 0=ky 0+b =2k 2+b ,所以,点M 的坐标为(2k 2+b,2k),①若k =0,则直线l 的方程为x =b ,则点A 、B 关于x 轴对称,而圆(x −5)2+y 2=16与x 轴的交点为(1,0)和(9,0),点M 的坐标为(b,0),则b =1或b =9,此时,直线l 的方程为x =1或x =9,符合题意!②若k ≠0,由CM ⊥AB 可知,k CM ⋅k AB =−1,即k2k 2+b−5⋅1k =−1,则有2k 2+b −5=−1,所以,2k 2+b −4=0. 另一方面,点M 在圆上,则有(2k 2+b −5)2+k 2=16,所以,1+k 2=16,则k 2=15,b =−26,此时,△=16k 2+16b =16(k 2+b)=−176<0,不合乎题意! 综上所述,直线l 的方程为x =1和x =9; (3)当r ≥5时,1条;当0<r ≤2或4≤r <5时,2条; 当2<r <4时,4条.【解析】(1)将直线l 的方程与抛物线的方程联立,利用韦达定理求出线段AB 的中点的纵坐标,从而求出k 的值,再利用弦长公式结合韦达定理可求出线段AB 的长度; (2)对k 是否为零进行分类讨论.当k =0时,得出点A 、B 关于x 轴对称,得知,点M 为圆与x 轴的交点,求出b 的值,可得出直线l的方程;当k≠0时,将直线l的方程与抛物线的方程联立,利用韦达定理求出点M的坐标,然后利用CM⊥AB,转化为这两条直线的斜率之积为−1《以及点M在圆上,求出k和b 的值,但同时还需满足△>0.结合这两种情况求出直线l的方程.(3)结合图形充分利用对称性可写出相应的结论.本题考查直线与抛物线的综合问题,考查韦达定理在抛物线综合问题中的应用,同时也考查了对称性思想的应用,属于难题.21.在平面直坐标系xOy中有曲线Γ:x2+y2=1(y>0).(1)如图1,点B为曲线Γ上的动点,点A(2,0),求线段AB的中点的轨迹方程;(2)如图2,点B为曲线Γ上的动点,点A(2,0),将△OAB绕点A顺时针旋转90∘得到△DAC,求线段OC长度的最大值.(3)如图3,点C为曲线Γ上的动点,点A(−1,0),B(1,0),延长AC到P,使CP=CB,求动点P的轨迹长度.【答案】解:(1)设点B的坐标为(x0,y0),则y0>0,设线段AB的中点为点M(x,y),由于点B在曲线Γ上,则x02+y02=1,①因为点M为线段AB的中点,则{x=x0+22y=y02,得{y=2yx0=2x−2,代入①式得(2x−2)2+y2=1,化简得(x−1)2+y2=14,其中y>0;(2)如下图所示,易知点D(2,2),结合图形可知,点C在右半圆D:(x−2)2+y2=1上运动,问题转化为,原点O到右半圆D上一点C的距离的最大值,连接OD并延长交右半圆D于点,当点C与点重合时,|OC|取最大值,且|OC|max=|OD|+1=2√2+1;(3)如下图所示,由于点C是曲线Γ上一点,则∠ACB=90∘,则∠BCP=90∘,由于CP=CB,所以,∠APB= 45∘,由于AB=2,由正弦定理可知,△ABP的外接圆的直径为2R=ABsin∠APB=2√2,∴R=√2,设曲线Γ与y轴交于点D(0,1),则AD=BD=√2,则△ABP的外接圆即为圆D:x2+(y−1)2=2,弦AB在圆D中所对的圆心角为∠ADB= 90∘,所以点P的轨迹是以√2为半径且圆心角为3π2的扇形,所以,点P的轨迹的长度为3π2R=3√22π.【解析】(1)设点B的坐标为(x0,y0),设线段AB的中点为点M(x,y),先将点M的坐标代入曲线Γ的方程,得出一个等式,由中点坐标公式得{x=x0+22y=y02,解得{y=2yx0=2x−2,代入等式可得出点M的轨迹方程,化简,同时标出相应变量的取值范围;(2)作出曲线Γ绕点A顺时针旋转90∘后得到的右半圆D,然后线段OC长度的最大值就转化为点O到右半圆D上一点距离的最大值,利用圆的性质可得出答案;(3)先求出∠APB=π4,由正弦定理知,动点P的轨迹在圆D上,求出弦AB所对的圆心角,可求出动点P轨迹在圆D中所对的圆心角,即可算出相应扇形的弧长.本题考查曲线方程的求法,考查数形结合思想在解题中的应用,属于中等题.。
上海市七宝中学2019-2020学年高二第一学期数学期中考试卷(简答)
七宝中学高二期中数学卷2019.11一. 填空题1. 已知向量,,若,则(1,1)m λ=+u r (2,2)n λ=+r ()()m n m n +⊥-u r r u r r λ=2. 把表示成一个三阶行列式是 22111133332223x y x y x y x y x y x y ++3. 已知向量,,则向量在向量上的投影为(1,2)a =r (3,4)b =-a r b r 4. 若,,则过、两点的直线的方程为11342x y -=22342x y -=11(,)A x y 22(,)B x y l 5. 点和在直线的两侧,则的取值范围是(3,1)(4,6)-320x y a -+=a 6. 直线过点,且在两坐标轴上的截距相等,则直线的方程为l (5,3)A --l 7. 点关于直线的对称点坐标为(1,5)A -90x y -+=8. 已知是△内部一点,记△、△、△的面P ABC 230PA PB PC ++=u u r u u r PBC PAC PAB 积分别为、、,则1S 2S 3S 123::S S S =9. 在△中,,,是边的中点,则ABC 5AB =7AC =D BC AD BC ⋅=u u u r u u u r 10. 在平面直角坐标系中,是坐标原点,两定点、满足,O A B ||||2OA OB OA OB ==⋅=u u r u u u r u u r u u u r 则点集所表示的区域的面积为{|,||||1,,}P OP OA OB λμλμλμ=++≤∈R u u u r u u r u u u r 11. 在平面上,,,,若,则12AB AB ⊥u u u r u u u r 12||||1OB OB ==u u u r u u u r 12AP AB AB =+u u u r u u u r u u u r 1||2OP <u u u r 的||OA u u r 取值范围是12. 已知集合(),,若{(,)|||||}A x y x y a =+=0a >{(,)|||1||||}B x y xy x y =+=+是平面上正八边形的顶点所构成的集合,则A B I a =二. 选择题13. 关于、的二元一次方程组的增广矩阵为( )x y 341310x y x y +=⎧⎨-=⎩A. B. C. D. 3411310-⎛⎫ ⎪-⎝⎭3411310-⎛⎫ ⎪--⎝⎭3411310⎛⎫ ⎪-⎝⎭3411310⎛⎫ ⎪⎝⎭14. 若、满足约束条件,则的取值范围是( )x y 222x y x y ≤⎧⎪≤⎨⎪+≥⎩2z x y =+A. B. C. D. [2,6][2,5][3,6][3,5]15. 点,到直线的距离都是4,满足条件的直线有()(5,0)A (1,B -A. 一条 B. 两条 C. 三条 D. 四条16. 已知是△所在平面上的一点,若(其中是△O ABC aPA bPB cPC PO a b c++=++u u r u u r u u u r u u u r P 所ABC 在平面内任意一点),则点是△的( )O ABC A. 外心 B. 内心 C. 重心 D. 垂心三. 解答题17. 用行列式的方法解关于、的方程组,并对解的情况x y (31)(14)54(1)(12)3k x k y k k x k y k ++-=+⎧⎨++-=⎩进行讨论.18. △的顶点,的中线方程为,的平分线方程为ABC (3,1)A -AB 610590x y +-=B ∠,求:4100x y -+=(1)点的坐标;B (2)边所在的直线方程.BC 19. 已知△的三边长,,.ABC 8AB =7BC =3AC =(1)求;AB AC ⋅u u u r u u u r (2)的半径为3,设是的一条直径,A e PQ A e 求的最大值和最小值.BP CQ ⋅u u ru u u r20. 在△中,,,,点为△所在平面上一点,ABC 2AC =6BC =60ACB ∠=︒O ABC 满足(且).OC mOA nOB =+u u u r u u r u u u r ,m n ∈R 1m n +≠(1)证明:;11m n CO CA CB m n m n =++-+-u u u r u u r u u r (2)若点为△的重心,求、的值;O ABC m n (3)若点为△的外心,求、的值.O ABC m n 21.(1)已知直线过点,它的一个方向向量为.l 53(,)22P -(3,3)m =u r ① 求直线的方程;l ② 一组直线()都与直线平行,它们到直线的距离依次为122,,,,,n n l l l l ⋅⋅⋅⋅⋅⋅*n ∈N l l (),且直线恰好经过原点,试用表示的关系式,并,2,,,,2d d nd nd ⋅⋅⋅⋅⋅⋅0d >n l n d 求出直线()的方程(用、表示);1l 1,2,,2i n =⋅⋅⋅n i (2)在坐标平面上,是否存在一个含有无穷多条直线的直线簇,使它同时12,,,,n L L L ⋅⋅⋅⋅⋅⋅满足以下三个条件:① 点;② ,其中是直线的斜率,(1,1)n L ∈1n n n k a b +=-1n k +1n L +和分别为直线在轴和轴上的截距;③ .()n a n b n l x y 10n n k k +>*n ∈N参考答案一. 填空题1. 2. 3. 4. 3-112233213x y x y x y -1-3420x y --=5.6. 或7.8. 724a -<<350x y -=80x y ++=(4,8)-1:2:39. 12 10.11. 12.2二. 选择题13. C14. A 15. C 16. B三. 解答题。
2018-2019学年七宝中学高二上学期期中卷学生版
2018-2019学年上海市闵行区七宝中学高二第一学期期中卷Ⅱ. Grammar (10)Directions: After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given words;for the other blanks, use one word that best fits each blank.As a man who has already stepped into the tomb of love, I always have a feeling that it is my obligation to warn those guys who (21)__________(not; marry) so far or are about to do that of the danger involved in marriage. Those simple-minded men can easily get (22)__________ “killed”, if they don’t know that comes with marriage.Since I am a quick learner, it didn’t take me long to find out (23)__________ a man needs to do to survive a marriage. Here are the 7 golden principles, which I would like t o call “Marriage Survival Tips for Man”. Ready to take notes?No.1: Anything (24)__________(request) by your wife should be doubtless taken as your priority in your to-do list. Whether it is to mop the floor or babysit your son, do it immediately and do it for life, idiot! Because the more times she has to ask you to do that, the (25)__________(patient) she will be.No.2: Caution! Do not argue with your wife! (26)__________ reasonable you think you are, you will always end up saying sorry to her. Trust me. As a man, the last thing you want to do is to get into an argument with a woman. Women are not supposed (27)_______(reason) with. They are not designed that way. (Pardon me for being such a male chauvinist(直男炎). Deep down, I totally support feminism.) So unless you want to start a full-scale war you can never win, otherwise apologize to her the moment you two make eye contact.No.3: Always give positive feedbacks (28)________ ________ any questions raised by her. For example, if she wants to buy a pair of high-heels or a fancy coat, say yes! Of course, always saying yes will cost you a fortune but at least it can save your life! And (29)_______ _______ ______ you are alive, money does grow on trees! But when the questions are related to her weight, stay with a simple life-saving rule: for God’s sake, she is not even a little bit fat! Calling her fat would be viewed as a horrible crime, which deserves a death sentence!No.4: Sorry man! I can’t make this up anymore. Her Majesty (30)__________(call) me again. Ihave to go now! Pray for me!III.Vocabulary(20)Directions:Complete the following passages by using the words in the box.. Each word can only be used once. Do note that there is one word more than you need.(A)What’s the point of reading if we all forget what we read sooner or later? Being a liberal arts(文科) teacher, more often than not, I will find myself confronted with this sort of question. And knowing that I am not in a good position to (31)_________ this question which can be simple and hard at the same time, I always quote what William Johnson Cory has written in a letter in which he articulates(清晰表达) the purpose of education. From my perspective, his words can also be taken as a (32)_______ argument for reading. The full quotation goes:“At school you are (33)________ not so much in acquiring knowledge as in making mental efforts under criticism. A certain amount of knowledge you can indeed with average faculties acquire so as to retain; nor need you (34)_________ the hours you spent on much that is forgotten, for the (35)_________ of lost knowledge at least protects you from many illusions. But you go to a great school not so much for knowledge as for arts and habits; for the habit of attention, for the art of entering (36)_________ into another person’s thoughts, for the habit of submitting to censure(批判) and refutation(反驳), for the art if indicating assent(附议) or dissent(异议) in gradated terms, for the habit of regarding minute points of accuracy, for the art of working out what is possible in a given time, for taste, for discrimination, for mental courage, and for mental soberness(清醒).”Besides, even though we eventually forget most of what we read, there can still be (37)_________ in the moments of reading a particular book in a particular place. What I remember most about Hemingway’s short-story (38)________ Men Without Women is the persistent cough I had due to a heavy cold, which at that time did give me a strong feeling that Iwould also like to die for my beloved.So the purpose of reading is not to memorize (39)_________ everything we read but to keep reading. After all, if there are no (40)________ opinions on the idea that we are what we eat, then we are also what we read.(B)You are my Solskjaer,My Ole SolskjaerYou make me happy when skies are grey.You’ll never know dear,How much I love you.Please don’t take my Solskjaer away.You are my SolskjaerMy only SolskjaerYou make my happy when skies are greyCoz when it’s pouring you just keep scoringPlease don’t take our Solskjaer awayThe above song is (41)_________ from Jimmie Davis’ You Are My Sunshine by Manchester United(曼联) fans to express their overwhelming love towards Ole Gunnar Solskjaer(索尔斯克亚), their favourite striker - Babyface Killer. Although the fans never want to bid (42)__________ to their beloved Ole, every journey always has an end, no matter how wonderful it is.Finally, the story of Ole as a United player ended. On 28 Aughst 2007 Ole Gunnar Solskjaer (43)__________ called time on his United playing career - eleven years and three days after it began with a goalscoring debut.On the field, the Norwegian became a perfect (44)_______ of United’s never-say-die spirit in their most successful decade, often scoring late goals when points or trophies looked to have been lost. But off the field, he finally had to admit one defeat - in his battle with his (45)________ knee injury.On retiring, Ole said,” I would like to tha nk the manager, the coaching and medical staff and most of all the supporters. The supporters have been fantastic ad were a real (46)_________ to me when I was out injuried. And their tremendous support will always be in my heart. I feel proud to have (47)________ United for 11 years and have some very special memories.”Ole’s many career (48)________ include his four goals in 12 minutes as a substitute at Nottingham Forest in United’s recrod away win(8-1). But surely the most special memory of all would be the Champions League final in 1999 when he best (49)________ his ability to seize the whisker of a chance. His injury-time-toe-poke past Bayern Munich keeper Oliver Kahn(卡恩) completed United’s Treble(三冠王)and (50)__________ himself an irreplaceable place in Reds history.IV.Cloze (15)Directions: For each blank in the following passage there are four words or phrases marked A,B, C and D.Fill in each blank with the word or phrase that best fits the context.It did take me quite a while to start noticing Dr. Y uval Noah Harari’s well-received book: Sapiens: A Brief History of Humankind(人类简史). I bought the book after I heard Dou Wentao (a renowned TV hose) mention it on his podcast, and to be frank, I read the first chapter with little (51)______. But it turned out to be the best book I read in 2017.While I was immediately (52)________, the book kept evolving as I read it. The book began with a brief introduction of the lives and activities of the earliest proto-humans(原始人) --- Neanderthals(尼安德特人), Homo erectus(直立人) and early Homo Sapiens(智人) --- and then (53)_______ an examination of why it was the Homo sapiens, after hundreds of thousands of years of surviving but pretty much existing in the middle of the food chain, (54)________ rocketed to the top of it without ant significant genetic changes, conquered multiple climates, and eventually domesticated the world around them, from farm animals to crops. And Harari includesan interesting but fairly (55)_______ argument about the true nature of our relationship to our most necessary crop ---wheat.Think for a moment about the (56)_______ Revolution from the viewpoint of wheat. Ten thousand years ago wheat was just a wild grass, one of many, confined to a small range in the Middle East. All of a sudden, within just a few short millennia, it was growing everywhere.So how did this grass turn from insignificant to ubiquitous(到处存在的)? Wheat did it by manipulating(操纵) Homo sapiens to its advantage. This ape had been living a fairly comfortable life (57)______ and gathering until about 10,000 years ago, but then began to (58)_________ more and more effort in cultivating wheat. Then, humans in many parts of the world were doing little from dawn to dusk (59)_______ taking care of wheat plants.However, the body of Homo sapiens had not evolved for cultivating wheat. Therefore human spines, knees, necks and arches paid the price. Moreover, the new agricultural tasks demanded so much time that people were forced to settle (60)_______ next to their wheat fields. This completely changed their way of life. We did not (61)_______ wheat. It’s the other way around. One of the most important and sustained ideas running through the book is that what ultimately (62)______ Homo sapiens from all other creatures --- other mammals, other apes, and even other “humans” like Neanderthals ---was not our opposable thumbs or some other (63)________ standards, but instead it was our ability to generate(生成), believe in and act upon what Yuval Noah Narari c alls “myths” or “(64)________” (essentially ideas and cultural institutions), particularly on a large scale and collective basis, which eventually transformed us from creatures that lived in small, loosely-organized groups (the typical feature of most apes) to our modern status.Sapiens: A Brief History of Humankind is a fascinating, ambitious and difficult-to-summarize book that is also just highly (65)_______. So as senior high school students, you won’t experience too much difficulty following the author’s train of thoughts. And I strongly recommend you to indulge(纵情于) yourself in this well-written book.Inspired by Yuval Noah Harari’s mind-blowing book:Sapiens: A Brief History of Humankind51.A.attention B.evaluation C.illustration D.expectation52.A.fascinated B.confused C.distracted D.uninterested53.A.turned to B.gave away C.prepared for D.went after54.A.naturally B.randomly C.suddenly D.hardly55.A.annoying B.touching C.embarrassing D.depressing56.A.Industrial B.Agricultural C.Cultural D.Political57.A.planting B.hunting C.trading D.wondering58.A.spare B.resist C.invest D.demand59.A.regardless of B.contrary to C.together with D.other than60.A.permanently B.delightfully C.temporarily D.instantly61.A.consume B.grow C.domesticate D.harvest62.A.distinguished B.disqualified C.discouraged D.dissatisfied63.A.intellectual B.physical C.psychological D.moral64.A.poems B.reports C.documents D.fictionsplex B.overrated C.readable D.appreciatedV.Reading Comprehension (32)Section A (24)Directions: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A,B,C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)Are you the peace-loving type or the ambitious go-getter? Your finger length might be a clue.Take a look at your fingers. Would you believe me if I told you that the length of your fingers in relation to one another can predict your personality? I know that it sounds like one of those silly tests you see on Wechat, but I have to admit that it is exactly correct for me. Here's how it works. Look at your three middle fingers of your hand.Is your index finger longer than your ringfinger?Is your ring finger longer than your index finger?Or are the two of the same length?Find the situation that best matches your hand and see if what follows is an accurate description of your personality.●Longer ring finger:The Sweet TalkerIf your ring finger is longer than your index finger,research shows that you're likely a charming type,and you can probably talk yourself out of any situation.Others often find this personality type irresistible and will go to great lengths to help you.You're more likely to take risks,and you're good at problem-solving.People in this category make great engineers,soldiers and crossword-puzzle solvers.●Longer index finger:The (Over)Confident OneIf your index finger is longer than your ring finger,chances are you are full of confidence-possibly even to the point of being over-confident and proud.You are not necessarily introverted(内向的),but you do enjoy Lime to yourself,especially when you're trying to complete a project.You are a dream pursuer who can make things happen,but you may be shy when it comes to taking the first step in a relationship.You are also probably happy with what you have,but you're always wishing for more.●Index and ring finger finger of the same length:The PeacemakerIf your index finger and ring finger are roughly of the same length,you are likely to avoid conflict at all costs and seek to keep the peace in your relationships.You are well-organized,faithful and sympathetic.But deep down under all of that caring and peace-loving,you also have a hot-tempered side that can appear unexpectedly when you're pushed too far.You will try your hardest to avoid a fight.How well do these descriptions match your personality?I was surprised at the accuracy in my case,but I'm not going to tell you which one it was!66.Which of the following hand could be an engineer's hand?67.We can learn from the passage that the author_______.A.is very charming and helpfulB.always tries hard to avoid a fightC.often takes silly tests on FacebookD.thinks the descriptions suit him well68.What does the passage intend to tell us?A.How accurate the description of the test is.B.The length of fingers can reveal personality.C.Why your fingers are not of the same length.D.Personality is determined by your finger length.69.Where can you probably find the text?A.A popular magazine.B.An official report.C.A public advertisement.D.A biology textbook.(B)In a time when too much emphasis has been attached to utilitarianism(功力主义), it doesn’t come as a surprise to me that we people no longer believe in poetry. Utility is now often the standard of what one should devote his or her time and energy to and what he or she shouldn’t. but don’t get me wrong. I am not saying or stating that one should not consider utility at all when making crucial decisions. On the contrary, everyone should take it into account. Medicine, law, business and engineering, these are noble pursuits and necessary to sustain life. But poetry, beauty, romance and love, these are what we stay alive for.My dad once told me that “life is suffering”, which now I think is probably a unanimous(无异议的) opinion among most people. With a tough life inevitably comes a tough language --- and that is what poetry is and why poetry is needed. It offers people a way to talk about the difficult-to-describe things in life, like death, suffering, profound joy and transformation. So whenever I hear someon e say that poetry is not a must, or it’s an option, or it’s only for the educated middle classed, I suspect that he or she must have had things pretty easy.But how can people live without poetry when poetry is nothing but a ceaseless flow of genuine human emotions?/ unlike what the academics may tell you, my suggestion about bring poetry into your lives is that don’t analyze it and don’t ask others to analyze it. Don’t deconstruct it or try to make meaning of it. Just find the poems that wake you up, tha t make you feel as if you’ve submerged(沉浸) yourself in a mineral hot spring or an ice bath. Find the poems that make you feel almost irrational joy or sadness. Find the poems that make you want to roll around in them or paint their colors all over your bedroom ceiling. Those are the poems you want to play with. Find the poems that communicate with the deepest parts your being and welcome them in.If finding others’ poems no longer satisfies you, compose your own! You don’t have yo be a poet to compose poetry. After all, poetry is not a luxury only for members desire or genuine willingness to vent feelings. It’s about searching for the real meaning of life and also about giving meaning to life. Just like Walt Whitman’s “O Me! O Life!”, a poem that ends by s peaking directly to its readers:” the powerful play goes on and you may contribute a verse,” poetry is about everyone. And now I want to ask you the mother of all inspirational questions:” What will your verse be?”Inspired by Jeanette Winterson’s book:Why Be Happy When Y ou Could Be Normal?70.The author thinks that poetry is very important for human beings because it _______.A.has nothing to do with the shared belief in utilityB.symbolizes a status as an educated middle classC.offers a medium for describing hard experiencesD.reveals a fundamental truth that life isn’t all roses71.To appreciate poetry by academic standards, one should first _______.A.deconstruct and making meaning of itB.enjoy a mineral hot spring or an ice bathC.paint it over his or her bedroom ceilingD.get consumed by irrational joy or sadness72.The author quoted Walt Whitman’s poem “O Me! O Life!” to emphasize that ______.A.only a true poet can write poemsB.no poem can appeal to everyoneC.life feels just like a powerful playD.every human is entitled to poetry73.The best title for the text may be ______.posing PoetryB.Poetry for LifeC.The Art of PoetryD.Analyzing Poetry(C)A scene from China’s biggest soap opera last year, Ode to Joy Season 2, has triggered much discussion on whether virginity is still a prized asset for women in “modern” China. In the scene, Qiu Yingying, a young but naive character, holds a dinner party to introduce her boyfriend to her friends, but the evening turns dramatic after her neighbour accidentally exposes her past relationship. Her boyfriend, Ying Qin, storms off, furious, expressing his thought that having sex before marriage is a moral spot for a woman.However controversial the show is, it does reveal a universal phenomenon that sex education in China is often relatively backward and of minor significance. Parents very rarely share information about sex with children. What almost all parents do, however, is to warn their children, especially daughters not to have any dates before they graduate from university, fearing that they get hurt or become pregnant. There are stories of biology teachers skipping details of the reproductive system in class and ask students to learn it by themselves. Chinese communities also traditionally expect virginity before marriage.Ren Yi, a student from East China Normal University, said it was perfectly reasonable for a boyfriend who was a virgin himself to require his girlfriend to be so. The only problem, she said, was if it was a double standard and the man expected virginity from a woman but not himself.The issue was also being discussed by her university classmates in WeChat. Someone questioned why ther e wasn’t gender equality when it came to virginity, but another said she couldn’t accept Ying Qin’s comment on women who had sex before marriage.While some still concentrate on virginity itself, others have shifted the focus of this debate.Zhu Pingping, a Shanghai-based English teacher, commented that even debating the topic was old-fashioned in the 21st Century. She thought it was “disgusting” for the show to make an issue of it. In modern cities, women are more independent and liberated, she said.Her husband, Shen Peng, also a teacher, added that not everyone will and can advance with thetimes as he or she ages, and some even want to restore the so-called traditional Chinese virtues of being a woman(女德), which from his perspective has long been used as a means to oppress women both physically and mentally throughout the history.“Teaching students knowledge about sex doesn’t mean we encourage them to eat the forbidden fruit. Actually, it is just the opposite, because only when sex --- like other aspects of lives ---is dealt with frankly and appropriately in the educational process can the healthy growth of children raised in it be truly encouraged,” he said.74.Which one of the following might be WRONG according to the first two paragraphs?A.The TV series Ode to Joy Season 2 has brought female virginity to attention.B.Qiu Yingying’s relationship has been buried with her previous romance revealed.C.Parents and teachers tend to avoid answering questions about sex from children.D.Sex education in China has weighed strongly in the current educational system.75.It can be learned from the three interviewees’ opinions that ______>A.a double standard encouraged in attitudes towards virginity is allowedB.the topic on virginity is still worth public discussion in the present dayC.keeping pace with the times is not necessarily a natural outcome of ageD.teaching students sex is likely to lead them to have sex at an early age.76.What is most likely to be discussed in the paragraph that follows?A.Measures to advocate sex educationB.Historical evidence of female oppressionC.Professional comments on the showD.Examples to illustrate the virginity issue77.The author’s purpose for writing this article is to ______.A.introduce a popular TV dramaB.explain a universal phenomenonC.criticize the current educationD.arouse people’s food for thoughtSection B (8)Directions: In the article, 4 sentences have been removed. Choose the most suitable ones from the list A-AC to fit into each of the numbered gaps. There are TWO which do not fit in any of the gaps.(D)The Museum of InnocenceLove and Madness in IstanbulIt’s all an elaborate(详尽的) and precisely constructed lie: a museum built to tell the story of a self-referential book created around the contents of the soon-to-finish museum. A tale of love and loss, but mainly madness, which is entirely fictional but in which the very real author plays a central narrative role. It is the Museum of Innocence.The story focuses on the character of Kemal(凯末尔) and his love for a beautiful shop girl and distant cousin Fusun(芙颂). Kemal’s story is one of lust, obsession, and eventual ruin. Were the character real, I expect that most of us would distance ourselves from him in the same way that Istanbul’s highh society do in the novel. (78)_________. And this is especially the case when one reads through Kemal’s story of madness and presents himself or herself in the collection of objects that repr esent Kemal’s love and loss.The museum follows the narrative of the book precisely presenting 83 displays that each reflects one of the book’s 83 chapters to accompany the storyline. (79)__________. In this case, as in many, the items that we accumulate in our lives are the reminders to help us recall those memories.(80)__________. Mounted on the ground floor is a display of 4,213 cigarette stubs, each of them smoked by Fusun and each noted with a time or date or comments speaking to the very moment it touched her hands, her lips, and so became a part of her story.Though it’s hard to match the visual appeal of that first display, representing chapter 68 of the book, the combination of sights and sounds throughout the three-story museum evokes(换起) the s tory in a way that reading alone couldn’t hope to do. (81)___________. You could surely visit the museum without having read the book or give it a read with no intention to visit the museum but, frankly, you shouldn’t. pick up a copy a month or two before a trip to Istanbul, and give yourself the pleasure of experiencing the story twice in different forms.If, like Kemal, you find yourself longing for some tangible physical reminder of the experience? Not to worry, there’s a well-stocked gift shop full of small souvenirs and Orhan Pamuk’s otherworks.VI. Translation(18分)Directions: Translate the following sentences into English, using the words given in the brackets.82.您什么时间方面来取一下您的体检报告?(convenient)83.一个合格的大学生应当有着良好的英语运用能力。
精品解析:【全国百强校】上海市七宝中学2018-2019学年高二上学期期末考试数学试题(原卷版)
2018-2019学年上海市闵行区七宝中学高二(上)期末数学试卷一、选择题(本大题共4小题)1.与圆相切,且横截距与纵截距相等的直线条数是( )A. 2B. 4C. 6D. 以上说法都不对2.直线:与直线:的夹角为( )A. B.C. D. 以上说法都不对3.下列说法正确的是()A. 平面中两个定点A,B,k为非零常数,若,则动点P的轨迹是双曲线B. 定圆C上有一定点A和一动点不与A重合,O为坐标原点,若,则动点P的轨迹是椭圆C. 斜率为定值的动直线与抛物线相交于A,B两点,O为坐标原点,,则动点P的轨迹是直线D. 以上说法都不对4.点A为椭圆C:的右顶点,P为椭圆C上一点(不与A重合),若(是坐标原点),则(c为半焦距)的取值范围是()A. B.C. D. 以上说法都不对二、填空题(本大题共12小题)5.抛物线的准线方程为______.6.直线的倾斜角范围是______.7.直线:,直线:,若,则______.8.直线被圆所截得的弦长等于______.9.是双曲线上的一点,,为焦点,若,则______.10.过点且与原点距离为2的直线方程是______.11.已知p:曲线C上的点的坐标都是方程的解,q:曲线C是方程的曲线,则p成立是q成立的______条件.12.已知是椭圆上的一点,,为焦点,若,,则椭圆的焦距与长轴的比值为______.13.直线与双曲线有且仅有一个公共点,则______.14.若x,y满足约束条件,则的取值范围是______.15.已知曲线的参数方程为(为参数),则以下曲线的说法中:①关于原点对称;②在直线下方;③关于轴对称;④是封闭图形,正确的有______.16.已知,分别为双曲线C:的左、右焦点,过的直线l与双曲线的右支分别交于A,B两点,的内切圆半径为,的内切圆半径为,若,则直线l的斜率为______.三、解答题(本大题共5小题)17.已知两点,.(1)求直线AB的方程;(2)若A,B到直线l的距离都是2,求直线l的方程.18.双曲线M:过点,且它的渐近线方程是.(1)求双曲线M的方程;(2)设椭圆N的中心在原点,它的短轴是双曲线M的实轴,且椭圆N中斜率为的弦的中点轨迹恰好是M的一条渐近线截在椭圆N内的部分,试求椭圆N的方程.19.已知椭圆的两个焦点为,,且椭圆过点(1)求椭圆的方程.(2)已知斜率为的直线过,与椭圆分别交于P,Q;直线过,与直线垂直,与椭圆分别交于M,N,求四边形PMQN面积的函数解析式.20.设直线l:与抛物线相交于不同的两点A,B,M为线段AB中点,(1)若,且,求线段AB的长;(2)若直线l与圆C:相切于点M,求直线l的方程;(3)若直线l与圆C:相切于点M,写出符合条件的直线l的条数直接写出结论即可21.在平面直坐标系xOy中有曲线:.(1)如图1,点B为曲线上的动点,点,求线段AB的中点的轨迹方程;(2)如图2,点B为曲线上的动点,点,将绕点A顺时针旋转得到,求线段OC 长度的最大值.(3)如图3,点C为曲线上的动点,点,,延长AC到P,使,求动点P的轨迹长度.。
上海市七宝中学2019届高三数学上学期期中试题(含解析)
如果您喜欢这份文档,欢迎下载!祝您成绩进步,学习愉快!上海市七宝中学2019届高三数学上学期期中试题(含解析)一. 填空题1.集合的真子集有________个【答案】【解析】【分析】直接写出集合A的真子集即得解.【详解】集合A的真子集有,{0},{1},{2018},{0,1},{0,2018},{1,2018},所以集合A 的真子集个数为7,故答案为:7【点睛】本题主要考查集合的真子集及其个数,意在考查学生对这些知识的掌握水平和分析推理能力.2.设全集,,,则图中阴影部分所表示的集合是________(用区间表示)【答案】【解析】【分析】先化简集合M和N,再求M∩N,再求即得阴影部分所表示的集合.【详解】由题得M={x|x>2或x<-2},N={x|x≥0},所以M∩N={x|x>2},所以.所以阴影部分所表示的集合为[0,2].故答案为:【点睛】本题主要考查韦恩图和集合的运算,意在考查学生对这些知识的掌握水平和分析推理能力.3.命题“若实数、满足,则或”是________命题(填“真”或“假”)【答案】真【解析】【分析】先考虑其逆否命题“a>2且b>3则a+b>5”的真假,即得原命题的真假.【详解】由题得原命题的逆否命题为“a>2且b>3则a+b>5”,由不等式同向可加的性质得其逆否命题为真命题,所以原命题是真命题.故答案为:真【点睛】(1)本题主要考查原命题及其逆否命题,考查命题真假性的判断,意在考查学生对这些知识的掌握水平和分析推理能力.(2) 互为逆否关系的命题同真同假,即原命题与逆否命题的真假性相同,原命题的逆命题和否命题的真假性相同.所以,如果某些命题(特别是含有否定概念的命题)的真假性难以判断,一般可以判断它的逆否命题的真假性.4.某个时钟时针长6,则在本场考试时间内,该时针扫过的面积是________【答案】【解析】【分析】直接利用扇形的面积公式求解.【详解】由题得该时针扫过的面积为故答案为:【点睛】本题主要考查扇形面积的计算,意在考查学生对这些知识的掌握水平和分析推理计算能力.5.函数是奇函数,则实数的值为________【答案】【解析】【分析】化简f(-x)+f(x)=0即得a=±1,再检验得a=-1.【详解】由题得,所以所以,经检验a=1不符合题意,所以舍去,故答案为:-1【点睛】本题主要考查奇函数的性质和对数的运算,意在考查学生对这些知识的掌握水平和分析推理计算能力.6.函数在上单调递增,则实数的取值范围为________【答案】【解析】【分析】先对函数求导得在(1,2)上恒成立,再分离参数求出a的范围.【详解】由题得在(1,2)上恒成立,所以.故答案为:【点睛】(1)本题主要考查利用导数研究不等式的单调性和恒成立问题,意在考查学生对这些知识的掌握水平和分析推理能力.(2) 一般地,函数在某个区间可导,在某个区间是增函数≥0 .7.在△中,角、、所对的边分别为、、,若,,,则△的面积为________【答案】【解析】【分析】利用余弦定理可得b,再利用三角形面积计算公式即可得出.【详解】∵a=,∴a2=b2+c2﹣2bccosA,∴3=4+b2﹣4b×,化为b2﹣2b+1=0,解得b=1.∴S△ABC===.故答案为:.【点睛】本题主要考查了余弦定理、三角形面积计算公式,意在考查学生对这些知识的掌握水平和分析推理能力与计算能力.8.已知函数,则的解集是________【答案】【解析】【分析】由于函数是定义域在上的增函数,所以,解不等式即得解.【详解】由于函数是定义域在上的增函数,所以故答案为:【点睛】(1)本题主要考查幂函数的图像和性质,意在考查学生对这些知识的掌握水平和分析推理能力.(2)处理函数的问题,一定要注意“定义域优先的原则”,本题不要漏了3x-1≥0.9.若关于的不等式在上恒成立,则正实数的取值范围为________【答案】【解析】【分析】由题得|2x-a|>-x+1,再分1<x≤2和0≤x≤1两种情况讨论恒成立问题,即得解.【详解】由题得|2x-a|>-x+1,当1<x≤2时,-x+1<0,所以不等式恒成立.当0≤x≤1时,-x+1≥0,所以2x-a>-x+1或2x-a<x-1,所以a<3x-1或a>x+1在[0,1]上恒成立,所以a<-1或a>2,因为a>0,综合得a>2.故答案为:a>2【点睛】本题主要考查绝对值不等式的恒成立问题,意在考查学生对这些知识的掌握水平和分析推理能力.10.已知常数,函数的图像经过点、,若,则________ 【答案】【解析】【分析】直接利用函数的关系式,利用恒等变换求出相应的a值.【详解】函数f(x)=的图象经过点P(p,),Q(q,).则:,整理得:=1,解得:2p+q=a2pq,由于:2p+q=16pq,所以:a2=16,由于a>0,故:a=4.故答案为:4【点睛】本题主要考查函数的性质和指数幂的运算,意在考查学生对这些知识的掌握水平和分析推理计算能力.11.已知函数,若,则的最大值是________ 【答案】【解析】【分析】设g(x)=f(x)-3,再判断函数g(x)的奇偶性和单调性,再由得,再利用三角换元求的最大值.【详解】设g(x)=f(x)-3,所以g(x)=,所以所以g(-x)=-g(x),所以函数g(x)是奇函数,由题得,所以函数g(x)是减函数,因为,所以,所以g=0,所以g=g(1-,所以不妨设,所以==,所以的最大值为.故答案为:【点睛】(1)本题主要考查函数的奇偶性和单调性,考查函数的图像和性质,考查三角函数的图像和性质,意在考查学生对这些知识的掌握水平和分析推理计算能力.(2)本题的解题关键有三点,其一是构造函数g(x)得到函数g(x)的奇偶性和单调性,其二是由得,其三是利用三角换元求的最大值.12.已知函数,如果函数恰有三个不同的零点,那么实数的取值范围是________【答案】【解析】【分析】先求出函数的解析式,作出函数的图像,由题得有三个不同的实根,数形结合分析得到实数k的取值范围.【详解】当1<x≤2时,f(x)=-x+2,当时,1<2x≤2,所以f(x)=,当时,<2x≤1,所以f(x)=,当时,<2x≤,所以f(x)=,当时,<2x≤,所以f(x)=,所以函数的图像为:其图像为线段PA,EB,GC,HD,,(不包括上端点A,B,C,D,)直线y=k(x-1)表示过定点P(1,0)的直线系,由题得C(),D(),当直线在PD(可以取到)和直线PC(不能取到)之间时,直线和函数f(x)的图像有三个不同的交点,由题得.所以k的取值范围为.故答案为:【点睛】(1)本题主要考查函数的图像和性质,考查求函数的解析式,考查函数的零点问题,意在考查学生读这些知识的掌握水平和数形结合分析推理能力.(2)解答本题的关键是求出函数f(x)的解析式作出函数的图像.(3)函数的零点问题常用的方法有:方程法、图像法、方程+图像法.二. 选择题13.“函数存在反函数”是“函数在上为增函数”的()A. 充分而不必要条件B. 必要而不充分条件C. 充分必要条件D. 既不充分也不必要条件【答案】B【解析】函数存在反函数,至少还有可能函数在上为减函数,充分条件不成立;而必有条件显然成立。
七宝中学高二期中(2019.04
七宝中学高二期中数学试卷2019.04一. 填空题1. 复数(12i)(3i)-+的虚部为2. 如图所示,在复平面内,网格中的每个小正方形的边长都为1, 点A 、B 对应的复数分别是1z 、2z ,则221||z z = 3. 复数352019i i i i +++⋅⋅⋅+=4. 一个圆柱侧面展开是正方形,它的高与底面直径的比值是5. 如果复数z 满足|i ||i |2z z ++-=,那么|i 1|z ++的最小值是6. 在复数范围内分解因式:42625x x -+=7. 设z 是复数,()a z 表示满足1n z =是最小正整数n ,则1i ()1ia +=- 8. 已知α是实系数一元二次方程22(21)10x m x m --++=的一个虚数根,且||2α≤,则 实数m 的取值范围是9. 圆锥底面半径为10,母线长为30,从底面圆周上一点,绕侧面一周再回到该点的最短路 线的长度是10. 已知直三棱柱111ABC A B C -中,120ABC ∠=︒,2AB =,11BC CC ==,则异面直线1AB 与1BC 所成角的余弦值为11. 设m 、n 是两条不同的直线,α、β是两个不同的平面,对于以下命题:(1)若m ∥n ,α∥β,那么m 与α所成的角和n 与β所成的角相等;(2)若m n ⊥,m α⊥,n β⊥,则αβ⊥;(3)若m β⊥,αβ⊥,则m ∥α;(4)若m ∥α,αβ⊥,则m β⊥.其中正确命题的序号是12. 在四面体ABCD 中,面ABC 与面BCD 成60°的二面角,顶点A 在面BCD 上的射影H 为△BCD 的垂心,G 为△ABC 的重心,若4AH =,AB AC =,则GH =二. 选择题13. 设1z 、2z 是复数,则下列命题中假命题是( )A. 若12||0z z -=,则12z z =B. 若12z z =,则12z z =C. 若12||||z z =,则1122z z z z ⋅=⋅D. 若12||||z z =,则2212z z =14. 设α、β是两个不同的两个平面,m 是直线且m α⊆,则“m ∥β”是“α∥β”的( )A. 充分不必要条件B. 必要不充分条件C. 充要条件D. 既不充分也不必要条件15. 过平面α外一点A 引斜线段AB 、AC 以及垂线段AO ,若AB 与α所成角是30°,6AO =,AC BC ⊥,则线段BC 长的取值范围是( )A. (0,6)B. (6,)+∞C. (0,63)D. (63,)+∞16. 如图,已知正四面体D ABC -中,P 、Q 、R 分别为AB 、BC 、CA 上的点,AP PB =,2BQ CR QC RA==,分别记二面角 D PR Q --、D PQ R --、D QR P --的平面角为α、β、γ,则α、β、γ的大小关系为( )A. αβγ<<B. αγβ<<C. βαγ<<D. γαβ<<三. 解答题17. 已知复数213(i)22z =-是一元二次方程210mx nx ++=(,)m n ∈R 的一个根. (1)求m 和n 的值;(2)若1(2i)z a z =-,a ∈R ,1z 为纯虚数,求|2i |a +的值.18. 如图,已知正方体1111ABCD A B C D -的棱长为1.(1)正方体1111ABCD A B C D -中哪些棱所在的直线与直线1A B 是异面直线;(2)若M 、N 分别是1A B 、1BC 的中点,求异面直线MN 与BC 所成角的大小.19. 《九章算术》中,将底面为长方形且有一条侧棱与底面垂直的四棱锥称之为阳马,将四个面都是直角三角形的四面体称之为鳖臑,首届中国国际进门博览会是某展馆棚顶一角的钢结构可以抽象为空间图形阳马,如图所示,在阳马P ABCD -中,PD ⊥底面ABCD .(1)若4AD CD m ==,斜梁PB 与底面ABCD 所成角为15°,求立柱PD 的长; (精确到0.01m )(2)请证明四面体PDBC 为鳖臑;若2PD =,2CD =,1BC =,点E 为线段PB 上一个动点,求△ECD 面积的最小值.20. 如图,四棱柱1111ABCD A B C D -中,侧棱1AA ⊥底面ABCD ,AB ∥CD ,AB AD ⊥,1AD DC ==,12AA AB ==,E 为棱1AA 的中点.(1)证明:11B C CE ⊥;(2)求二面角11B CE C --的正弦值;(3)设点M 为线段1C E 上,且直线AM 与平面 11ADD A 所成角正弦值为2,求线段AM 的长.21. 设z C ∈,且(Re 0)()(Re 0)z z f z z z ≥⎧=⎨-<⎩. (1)已知2()()429i f z f z z +-=-+(z C ∈),求z 的值;(2)若Re 0z ≥,设集合1{|()()2i ()2i ()120,}P z f z f z f z f z z C =⋅-⋅+⋅-=∈,21{|i ,}P z z P ωω==∈,求复平面内2P 对应的点集表示的曲线的对称轴;(3)若1()z u u C =∈,21(1)()n n n z f z z n *+=++∈N ,是否存在u ,使得数列12,,z z ⋅⋅⋅满足n m n z z +=(m 为常数,且m *∈N )对一切正整数n 均成立?若存在,试求出所有的u ,若不存在,请说明理由.参考答案一. 填空题1. 5-2. 53. 04. π5. 16. (2i)(2i)(2i)(2i)x x x x ++--+--+7. 48. 3(4-9. 10. 5 11. (1)(2)(4)12.9二. 选择题13. D 14. B 15. C 16. B三. 解答题17.(1)1m n ==;(2)4.18.(1)AD ,DC ,1CC ,1DD ,11C D ,11B C ;(2)4π.19.(1)1.52;(2.20.(1)略;(2)7;(321.(1)23i z =-;(2)2x =;(3)i u =±.。
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2018学年七宝中学高二上期中考试
一、填空题(本大题共有12题,满分54分,第1-6题每题4分,第7-12题每题5分)
1. 若线性方程组的增广矩阵是122301c c ⎛⎫ ⎪
⎝⎭,其解为1
1x y =⎧⎨=⎩
,则12c c +=____________.
2. 已知行列式451
3732
x
x 中元素4的代数余子式是1,则实数x 的值是
____________.
3. 求()2
*
6100lim 31100
45n
n n n n n n →∞⎧≤⎪⎪∈⎨+⎪
>⎪+⎩
N =____________.
4. 在ABC 中,()()()0,0,3,5,4,4A B C ,则ABC 的面积为____________.
5. 已知()()()()
1,2,2,3,2//a b a b a kb ==-++
,则实数k 的值是
____________.
6. 秦九韶是我国南宋时期的数学家,普州(现四川省安岳县)人,他在所著的《数书九章》中提出的多项式求值的秦九韶算法,至今仍是比较先进的算法,如图所示的程序框图给出了利用秦九韶算法求某多项式值的一个实例。
若输入,n x 的值分别为3,2,则输出v 的值为____________.
7. 设()*111,122f n n n n n
=
+++∈++N ,若*k ∈N ,则()()1_________f k f k +=+. 8. 已知1,1,,a b a b == 的夹角是60°,若向量c 满足()
1c a b -+= ,则c
的最小值为
____________.
9. 设函数2y nx n =-+和()*11
,222
y x n n n =-
+∈≥N 的图像与坐标轴围成的封闭图形的面积为n S ,则lim n n S →∞
=____________.
10. 已知圆O 中,弦3,5AB AC ==,则AO BC ⋅
的值是____________.
11. 定义平面向量之间的一种运算“*”如下:对任意的()(),,,a m n b p q == ,令*m n
a b p q
= .给
出以下四个命题:
(1)若a 与b 共线,则*0a b = ;(2)**a b b a = ;(3)对任意的λ∈R ,有()()
**a b a b λλ=
;
(4)()()
22
22
*a b a b
a b ++=⋅ .(注:这里a b ⋅ 指a 与b 的数量积)
则其中所有真命题的序号是____________.
12. 已知无穷等比数列{}n a 的公比为q ,前
n 项和为n S ,且lim n n S S →∞
=,若对于任意
*,2n n S S ∈<N 恒成立,则公比q 的取值范围是____________.
二、选择题(本大题共有4题,满分20分,每题5分) 13. 用数学归纳法证明()*1111,22321
n n n n ++++<∈≥-N ,第一步验证( )
A. 1
122
+
<
B. 111223++<
C. 111323++<
D. 11113234
+++<
14. 已知,,A B C 是平面不同三点,则“0AB BC CA ++=”是“,,A B C 三点能构成三角形”的( )条件
A. 充分非必要
B. 必要非充分
C. 充要
D. 既非充分也非必要
15. 若等比数列{}n a 的公比为q ,则关于x 的二元一次方程组1324
23a x a y a x a y +=⎧⎨+=⎩的解的情况,下列说
法正确的是( )
A. 对任意,0q q ∈≠R ,方程组都有唯一解
B. 对任意,0q q ∈≠R ,方程组都无解
C. 当且仅当3
2q =
,方程组无解 D. 当且仅当3
2
q =,方程组无穷多解
16. 在正六边形ABCDEF 中,令,AB a AF b ==
,P 是CDE 内含边界的动点(如图),
A P x a y b =+ ,则x y +的最大值是( )
A. 1
B. 4
C. 4
D. 5
三、解答题(本题共有5题,满分76分) 17. (14分)
求证:对任意的*22,389n n n +∈--N 能被64整除.
18. (本题满分14分,第1小题满分6分,第2小题满分8分) 上海市旅游节刚落下帷幕,在旅游节期间。
甲、乙、丙三位市民顾客分别获得一些景区门票的
2(单位:元)。
表1 表2
(1)按照上述表格的行列次序分别写出这三位市民获得的折扣消费券数量矩阵A 和三个景区的门票折扣后价格矩阵B ;
(2)利用你所学的矩阵知识,计算三位市民各获得多少元折扣?
19. (本题满分14分,第1小题满分6分,第2小题满分8分)
已知平面直角坐标系内三点A 、B 、C 在一条直线上,满足
()()()3,1,,3,7,4O A m O B n O C =-+==
,且OA
OB ⊥ ,其中O 为坐标原点. (1)求实数m 、n 的值;
(2)设A O C 的重心G ,且23
OG OB =
,且12,P P 为线段
AB 的三等分点,求12OA AB OP AB OP AB OB AB ⋅+⋅+⋅+⋅
的值.
20. (本题满分16分,第1小题满分4分,第2小题满分6分,第3小题满分6分)
已知一非零向量列n a 满足:)
()()111111
,,,2
n n n n n n n a a x y x y x y ----=
==-+
(1)写出数列{}
n a
的通项公式;
(2)求向量1n a - 与n a 的夹角θ,并将12,,,n a a a 中所有与1a
平行的向量取出来,按原来的顺序
排成一列,组成新的数列{}
12,n n n b OB b b b =+++
,O 为坐标原点,求点列{}n B 的坐标;
(3)令()*
8128n n S a a a n =+++∈N ,求{}
8n S 的极限点位置.
21. (本题满分18分,第1小题满分4份,第2小题满分6分,第3小题满分8分)
几位大学生响应国家的创业号召,开发,,A B C 三款应用软件.为激发大家学习数学的兴趣,他
们推出了“解数学题获取软件激活码”的活动。
这款软件的激活码分别为下面数学问题的答案:已知数列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,…,其中第一项是0
2,接下来的两项是012,2,再接下来的三项是0122,2,2,依此类推,试根据下列条件分别求三款软件的激活码。
(1)A 款应用软件的激活码是该数列中第四个三位数的项数的平方; (2)B 款应用软件的激活码是该数列中第一个四位数及其前所有项的和;
(3)C 款应用软件的激活码是满足如下条件的最小整数0N ;①0100N >,②该数列的前0N 项和为2的整数幂.。