高二月考1
辽宁省名校联盟2024-2025学年高二上学期第一次月考数学试卷

辽宁省名校联盟2024-2025学年高二上学期第一次月考数学试卷一、单选题1.已知直线(20a x y ++=的倾斜角为30o ,则a =( )A .BCD .02.若()1,2,1a =--r,()1,3,2b =-r ,则()()2a b a b +⋅-=r r r r ( )A .22B .22-C .29-D .293.如果0AB >且0BC <,那么直线0Ax By C ++=不经过( ) A .第一象限B .第二象限C .第三象限D .第四象限4.如图,在四棱锥P -ABCD 中,底面ABCD 是平行四边形,点E 在侧棱PC 上,且12PE EC =,若AB a u u u r r=,AD b =u u u r r ,AP c =u u u r r ,则AE =u u u r ( )A .112333a b c ---r r rB .112333a b c ++r r rC .221333a b c ++r r rD .221333a b c ---r r r5.已知m 为实数,直线()()12:220,:5210l m x y l x m y ++-=+-+=,则“12l l //”是“3m =-”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6.已知空间中三点()0,0,0A ,()1,1,2B -,()1,2,1C --,则以AB ,AC 为邻边的平行四边形的面积为( )A .32B C .3 D .7.点()2,4A -到直线()():131440l m x m y m -+-++=(m 为任意实数)的距离的取值范围是( )A .[]0,5B .⎡⎣C .[]0,4D .⎡⎣8.在正三棱锥P ABC -中,4PA AB ==,点,D E 分别是棱,PC AB 的中点,则AD PE ⋅=u u u r u u u r( ) A .2-B .4-C .6-D .8-二、多选题9.下列说法正确的是( )A .直线10x y -+=与直线10x y --=B .直线240x y --=在两坐标轴上的截距之和为6C .将直线y x =绕原点逆时针旋转75o ,所得到的直线为y =D .若直线l 向左平移3个单位长度,再向上平移2个单位长度后,回到原来的位置,则直线l 的斜率为23-10.在正方体1111ABCD A B C D -中,能作为空间的一个基底的一组向量有( )A .1AA u u u r ,AB u u u r,AC u u u r B .BA u u u r ,BC u u ur ,BD u u u rC .1AC uuu r ,1BD u u u u r,1CB u u u rD .1AD uuu r ,1BA u u u r ,AC u u u r11.如图,在棱长均为1的平行六面体1111ABCD A B C D -中,1BB ⊥平面,60ABCD ABC ∠=o ,,P Q 分别是线段AC 和线段1A B 上的动点,且满足()1,1BQ BA CP CA λλ==-u u u r u u u r u u u r u u u r,则下列说法正确的是( )A .当12λ=时,PQ //1A D B .当12λ=时,若()1,,PQ xAB yAD z AA x y z =++∈R u u u r u u u r u u u r u u u r ,则0x y z ++=C .当13λ=时,直线PQ 与直线1CC 所成角的大小为π6D .当()0,1λ∈时,三棱锥Q BCP -三、填空题12.已知直线l 过点()1,2,且在y 轴上的截距为在x 轴上的截距的两倍,则直线l 的方程是. 13.在空间直角坐标系O xyz -中,已知()()()2,2,0,2,1,3,0,2,0A B C -,则三棱锥O ABC -的体积为.14.在棱长为4的正方体1111ABCD A B C D -中,点E ,F 分别为棱DA ,1BB 的中点,M ,N 分别为线段11D A ,11A B 上的动点(不包括端点),且EN FM ⊥,则线段MN 的长度的最小值为.四、解答题15.如图,正方体1111ABCD A B C D -的棱长为2.(1)用空间向量方法证明:11//AC 平面1ACD ;(2)求直线BD 与平面1ACD 所成角的正弦值.16.已知点()1,3P ,点()3,1N --,直线1l 过点()2,4-且与直线PN 垂直. (1)求直线1l 的方程;(2)求直线2:250+-=l x y 关于直线1l 的对称直线的方程. 17.平行六面体ABCD A B C D -'''',(1)若4AB =,3AD =,3AA '=,90BAD ∠=︒,60BAA '∠=︒,60DAA '∠=︒,求AC '长; (2)若以顶点A 为端点的三条棱长均为2,且它们彼此的夹角都是60°,则AC 与BD '所成角的余弦值.18.如图,四边形ABCD 是直角梯形,//,,22,AB CD AB BC AB BC CD E ⊥===为BC 的中点,P 是平面ABCD 外一点,1,,PA PB PE BD M ==⊥是线段PB 上一点,三棱锥M BDE -的体积是19.(1)求证:PA ⊥平面ABCD ; (2)求二面角M DE A --的余弦值.19.图,在三棱台111ABC A B C -中,ABC V 是等边三角形,11124,2AB A B CC ===,侧棱1CC ⊥平面ABC ,点D 是棱AB 的中点,点E 是棱1BB 上的动点(不含端点B ).(1)证明:平面AA B B 平面11DCC;1(2)求平面ABE与平面ACE的夹角的余弦值的最小值.。
高二数学月考卷1

高二数学月考卷1一、选择题(每题1分,共5分)1. 函数f(x) = (x² 1)/(x 1)的定义域是()A. RB. {x | x ≠ 1}C. {x | x ≠ 0}D. {x | x ≠ 1}2. 若向量a = (2, 3),向量b = (1, 2),则2a 3b = ()A. (8, 1)B. (8, 1)C. (8, 1)D. (8, 1)3. 二项式展开式(x + y)⁵中x²y³的系数是()A. 5B. 10C. 20D. 304. 已知等差数列{an}中,a1 = 3,a3 = 9,则公差d为()A. 2B. 3C. 4D. 65. 若复数z满足|z 1| = |z + 1|,则z在复平面上的对应点位于()A. 实轴上B. 虚轴上C. y = x上D. y = x上二、判断题(每题1分,共5分)1. 任何两个实数的和都是实数。
()2. 若矩阵A的行列式为0,则A不可逆。
()3. 两条平行线上的任意一对对应线段比例相等。
()4. 双曲线的渐近线一定经过原点。
()5. 若函数f(x)在区间[a, b]上单调递增,则f'(x) > 0。
()三、填空题(每题1分,共5分)1. 若log₂x = 3,则x = ______。
2. 若等差数列{an}中,a4 = 8,a7 = 19,则a10 = ______。
3. 圆的标准方程(x h)² + (y k)² = r²中,(h, k)表示圆的______。
4. 若sinθ = 1/2,且θ是第二象限的角,则cosθ = ______。
5. 矩阵A = [[1, 2], [3, 4]]的行列式|A| = ______。
四、简答题(每题2分,共10分)1. 简述矩阵乘法的定义。
2. 请解释什么是反函数。
3. 简述等差数列的通项公式。
4. 请说明直线的斜率的意义。
5. 简述三角函数的周期性。
四川省成都市2024-2025学年高二上学期月考(一)数学试题含答案

高二上数学月考(一)(答案在最后)一、单项选择题:本题共8个小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.某高校对中文系新生进行体测,利用随机数表对650名学生进行抽样,先将650名学生进行编号,001,002,…,649,650.从中抽取50个样本,下图提供随机数表的第4行到第6行,若从表中第5行第6列开始向右读取数据,则得到的第6个样本编号是()32211834297864540732524206443812234356773578905642 84421253313457860736253007328623457889072368960804 32567808436789535577348994837522535578324577892345A.623B.328C.072D.457【答案】A【解析】【分析】按照随机数表提供的数据,三位一组的读数,并取001到650内的数,重复的只取一次即可【详解】从第5行第6列开始向右读取数据,第一个数为253,第二个数是313,第三个数是457,下一个数是860,不符合要求,下一个数是736,不符合要求,下一个是253,重复,第四个是007,第五个是328,第六个数是623,,故A正确.故选:A.2.某校高一共有10个班,编号1至10,某项调查要从中抽取三个班作为样本,现用抽签法抽取样本,每次抽取一个号码,共抽3次,设五班第二次被抽到的可能性为b,则()A.19b= B.29b= C.310b= D.110b=【答案】D【解析】【分析】根据题意,在抽样过程中每个个体被抽到的概率相等即可求解.【详解】因为总体中共有10个个体,所以五班第一次没被抽到,第二次被抽到的可能性为91110910b=⨯=.故选:D.3.已知向量1,22AB ⎛⎫=- ⎪ ⎪⎝⎭,122BC ⎛⎫=- ⎪ ⎪⎝⎭,则ABC ∠=()A.30°B.150°C.60°D.120°【答案】B 【解析】【分析】根据向量夹角的坐标表示求出向量夹角,进而求解几何角.【详解】因为向量13,22AB ⎛⎫=- ⎪ ⎪⎝⎭ ,31,22BC ⎛⎫=- ⎪ ⎪⎝⎭,所以13312222cos ,2AB BC AB BC AB BC⎛⎫⎛⎫⨯+-⨯- ⎪ ⎪⋅==⋅,又0,180AB BC ≤≤,所以,30AB BC =,所以,18030150BA BC =-= ,所以150ABC ∠=o .故选:B.4.已知,a b 为两条不同的直线,,αβ为两个不同的平面,则下列说法错误的是()A.若//a b ,,b a αα⊂⊄,则//a αB.若,a b αα⊥⊥,则//a bC.若,,b a b αβαβ⊥⋂=⊥,则a β⊥D.若,a b 为异面直线,,a b αβ⊂⊂,//a β,//b α,则//αβ【答案】C 【解析】【分析】根据线面平行的判定定理判断A ,根据线面垂直的性质判断B ,当a α⊄时即可判断C ,根据异面直线的定义及线面平行的性质定理判断D.【详解】对于A :若//a b ,,b a αα⊂⊄,根据线面平行的判定定理可知//a α,故A 正确;对于B :若,a b αα⊥⊥,则//a b ,故B 正确;对于C :当a α⊂时,,,b a b αβαβ⊥⋂=⊥,由面面垂直的性质定理可得a β⊥,当a α⊄时,,,b a b αβαβ⊥⋂=⊥,则//a β或a β⊂或a 与β相交,故C 错误;对于D :因为a α⊂,//b α,所以存在b α'⊂使得//b b ',又b β⊂,b β'⊄,所以//b β',又//a β且,a b 为异面直线,所以平面α内的两直线b '、a 必相交,所以//αβ,故D 正确.故选:C5.下列说法正确的是()A.互斥的事件一定是对立事件,对立事件不一定是互斥事件B.若()()1P A P B +=,则事件A 与事件B 是对立事件C.从长度为1,3,5,7,9的5条线段中任取3条,则这三条线段能构成一个三角形的概率为25D.事件A 与事件B 中至少有一个发生的概率不一定比A 与B 中恰有一个发生的概率大【答案】D 【解析】【分析】根据互斥事件、对立事件和古典概型及其计算逐一判定即可.【详解】对于A ,由互斥事件和对立事件的关系可判断,对立事件一定是互斥事件,互斥事件不一定是对立事件,故A 错误;对于B ,由()()1P A P B +=,并不能得出A 与B 是对立事件,举例说明:现从a ,b ,c ,d 四个小球中选取一个小球,已知选中每个小球的概率是相同的,设事件A 表示选中a 球或b 球,则1()2P A =,事件B 表示选中b 球或c 球,则1()2P B =,所以()()1P A P B +=,但A ,B 不是对立事件,故B 错误;对于C ,该试验的样本空间可表示为:{(1,3,5),(1,3,7),(1,3,9),(1,5,7),(1,5,9),(1,7,9),(3,5,7),(3,5,9),(3,7,9)(5,7,9)}Ω=,共有10个样本点,其中能构成三角形的样本点有(3,5,7),(3,7,9),(5,7,9),共3个,故所求概率310P =,故C 错误;对于D ,若A ,B 是互斥事件,事件A ,B 中至少有一个发生的概率等于A ,B 中恰有一个发生的概率,故D 正确.故选:D.6.一组数据:53,57,45,61,79,49,x ,若这组数据的第80百分位数与第60百分位数的差为3,则x =().A.58或64B.58C.59或64D.59【答案】A 【解析】【分析】先对数据从小到大排序,分57x ≤,79x ≥,5779x <<三种情况,舍去不合要求的情况,列出方程,求出答案,【详解】将已知的6个数从小到大排序为45,49,53,57,61,79.若57x ≤,则这组数据的第80百分位数与第60百分位数分别为61和57,他们的差为4,不符合条件;若79x ≥,则这组数据的第80百分位数与第60百分位数分别为79和61,它们的差为18,不符合条件;若5779x <<,则这组数据的第80百分位数与第60百分位数分别为x 和61(或61和x ),则613x -=,解得58x =或64x =故选:A7.如图,四边形ABCD 为正方形,ED ⊥平面,,2ABCD FB ED AB ED FB ==∥,记三棱锥,,E ACD F ABC F ACE ---的体积分别为123,,V V V ,则()A.322V V =B.31V V =C.3123V V V =-D.3123V V =【答案】D 【解析】【分析】结合线面垂直的性质,确定相应三棱锥的高,求出123,,V V V 的值,结合选项,即可判断出答案.【详解】连接BD 交AC 于O ,连接,OE OF ,设22AB ED FB ===,由于ED ⊥平面,ABCD FB ED ∥,则FB ⊥平面ABCD ,则1211141112222,22133233323ACD ABC V S ED V S FB =⨯⨯=⨯⨯⨯⨯==⨯⨯=⨯⨯⨯⨯= ;ED ⊥平面,ABCD AC Ì平面ABCD ,故ED AC ⊥,又四边形ABCD 为正方形,则AC BD ⊥,而,,ED BD D ED BD =⊂ 平面BDEF ,故AC ⊥平面BDEF ,OF ⊂平面BDEF ,故AC OF ⊥,又ED ⊥平面ABCD ,FB ⊥平面ABCD ,BD ⊂平面ABCD ,故,ED BD FB BD ⊥⊥,222222,26,3,BD OD OB OE OD ED OF OB BF =∴===+==+=而()223EF BD ED FB =+-=,所以222EF OF OE +=,即得OE OF ⊥,而,,OE AC O OE AC =⊂ 平面ACE ,故OF ⊥平面ACE ,又22222AC AE CE ===+=,故(2231131323233434F ACE V V ACE S OF AC OF =-=⋅=⨯⋅=⨯= ,故323131231,2,,233V V V V V V V V V ≠≠≠-=,故ABC 错误,D 正确,故选:D8.已知平面向量a ,b ,e ,且1e = ,2a = .已知向量b 与e所成的角为60°,且b te b e -≥- 对任意实数t 恒成立,则12a e ab ++-的最小值为()A.31+ B.23C.35 D.25【答案】B【解析】【分析】b te b e -≥-对任意实数t 恒成立,两边平方,转化为二次函数的恒成立问题,用判别式来解,算出||2b =r ,借助2a =,得到122a e a e +=+ ,12a e a b ++- 的最小值转化为11222a e a b++- 的最小值,最后用绝对值的三角不等式来解即可【详解】根据题意,1cos 602b e b e b ⋅=⋅︒=,b te b e -≥- ,两边平方22222||2||2b t e tb e b e b e +-⋅≥+-⋅ ,整理得到210t b t b --+≥ ,对任意实数t 恒成立,则()2Δ||410b b =--+≤ ,解得2(2)0b -≤ ,则||2b =r .由于2a =,如上图,122a e a e +=+ ,则111112(2)()22222a e a b a e a b a e a b ++-=++-≥+--222843e b e b b e =+=++⋅12a e ab ++- 的最小值为23当且仅当12,,2e b a -终点在同一直线上时取等号.故选:B .二、多项选择题.本题共3个小题,每小题6分,共18分.在每个小题给出的选项中,有多项符合题目要求,部分选对的得部分,有选错的得0分.9.某保险公司为客户定制了5个险种:甲,一年期短期;乙,两全保险;丙,理财类保险;丁,定期寿险;戊,重大疾病保险.各种保险按相关约定进行参保与理赔.该保险公司对5个险种参保客户进行抽样调查,得到如图所示的统计图表.则()A.丁险种参保人数超过五成B.41岁以上参保人数超过总参保人数的五成C.18-29周岁人群参保的总费用最少D.人均参保费用不超过5000元【答案】ACD 【解析】【分析】根据统计图表逐个选项进行验证即可.【详解】由参保险种比例图可知,丁险种参保人数比例10.020.040.10.30.54----=,故A 正确;由参保人数比例图可知,41岁以上参保人数超过总参保人数的45%不到五成,B 错误;由不同年龄段人均参保费用图可知,1829~周岁人群人均参保费用最少()3000,4000,但是这类人所占比例为15%,54周岁以上参保人数最少比例为10%,54周岁以上人群人均参保费用6000,所以18-29周岁人群参保的总费用最少,故C 正确.由不同年龄段人均参保费用图可知,人均参保费用不超过5000元,故D 正确;故选:ACD .10.在发生公共卫生事件期间,有专业机构认为该事件在一段时间内没有发生大规模群体感染的标志为“连续10天,每天新增疑似病例不超过7人”过去10日,甲、乙、丙、丁四地新增疑似病例数据信息如下:甲地:中位数为2,极差为5;乙地:总体平均数为2,众数为2;丙地:总体平均数为1,总体方差大于0;丁地:总体平均数为2,总体方差为3.则甲、乙、丙、丁四地中,一定没有发生大规模群体感染的有()A.甲地B.乙地C.丙地D.丁地【答案】AD 【解析】【分析】假设最多一天疑似病例超过7人,根据极差可判断AD ;根据平均数可算出10天疑似病例总人数,可判断BC .【详解】解:假设甲地最多一天疑似病例超过7人,甲地中位数为2,说明有一天疑似病例小于2,极差会超过5,∴甲地每天疑似病例不会超过7,∴选A .根据乙、丙两地疑似病例平均数可算出10天疑似病例总人数,可推断最多一天疑似病例可能超过7人,由此不能断定一定没有发生大规模群体感染,∴不选BC ;假设丁地最多一天疑似病例超过7人,丁地总体平均数为2,说明极差会超过3,∴丁地每天疑似病例不会超过7,∴选D .故选:AD .11.勒洛四面体是一个非常神奇的“四面体”,它能像球一样来回滚动.勒洛四面体是以正四面体的四个顶点为球心,以正四面体的棱长为半径的四个球的相交部分围成的几何体.如图所示,设正四面体ABCD 的棱长为2,则下列说法正确的是()A.勒洛四面体能够容纳的最大球的半径为22-B.勒洛四面体被平面ABC 截得的截面面积是(2π-C.勒洛四面体表面上交线AC 的长度为2π3D.勒洛四面体表面上任意两点间的距离可能大于2【答案】ABD 【解析】【分析】A 选项:求出正四面体ABCD 的外接球半径,进而得到勒洛四面体的内切球半径,得到答案;B 选项,作出截面图形,求出截面面积;C 选项,根据对称性得到交线AC 所在圆的圆心和半径,求出长度;D 选项,作出正四面体对棱中点连线,在C 选项的基础上求出长度.【详解】A 选项,先求解出正四面体ABCD 的外接球,如图所示:取CD 的中点G ,连接,BG AG ,过点A 作AF BG ⊥于点F ,则F 为等边ABC V 的中心,外接球球心为O ,连接OB ,则,OA OB 为外接球半径,设OA OB R ==,由正四面体的棱长为2,则1CG DG ==,BG AG ==133FG BG ==,233BF BG ==3AF ===,3OF AF R R =-=-,由勾股定理得:222OF BF OB +=,即22233R R ⎛⎫⎛-+= ⎪ ⎪ ⎪⎝⎭⎝⎭,解得:2R =,此时我们再次完整的抽取部分勒洛四面体,如图所示:图中取正四面体ABCD 中心为O ,连接BO 交平面ACD 于点E ,交 AD 于点F ,其中 AD 与ABD △共面,其中BO 即为正四面体外接球半径2R =,设勒洛四面体内切球半径为r ,则22r OF BF BO ==-=-,故A 正确;B 选项,勒洛四面体截面面积的最大值为经过正四面体某三个顶点的截面,如图所示:面积为(2221π333322222344⎛⎫⨯⨯⨯-⨯+⨯= ⎪ ⎪⎭⎝,B 正确;C 选项,由对称性可知:勒洛四面体表面上交线AC 所在圆的圆心为BD 的中点M ,故3MA MC ==2AC =,由余弦定理得:2221cos 23233AM MC AC AMC AM MC +-∠===⋅⨯⨯,故1arccos3AMC ∠=3AC 133,C 错误;D 选项,将正四面体对棱所在的弧中点连接,此时连线长度最大,如图所示:连接GH ,交AB 于中点S ,交CD 于中点T ,连接AT ,则22312ST AT AS =-=-=则由C 选项的分析知:3TG SH ==,所以323322GH =+=,故勒洛四面体表面上两点间的距离可能大于2,D 正确.故选:ABD.【点睛】结论点睛:勒洛四面体考试中经常考查,下面是一些它的性质:①勒洛四面体上两点间的最大距离比四面体的棱长大,是对棱弧中点连线,最大长度为232a a ⎫->⎪⎪⎭,②表面6个弧长之和不是6个圆心角为60︒的扇形弧长之和,其圆心角为1arccos 3,半径为32a .三、填空题:本题共3个小题,每小题5分,共15分.12.某工厂生产A 、B 、C 三种不同型号的产品,产品数量之比依次为3:4:7,现在用分层抽样的方法抽出容量为n 的样本,样本中的A 型号产品有15件,那么样本容量n 为________.【答案】70【解析】【分析】利用分层抽样的定义得到方程,求出70n =.【详解】由题意得315347n=++,解得70n =.故答案为:7013.平面四边形ABCD 中,AB =AD =CD =1,BD =BD ⊥CD ,将其沿对角线BD 折成四面体A ′﹣BCD ,使平面A ′BD ⊥平面BCD ,若四面体A ′﹣BCD 顶点在同一个球面上,则该球的表面积_____.【答案】3π【解析】【分析】根据BD ⊥CD ,BA ⊥AC ,BC 的中点就是球心,求出球的半径,即可得到球的表面积.【详解】因为平面A′BD ⊥平面BCD ,BD ⊥CD ,所以CD ⊥平面ABD ,∴CD ⊥BA ,又BA ⊥AD ,∴BA ⊥面ADC ,所以BA ⊥AC ,所以△BCD 和△ABC 都是直角三角形,由题意,四面体A ﹣BCD 顶点在同一个球面上,所以BC 的中点就是球心,所以BC =2所以球的表面积为:242π⋅=3π.故答案为:3π.【点睛】本题主要考查面面垂直的性质定理和球的外接问题,还考查空间想象和运算求解的能力,属于中档题.14.若一组样本数据12,,n x x x 的平均数为10,另一组样本数据1224,24,,24n x x x +++ 的方差为8,则两组样本数据合并为一组样本数据后的方差是__________.【答案】54【解析】【分析】计算出1n ii x =∑、21nii x=∑的值,再利用平均数和方差公式可求得合并后的新数据的方差.【详解】由题意可知,数据12,n x x x 的平均数为10,所以12)101(n x x x x n =+++= ,则110ni i x n ==∑,所以数据1224,24,,24n x x x +++ 的平均数为121(242424)210424n x x x x n'=++++++=⨯+= ,方差为()(()222221111444[24241010n n n i i i i i i s x x x x n n n n n ===⎤⎡⎤=+-+=-=-⨯⨯⎦⎣⎦∑∑∑2144008n i i x n ==-=∑,所以21102nii xn ==∑,将两组数据合并后,得到新数据1212,24,24,,24,n n x x x x x x +++ ,,则其平均数为11114)4)11113]4)[(2(3(222n i nn n i i i i i i i x x x x x n n n ====''=+=⨯+=⨯++∑∑∑∑()13104172=⨯⨯+=,方差为()()2222111111172417(586458)22n n n ni i i i i i i i s x x x x n n n ====⎡⎤=-++-=-+⎢⎥⎣⎦'∑∑∑∑1(51028610458)542n n n n=⨯-⨯+=.故答案为:54.四、解答题:本题共5个小题,共77分.解答应写出文字说明、证明过程或演算步骤.15.袋中有形状、大小都相同的4个小球,标号分别为1,2,3,4.(1)从袋中一次随机摸出2个球,求标号和为奇数的概率;(2)从袋中每次摸出一球,有放回地摸两次.甲、乙约定:若摸出的两个球标号和为奇数,则甲胜,反之,则乙胜.你认为此游戏是否公平?说明你的理由.【答案】(1)23(2)是公平的,理由见解析【解析】【分析】(1)利用列举法写出样本空间及事件的样本点,结合古典概型的计算公式即可求解;(2)利用列举法写出样本空间及事件的样本点,结合古典概型的计算公式及概率进行比较即可求解.【小问1详解】试验的样本空间{(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)}Ω=,共6个样本点,设标号和为奇数为事件B ,则B 包含的样本点为(1,2),(1,4),(2,3),(3,4),共4个,所以42().63P B ==【小问2详解】试验的样本空间Ω{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4),(4,1),(4,2),(4,3),(4,4)}=,共有16个,设标号和为奇数为事件C ,事件C 包含的样本点为(1,2),(1,4),(2,1),(2,3),(3,2),(3,4),(4,1),(4,3),共8个,故所求概率为81()162P C ==,即甲胜的概率为12,则乙胜的概率为12,所以甲、乙获胜的概率是公平的.16.(1)请利用已经学过的方差公式:()2211ni i s x xn ==-∑来证明方差第二公式22211n i i s x x n ==-∑;(2)如果事件A 与B 相互独立,那么A 与B 相互独立吗?请给予证明.【答案】(1)证明见解析;(2)独立,证明见解析【解析】【分析】(1)根据题意,对方差公式恒等变形,分析可得结论;(2)根据相互独立事件的定义,只需证明()()()P AB P A P B =即可.【详解】(1)()()()()2222212111n i n i s x xx x x x x x n n =⎡⎤=-=-+-++-⎢⎥⎣⎦∑ ()()2222121212n n x x x x x x x nx n ⎡⎤=+++-+++⎢⎥⎣⎦ ()22221212n x x x x nx nx n ⎡⎤=+++-⨯+⎢⎥⎣⎦ ()222121n x x x nx n ⎡⎤=+++-⎢⎥⎣⎦ 2211n i i x x n ==-∑;(2)因为事件A 与B 相互独立,所以()()()P AB P A P B =,因为()()()P AB P AB P A +=,所以()()()()()()P AB P A P AB P A P A P B =-=-()()()()()1P A P B P A P B =-=,所以事件A 与B 相互独立.17.如图,四棱锥P ABCD -的侧面PAD 是边长为2的正三角形,底面ABCD 为矩形,且平面PAD ⊥平面ABCD ,M ,N 分别为AB ,AD 的中点,二面角D PN C --的正切值为2.(1)求四棱锥P ABCD -的体积;(2)证明:DM PC⊥(3)求直线PM 与平面PNC 所成角的正弦值.【答案】(1)3(2)证明见解析(3)35【解析】【分析】(1)先证明DNC ∠为二面角D PN C --的平面角,可得底面ABCD 为正方形,利用锥体的体积公式计算即可;(2)利用线面垂直的判定定理证明DM ⊥平面PNC ,即可证明DM PC ⊥;(3)由DM⊥平面PNC 可得MPO ∠为直线PM 与平面PNC 所成的角,计算其正弦值即可.【小问1详解】解:∵PAD △是边长为2的正三角形,N 为AD 中点,∴PN AD ^,PN =又∵平面PAD ⊥平面ABCD ,平面PAD ⋂平面ABCD AD =∴PN ^平面ABCD又NC ⊂平面ABCD ,∴PN NC ⊥∴DNC ∠为二面角D PN C --的平面角,∴tan 2DC DNC DN∠==又1DN =,∴2DC =∴底面ABCD 为正方形.∴四棱P ABCD -的体积12233V =⨯⨯=.【小问2详解】证明:由(1)知,PN ^平面ABCD ,DM ⊂平面ABCD ,∴PN DM⊥在正方形ABCD 中,易知DAM CDN ≌△△∴ADM DCN ∠=∠而90ADM MDC ∠+∠=︒,∴90DCN MDC ∠+∠=︒∴DM CN ⊥∵PN CN N = ,∴DM ⊥平面PNC∵PC ⊂平面PNC ,∴DM PC ⊥.【小问3详解】设DM CN O ⋂=,连接PO ,MN .∵DM⊥平面PNC .∴MPO ∠为直线PM 与平面PNC 所成的角∵2,1AD AM ==,∴DM =5DO ==∴55MO ==又MN =PM ==∴35sin 5MO MPO PM ∠===∴直线PM 与平面PNC 所成角的正弦值为35.18.某市根据居民的月用电量实行三档阶梯电价,为了深入了解该市第二档居民用户的用电情况,该市统计局用比例分配的分层随机抽样方法,从该市所辖A ,B ,C 三个区域的第二档居民用户中按2:2:1的比例分配抽取了100户后,统计其去年一年的月均用电量(单位:kW h ⋅),进行适当分组后(每组为左闭右开的区间),频率分布直方图如下图所示.(1)求m 的值;(2)若去年小明家的月均用电量为234kW h ⋅,小明估计自己家的月均用电量超出了该市第二档用户中85%的用户,请判断小明的估计是否正确?(3)通过进一步计算抽样的样本数据,得到A 区样本数据的均值为213,方差为24.2;B 区样本数据的均值为223,方差为12.3;C 区样本数据的均值为233,方差为38.5,试估计该市去年第二档居民用户月均用电量的方差.(需先推导总样本方差计算公式,再利用数据计算)【答案】(1)0.016m =(2)不正确(3)78.26【解析】【分析】(1)利用频率和为1列式即可得解;(2)求出85%分位数后判断即可;(3)利用方差公式推导总样本方差计算公式,从而得解.【小问1详解】根据频率和为1,可知()0.0090.0220.0250.028101m ++++⨯=,可得0.016m =.【小问2详解】由题意,需要确定月均用电量的85%分位数,因为()0.0280.0220.025100.75++⨯=,()0.0280.0220.0250.016100.91+++⨯=,所以85%分位数位于[)230,240内,从而85%分位数为0.850.7523010236.252340.910.75-+⨯=>-.所以小明的估计不正确.【小问3详解】由题意,A 区的样本数为1000.440⨯=,样本记为1x ,2x ,L ,40x ,平均数记为x ;B 区的样本数1000.440⨯=,样本记为1y ,2y ,L ,40y ,平均数记为y ;C 区样本数为1000.220⨯=,样本记为1z ,2z ,L ,20z ,平均数记为z .记抽取的样本均值为ω,0.42130.42230.2233221ω=⨯+⨯+⨯=.设该市第二档用户的月均用电量方差为2s ,则根据方差定义,总体样本方差为()()()40402022221111100i j k i i i s x y z ωωω===⎡⎤=-+-+-⎢⎥⎣⎦∑∑∑()()()4040202221111100i j k i i i x x x y y y z z z ωωω===⎡⎤=-+-+-+-+-+-⎢⎥⎣⎦∑∑∑因为()4010ii x x =-=∑,所以()()()()404011220iii i x x x x x x ωω==--=--=∑∑,同理()()()()404011220jji i yyy y yy ωω==--=--=∑∑,()()()()202011220kki i zz z z zz ωω==--=--=∑∑,因此()()()()4040404022222111111100100i j i i i i s x x x y y y ωω====⎡⎤⎡⎤=-+-+-+-⎢⎥⎢⎥⎣⎦⎣⎦∑∑∑∑()()202022111100k i i z z z ω==⎡⎤+-+-⎢⎥⎣⎦∑∑,代入数据得()()222114024.2402132214012.340223221100100s ⎡⎤⎡⎤⎣⎦⎦=⨯+⨯-+⨯-⎣+⨯()212038.32023322178.26100⎡⎤+⨯+⨯-=⎣⎦.19.在世界杯小组赛阶段,每个小组内的四支球队进行循环比赛,共打6场,每场比赛中,胜、平、负分别积3,1,0分.每个小组积分的前两名球队出线,进入淘汰赛.若出现积分相同的情况,则需要通过净胜球数等规则决出前两名,每个小组前两名球队出线,进入淘汰赛.假定积分相同的球队,通过净胜球数等规则出线的概率相同(例如:若B ,C ,D 三支积分相同的球队同时争夺第二名,则每个球队夺得第二名的概率相同).已知某小组内的A ,B ,C ,D 四支球队实力相当,且每支球队在每场比赛中胜、平、负的概率都是13,每场比赛的结果相互独立.(1)求A 球队在小组赛的3场比赛中只积3分的概率;(2)已知在已结束的小组赛的3场比赛中,A 球队胜2场,负1场,求A 球队最终小组出线的概率.【答案】(1)427(2)7981【解析】【分析】(1)分类讨论只积3分的可能情况,结合独立事件概率乘法公式运算求解;(2)由题意,若A 球队参与的3场比赛中胜2场,负1场,根据获胜的三队通过净胜球数等规则决出前两名,分情况讨论结合独立事件概率乘法公式运算求解.【小问1详解】A 球队在小组赛的3场比赛中只积3分,有两种情况.第一种情况:A 球队在3场比赛中都是平局,其概率为111133327⨯⨯=.第二种情况:A球队在3场比赛中胜1场,负2场,其概率为11113 3339⨯⨯⨯=.故所求概率为114 27927+=.【小问2详解】不妨假设A球队参与的3场比赛的结果为A与B比赛,B胜;A与C比赛,A胜;A与D比赛,A胜.此情况下,A积6分,B积3分,C,D各积0分.在剩下的3场比赛中:若C与D比赛平局,则C,D每队最多只能加4分,此时C,D的积分都低于A的积分,A可以出线;若B与C比赛平局,后面2场比赛的结果无论如何,都有两队的积分低于A,A可以出线;若B与D比赛平局,同理可得A可以出线.故当剩下的3场比赛中有平局时,A一定可以出线.若剩下的3场比赛中没有平局,则当B,C,D各赢1场比赛时,A可以出线.当B,C,D中有一支队伍胜2场时,若C胜2场,B胜1场,A,B,C争夺第一、二名,则A淘汰的概率为11111 333381⨯⨯⨯=;若D胜2场,B胜1场,A,B,D争夺第一、二名,则A淘汰的概率为11111 333381⨯⨯⨯=.其他情况A均可以出线.综上,A球队最终小组出线的概率为1179 1818181⎛⎫-+=⎪⎝⎭.【点睛】关键点点睛:解题的关键在于分类讨论获胜的三队通过净胜球数等规则决出前两名,讨论要恰当划分,做到不重不漏,从而即可顺利得解.。
2022-2023学年安徽省桐城中学高二上学期月考(1)数学试题(解析版)

2022-2023学年安徽省桐城中学高二上学期月考(1)数学试卷一、单选题1.已知直线l的倾斜角为,且经过点,则直线l的方程为( )A. B.C. D.2.设点,,直线l过点且与线段AB相交,则l的斜率k的取值范围是( )A. 或B.C. D. 或3.与向量平行的一个向量的坐标是( )A. B.C. D.4.已知点,,则直线AB的斜率是( )A. B. C. 3 D.5.如图所示,在四面体中,,,,点M在OA上,且,N为BC的中点,则( )A. B.C. D.6.直三棱柱中,为等边三角形,,M是的中点,则AM与平面所成角的正弦值为( )A. B. C. D.7.已知正四面体ABCD,M为BC中点,N为AD中点,则直线BN与直线DM所成角的余弦值为( )A. B. C. D.8.如图,在直三棱柱中,,,,,则与所成的角的余弦值为( )A.B.C.D.9.如图,在平行六面体中,( )A.B.C.D.10.已知直线l过定点,且方向量为,则点到l的距离为( )A. B. C. D.11.已知空间向量,,满足,,,,则与的夹角为( )A. B. C. D.12.在平面直角坐标系xOy中,若双曲线的右焦点到一条渐近线的距离为,则其离心率的值为( )A. 4B. 2C.D.13.若直线:与直线:平行,则直线与之间的距离为______ .14.直线l:被圆O:截得的弦长最短,则实数______.15.在空间直角坐标系Oxyz中,,,,点Q在直线OP上运动,则当取得最小值时,点Q的坐标是______.16.已知向量,,若,则__________.17.在中,已知,,求边BC所在的直线方程;求的面积.18.已知三角形的三个顶点的坐标分别是、、求BC边所在直线的方程;求BC边上的中线所在直线的方程.19.如图,已知平面ABCD,底面ABCD为正方形,,M,N分别为AB,PC的中点.求证:平面PCD;求PD与平面PMC所成角的正弦值.20.已知直线经过点,,直线经过点,,且,求实数a的值.21.如图,在三棱柱中,四边形是边长为的正方形,,,证明:平面平面;在线段上是否存在点M,使得,若存在,求的值;若不存在,请说明理由.22.如图,在四棱锥中,底面ABCD,底面ABCD为梯形,,,且,若点F为PD上一点且,证明:平面PAB;求直线PA与平面BPD所成角的正弦答案和解析1.【答案】C【解析】解:由题意知:直线l的斜率为,则直线l的方程为故选:2.【答案】D【解析】解:,直线l过点且与线段AB相交,则l的斜率k的取值范围是或故选:3.【答案】C【解析】解:对于C中的向量:,因此与向量平行的一个向量的坐标是故选:4.【答案】D【解析】解:因为,,所以直线AB的斜率故选5.【答案】B【解析】解:连接ON,是BC的中点,,,,,故选:6.【答案】C【解析】解:因为M是的中点,为等边三角形,可得,又平面,平面,所以,而,,所以平面,以M为坐标原点,,所在直线分别为x,y轴,过M平行于的直线为z轴建立空间直角坐标系,设,则,,,又,所以,,,则,,设平面的法向量为,则,取,则,,所以,所以AM与平面所成角的正弦值为,故选:7.【答案】B【解析】解:设该正四面体的棱长为1,为BC中点,N为AD中点,,是BC中点,N为AD中点,,,,,根据异面直线所成角的定义知直线BN与直线DM所成角的余弦值为故选:8.【答案】A【解析】解:在直三棱柱中,,,,,建立以C为坐标原点,CA,CB,分别为x,y,z轴建立空间直角坐标系,则,,,,所以,,则,,,则,所以直线与所成角的余弦值为,故选:9.【答案】B【解析】解:为平行四面体,故选:10.【答案】A【解析】解:因为,,所以,又因为直线l的方向量为,所以点P到l的距离为,故选:11.【答案】C【解析】解:,,,,,,,,,,,,故选:12.【答案】B【解析】解:双曲线的右焦点到一条渐近线的距离为,可得:,可得,即,所以双曲线的离心率为:故选:13.【答案】【解析】解:直线与平行,所以,解得,所以直线:,直线:,所以直线与之间的距离为:故答案为:14.【答案】1【解析】解:直线MN的方程可化为,由,得,所以直线MN过定点,因为,即点A在圆内.当时,取最小值,由,得,,即故答案为:15.【答案】【解析】解:设,,,,由点Q在直线OP上,可得存在实数使得,则,根据二次函数的性质,得当时,取得最小值此时Q点的坐标为:故答案为:16.【答案】【解析】解:因为向量,,,由,则,解得故答案为:17.【答案】解:,,边BC所在的直线方程为,即;设B到AC的距离为d,则,,AC方程为:,即:,【解析】直接由两点式直线方程公式求解即可;求出B到AC的距离为d,再求AC的距离,然后利用面积公式求解即可.18.【答案】解:因为、,所以,所以直线BC的方程为,即;因为,、,所以BC的中点为,所以,所以中线AD的方程为,即;【解析】首先根据斜率公式求出,再由点斜式求出直线方程;求出BC的中点D的坐标,然后求出,再由点斜式求出直线方程;19.【答案】解:以A为原点建立如图所示空间直角坐标系,则,,,,则,,所以,,由于,所以平面,,设平面PMC的法向量为,则,令,则,,所以设直线PD与平面PMC所成角为,则【解析】建立空间直角坐标系,利用向量法证得平面利用直线PD的方向向量,平面PMC的法向量,计算线面角的正弦值.20.【答案】解:当直线的斜率不存在时,,解得,此时,,直线的斜率为0,满足,当直线的斜率存在时,直线的斜率,直线的斜率,,,解得,综上所述,实数a的值为0或【解析】根据已知条件,分直线的斜率存在和不存在两种情况讨论,即可求解.21.【答案】解:证明:在中,,,,有,可得,又,,可得平面,即有,由四边形是边长为的正方形,可得,而,可得平面,又平面,则平面平面;在线段上存在点M,使得,且理由如下:由可得,以C为原点,CA,CB,所在直线分别为x,y,z轴建立空间直角坐标系,如图所示,则,,,,,设,,所以,解得,,,所以,,要使,则需,即,解得故线段上存在点M,使得,且【解析】运用勾股定理和正方形的性质,推得平面,再由面面垂直的判定定理,即可得证;假设在线段上存在点M,使得,以C为原点,CA,CB,所在直线分别为x,y,z轴建立空间直角坐标系,设,,运用向量共线的坐标表示和向量垂直的数量积的坐标表示,可判断存在性.22.【答案】证明:作交PA于点H,连接BH,因为,则,又且,则且,所以四边形HFCB为平行四边形,故,又平面PAB,平面PAB,所以平面PAB;解:因为平面ABCD,平面ABCD,所以,又所以,则,以点B为坐标原点建立空间直角坐标系如图所示,则,,,,所以,设平面PBD的法向量为,则,即,令,则,,故,所以,故直线PA与平面BPD所成角的正弦值为【解析】作交PA于点H,连接BH,利用且,证明四边形HFCB 为平行四边形,从而得到,由线面平行的判定定理证明即可;建立合适的空间直角坐标系,求出所需点的坐标和向量的坐标,然后利用待定系数法求出平面PBD的法向量,由向量的夹角公式求解即可.。
2022~2023年高二第一次月考试卷完整版(贵州省思南中学)

选择题如今的消费者对手机的功能要求越来越高,如要有人脸识别功能,未来要支持AR应用等。
为此苹果公司计划2017年下半年,推出搭载AMOLED面板和人脸识别功能的机型,多家手机生产企业也有类似计划。
厂家之所以千方百计开发更新更好的产品行为( )A. 消费者对产品需求主要在于产品的质量B. 价值是使用价值的物质承担者C. 新产品开发能决定消费者的购买行为D. 商品是使用价值和价值的统一体【答案】D【解析】商品是使用价值和价值的统一体,使用价值是价值的物质承担者。
更新更好的产品能更好地满足消费者需要,实现商品价值,从而为生产者带来更多的利润,故D选项符合题意;A说法错误,消费者关注的是使用价值与价值的统一,排除;B说法错误,使用价值是价值的物质承担者,排除;C说法错误,新产品开发能影响消费者的购买行为,但不起决定性作用,排除。
故本题答案选D。
选择题2017年7月,河北容城县西瓜种植基地现场采摘的西瓜零售价仅0.35元每斤。
但此时千里之外的成都,本地大棚种植西瓜在超市、农贸市场的零售价竟达到1.5元每斤。
下列对成都西瓜价格高的原因推断正确的是( )①采用大棚种植,投入多成本高②流通环节多,流通成本较高③种植技术先进,劳动生产率高④西瓜品质好,决定其价格高A. ③④B. ①④C. ②③D. ①②【答案】D【解析】影响价格变动的因素主要有两个:一是价值决定价格,二是供求影响价格。
供不应求,价格上涨;供过于求,价格下降。
采用大棚种植且在超市、农贸市场出售的西瓜价格高,是因为种植投入多成本高,且流通环节多,流通成本较高,价值量高,所以价格高,故选项①②符合题意;选项③说法错误,个别劳动生产率的高低不会影响商品的价值量,不影响价格,排除;选项④说法错误,“品质好”强调的是西瓜的使用价值,决定价格的是商品的价值,排除。
故本题答案选D。
选择题下列两幅图反映了15年来手机的变化情况。
随着社会劳动生产率提高,手机的需求曲线出现了从左到右的变化。
四川省内江市2023-2024学年高二上学期第一次月考语文试题含解析

内江2023-2024学年度高二上期第一次月考语文试题(答案在最后)考试时间:150分钟满分:150分注意事项:1.答题前填写好自己的姓名、班级、考号等信息2.请将答案正确填写在答题卡上一、现代文阅读(35分)(一)现代文阅读Ⅰ(本题共5小题,17分)阅读下面的文字,完成小题。
材料一:先秦诸子百家中,儒、道、墨、法、阴阳、名六家属第一流的大学派。
汉以后,法、阴阳、名三家,其基本思想为儒、道所吸收,不再成为独立学派;墨家中绝;唯有儒、道两家长期共存,互相竞争,互相吸收,形成中国传统文化中一条纵贯始终的基本发展线索。
在中国传统文化的多元成分中,儒家和道家是主要的两极,形成鲜明的对立和有效的互补。
两者由于处处相反,因而能够相辅相成,给予整个中国传统文化以深刻的影响。
儒家的人生观,以成就道德人格和救世事业为价值取向,内以修身,充实仁德,外以济民,治国平天下,这便是内圣外王之道。
其人生态度是积极进取的,对社会现实强烈关切并有着历史使命感,以天下为己任,对同类和他人有不可自已的同情,“己所不欲,勿施于人”“己欲立而立人,己欲达而达人”“达则兼济天下,穷则独善其身”,不与浊俗同流合污,在生命与理想发生不可兼得的矛盾时,宁可杀身成仁,舍生取义,以成就自己的道德人生。
道家的人生观,以超越世俗人际关系网的羁绊,获得个人内心平静自在为价值取向,既反对心为形役,逐外物而不反,又不关心社会事业的奋斗成功,只要各自顺任自然之性而不相扰,必然自为而相因,成就和谐宁静的社会。
其人生态度消极自保,以免祸全生为最低目标,以各安其性命为最高目标。
或隐于山林,或陷于朗市,有明显的出世倾向。
儒家的出类拔萃者为志士仁人,道家的典型人物为清修隐者。
儒道两家的气象不同,大儒的气象似乎可以用“刚健中正”四字表示,就是道德高尚、仁慈亲和、彬彬有礼、忠贞弘毅、情理俱得、从容中道、和而不同、以权行经等等,凡事皆能观研深究,以求合理、合时、合情,可谓为曲践乎仁义,足以代表儒家的态度。
高二年级第一学期语文第一次月考试卷(附答案)

高二年级第一学期语文第一次月考试卷(附答案)一、现代文阅读(36 分)(一)论述类文本阅读(本题共 3 小题,9 分)阅读下面的文字,完成 1~3 题。
中国传统文化中的“和” 理念,具有丰富的内涵和深远的影响。
“和” 强调和谐、协调、平衡,既包括人与人之间的和谐相处,也包括人与自然的和谐共生。
在人与人的关系中,“和” 体现为一种包容、宽厚的态度。
孔子提出“君子和而不同”,强调在人际交往中,既要尊重他人的观点和差异,又要保持自己的独立思考和个性。
这种“和而不同” 的理念,有助于促进不同文化、不同思想之间的交流与融合,避免冲突和对抗。
在人与自然的关系中,“和” 则意味着尊重自然、顺应自然。
中国古代的思想家们认为,人类是自然的一部分,应该与自然和谐相处。
老子说:“人法地,地法天,天法道,道法自然。
” 强调人类应该遵循自然的规律,与自然保持一种和谐的关系。
这种理念对于我们今天处理人与自然的关系,具有重要的启示意义。
“和” 的理念还体现在社会治理方面。
中国古代的统治者们往往追求“政通人和” 的理想境界,通过推行仁政、德治等方式,促进社会的和谐稳定。
在现代社会,“和” 的理念也可以为我们构建和谐社会提供有益的借鉴。
我们可以通过加强民主法治建设、促进公平正义、弘扬社会主义核心价值观等方式,营造一个和谐、稳定、有序的社会环境。
1. 下列关于原文内容的理解和分析,正确的一项是(3 分)A.“和” 理念只强调人与人之间的和谐相处,不包括人与自然的和谐共生。
B. 孔子提出的“君子和而不同”,意味着在人际交往中要完全放弃自己的观点。
C. 中国古代思想家认为人类应该遵循自然规律,与自然和谐相处,这体现了“和” 的理念。
D.“和” 的理念在现代社会已经没有任何价值,不能为构建和谐社会提供借鉴。
2. 下列对原文论证的相关分析,不正确的一项是(3 分)A. 文章从人与人的关系、人与自然的关系、社会治理三个方面,论述了“和” 理念的内涵和影响。
高二第一学期政治第一次月考试卷(含答案)

高二第一学期政治第一次月考试卷(含答案)一、单项选择题(每题3 分,共60 分)1. 哲学是关于世界观的学说,世界观人人都有,但并非人人都是哲学家。
这是因为()A. 哲学是系统化理论化的世界观B. 哲学是对自然科学的概括和总结C. 世界观决定方法论,方法论体现世界观D. 哲学是关于整个世界最一般的本质和最普遍的规律2. 划分唯物主义和唯心主义的唯一标准是()A. 对思维和存在何者为第一性问题的不同回答B. 对思维和存在有没有同一性问题的不同回答C. 对世界是否具有统一性问题的不同回答D. 对世界是怎样存在的问题的不同回答3. 唯物主义在其历史发展中形成了三种基本形态,下列选项符合唯物主义发展顺序的是()①原子的属性就是物质的属性②阴阳二气充满太虚,此外更无他物③心外无物④物质是标志客观实在的哲学范畴,为我们的感觉所复写、摄影、反映A. ①→②→④B. ②→①→④C. ③→①→④D. ④→②→①4. 马克思主义哲学的产生实现了哲学史上的伟大变革。
马克思主义哲学的直接理论来源是()A. 德国古典哲学B. 古希腊哲学C. 近代形而上学唯物主义D. 空想社会主义5. 物质的唯一特性是()A. 客观实在性B. 运动C. 可知性D. 永恒性6. 世界的真正统一性在于它的()A. 物质性B. 主观性C. 实践性D. 规律性7. 运动是物质的根本属性和存在方式。
下列诗句中体现运动绝对性的是()A. 人生代代无穷已,江月年年只相似B. 博望沉埋不复旋,黄河依旧水茫然C. 无边落木萧萧下,不尽长江滚滚来D. 坐地日行八万里,巡天遥看一千河8. 规律是客观的,是不以人的意志为转移的。
下列说法能体现这一观点的是()A. 人有悲欢离合,月有阴晴圆缺B. 沉舟侧畔千帆过,病树前头万木春C. 忽如一夜春风来,千树万树梨花开D. 横看成岭侧成峰,远近高低各不同9. 意识是物质世界长期发展的产物。
从意识的起源看,意识是()A. 自然界长期发展的产物B. 社会发展的产物C. 物质世界长期发展的产物D. 人脑的产物10. 意识活动具有目的性、自觉选择性和主动创造性。
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高二下学期第一次月考(满分150分。
考试时间120分钟)第一部分:听力(共两节,满分30分)第一节:听下面5段对话。
每段对话后有一个小题, 从题中所给的A、B、C三个选项中选出最佳选项, 并标在试卷的相应位置。
听完每段对话后, 你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.What does the woman say about the World Cup?[A]She has no interest in it. [B]The ticket is too dear. [C]It's not worth seeing.2.Who will go to KTV?[A]The speakers. [B]Patty and Avery. [C]The woman and Avery.3.What is the woman?[A]A teacher. [B]A doctor. [C]A shop assistant.4.What are the speakers mainly talking about?[A]A team. [B]A game. [C]A goal.5.How much will the woman pay?[A]$800. [B]$500. [C]$30.第二节听下面5段对话或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听每段对话或独白前,你将有时间阅读各个小题, 每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听下面一段对话,回答第6和第7两个小题。
6.What's the relationship between the speakers?[A]Fellow students. [B]Co-workers. [C]A couple.7.What season is it now?[A]Spring. [B]Autumn. [C]Winter.听下面一段对话,回答第8和第9两个小题。
8.What do we know about the woman?[A]She stayed in London for 2 years.[B]She knows a lot about the newspaper.[C]She is good at sports.9.Why did the woman give up her first job?[A]She was poorly paid.[B]She wanted a higher position.[C]She went to live in another city.听下面一段对话,回答第10至第12三个小题。
10.Why can't the woman pick the man up at the station?[A]She has to work. [B]She can't find a taxi.[C]She will meet her teacher.11.Why does the woman ask the man to get off at the golf course?[A]They will play golf together.[B]The pathway is tough for the taxi to go.[C]It is the nearest place from her home.12.When will the speakers meet?[A]At 12:00 noon. [B]At about 12:30 pm. [C]At around 8:00 pm.听下面一段对话,回答第13至第16四个小题。
13.How often are Martin's wildlife programs on?[A]Once a day. [B]Once a week. [C]Once a month.14.Where were the wildlife programs filmed in the early 1960s?[A]In the mountains. [B]On the island. [C]In the local zoo.15.What did the Head of Programs think of Martin's idea of filming in Africa at first?[A]Great. [B]Boring. [C]Ridiculous.16.What is the feature of today's filming?[A]It is strictly planned.[B]The scenes are uncertain.[C]It takes less time to prepare.听下面一段独白,回答第17至第20四个小题。
17.What is the passage mainly about?[A]The history of hostels.[B]The development of hostels.[C]The reasons for the popularity of hostels.18.By whom were hostels mostly used years ago?[A]Hikers. [B]Business travelers. [C]Older tourists.19.How did the economic crisis influence hostels?[A]It brought down the price.[B]It increased the business.[C]It promoted the better services.20.What does the speaker say about modern hostels?[A]Their services are various.[B]Most of them are located in the countryside.[C]They provide cheap plane tickets for the customers.第二部分:阅读理解(共两节,满分 40分)第一节:(共15小题;每小题2分,满分30)AIn 1826, a Frenchman named Niepce needed pictures for his business .But he was not a good artist .So he invented a very simple camera .He put it in a window of his house and took a picture of his garden .That was the first photo.The next important date in the history of photography was in 1837. That year, Daguere, another Frenchman ,took a picture of his reading room .He used a new kind of camera in a different way. In his picture you could see everything very clearly ,even the smallest thing. This kind of photo was called a Daguerreotype.Soon, other people began to use Daguerre’s way. Travelers brought back wonderful photos from all around the world .people took picture of famous buildings, cities and mountains.In about 1840, photography was developed .Then photographers could take picture of people and moving things .That was not simple .The photographers had to carry a lot of film and other machines. But this did not stop them ,for example, some in the United States worked so hard.Mathew Brady was a famous American photographers. He took many picture of great people .The picture were unusual because they were very lifelike.Photographers also became one kind of art by the end of the 19th century .Some photos were nor just copies of the real world .They showed and feelings, like other kinds of art.21.The first photo taken by Niepce was a picture of ____________A. his businessB. his houseC. his gardenD. his window22.The Daguerreotype was____________.A. a FrenchmanB. a kind of pictureC. a kind of cameraD. a photographer23.Mathew Brady______________.A. was very lifelikeB. was famous for his unusual picturesC. was quite strongD. took many pictures of moving peopleBRobert Owen was born in Wales in 1771. At the age of ten he went to work. His employer had a large private library so Owen was able to educate himself. He read a lot in his spare time and at nineteen he was given the job of superintendent(监工) at a Manchester cotton mill. He was so successful there that he persuaded his employer to buy the New Lanark mill in Scotland.When he arrived at New Lanark it was a dirty little town with a population of 2,000 people. Nobody paid any attention to the workers' houses or their children's education. The conditions in the factories were very bad. There was a lot of crime and the men spent most of their wages on alcoholic drinks.Owen improved the houses. He encouraged people to be clean and save money. He opened a shop and sold the workers cheap, well-made goods to help them. He limited the sale of alcoholic drinks. Above all, he fixed his mind on the children's education. In 1816 he opened the first free primary school in Britain.People came from all over the country to visit Owen's factory. They saw that the workers were healthier and more efficient than in other towns. Their children were better fed and better educated. Owen tried the same experiment in the United States. He bought some land there in 1825,but the community was too far away. He could not keep it under control and lost most of his money.Owen never stopped fighting for his idea. Above all he believed that people are not born good or bad. He was a practical man and his ideas were practical. "If you give people good working conditions," he thought, "they will work well and, the most important thing of all, if you give them the chance to learn, they will be better people."24. For Owen, his greatest achievement in New Lanark was _____________.A. improving worker's housesB. helping people to save moneyC. preventing men from getting drunkD.providing the children with a good education25. From the passage we may infer that Owen was born ___________.A. into a rich familyB. into a noble familyC. into a poor familyD. into a middle class family26. Owen's experiment in the United States failed because _______.A. he lost all his moneyB. he did not buy enough landC. people who visited it were not impressedD. it was too far away for him to organize it properly27. We may infer form the passage that no children in Britain could enjoy free education until ____.A. 1771B. 1816C. 1825D. 1860CRead the advertisements,then choose the right answer.Driver Wanted①Clean driving license②Must be of smart appearance.③Age over 25.Apply to:Capes Taxi,17 Palace Road, Roston.Air Hostesses for International Flights Wanted.①Applicants must be between 20 and 33 years old.②Height 1.6m to 1.75m.③Education to GCSE standard(标准).④Two languages. Must be able to swim.Apply to: Recruitment(招聘)office, Southern Airlines,Heathrow Airport West.HR37KK.Teachers NeededFor private language school.Teaching experience unnecessary.Apply to :The Director of Studies, Instant Languages Ltd,279 Canal Street, Roston.28.What prevent Jack, an experienced taxi driver, working for Capes Taxis?A. Fond of sports and music.B. Punished for speeding and wrong parking.C. Unable to speak a foreign language.D .Not having college education.29.Ben,aged 22,fond of swimming and driving, has just graduated from a college. Which job might be given to him?A. Driving for Capes taxis.B. Working for Southern Airlines.C. Teaching at Instant languages Ltd.D. None of the three.30.What prevent Mary, aged 25,becoming an air hostess?A. She once broke a traffic law and was fined(罚款)B. She can’t speak Japanese very well.C. She has never worked as an air hostess before.D. She is 1.85m in height.31.Which of the following is not mentioned in the three advertisements?A. Marriage.B. Male or female.C. Education . D .Working experienceDWhen someone has deeply hurt you, it can be extremely difficult to let go of your anger. But forgiveness is possible - and it can be surprisingly helpful to your physical and mental health. Indeed, research has shown that people who forgive report more energy, better appetite (胃口) and better sleep patterns. "People who forgive show less anger and more hopefulness," says Dr. Frederic Luskin, who wrote the book Forgive for Good. "So it can help save on the wear and tear on our system and allow people to feel more energetic."So when someone has hurt you, calm yourself first. Take a couple of breaths and think of something that gives you pleasure: a beautiful scene in nature, someone you love. Don' t wait for an apology. "Many times the person who hurt you may never think of apologizing," says Dr. Luskin. "They may have wanted to hurt you or they just don't see things the same way. So if you wait for people to apologize, you could be waiting a very long time. " Keep in mind that forgiveness does not necessarily mean accepting the action of the person who upset you. Mentally going over your hurt gives power to the person who brought you pain. Instead, learn to look for the love, beauty and kindness around you. Finally, try to see things from the other person' s perspective 视角). You may realize that he or she-was acting out of ignorance , fear - even love. To gain perspective, you may want to write a letter to yourself from that person' s point of view.32. The text is mainly written to explain _______.A. how to keep yourself from being hurtB. how to stay mentally healthyC. how to stay calmD. how to get on well with others33. According to the writer, what is the right way to calm down after being hurt?A. Try to figure out why you get hurt.B. Write a letter to the person who hurt you.C. Persuade yourself to accept what others have done to you.D. Think about pleasant things and forget about the hurt.34. Dr. Luskin advises us not to wait for an apology after being hurt because ______.A. we are not patient enoughB. we' d feel worse accepting others' apologyC. people seldom want to apologizeD. people like to apologize to strangers.35.Where does this passage comes from?A. Geography magazineB. Science magazineC. Social skills magazineD. Travel and food magazine第二节:(共5小题;每小题2分,满分10分)Weight loss is a hard topic, Lots of people aren’t satisfied with their present weight, but most people aren’t sure how to change it. You may want to look like the models or actors in magazines or on TV, but those goals might not be healthy or realistic(现实的) for you. 36So what should you do about your weight control?37 The best way to find out if you are at a healthy weight or if you need to lose or gain weight is to talk to a doctor or dietitian(营养学家). 38 If it turns out that you can benefit from weight loss then you can follow a few of the simple suggestions listed below to get started.39 People who lose weight quickly by crash dieting or other extreme measures usually gain back all of the pounds they lost, because they haven’t permanently (永久地) changed their habits. Therefore, the best weight management ways are those that you can maintain for a lifetime.Small changes are a lot easier to stick with (坚持做) than large ones. Try reducing the size of what you eat. 40 Once you have that done, start gradually introducing healthier foods and exercise into your life.It’s a good idea to maintain a healthy weight because it’s just that: healthy.A.Try giving up regular soda for a week.B.Try to pay attention as you eat and stop when you’re full.C.Weight management is about long-term success.D.Besides, no magical diet will make you look like someone else.E.Being healthy is really about being at a weight that is right for you.F.Changing from whole to nonfat or low-fat milk is also a good idea.G.They will compare your weight with healthy standards and help you set goals.第三部分:英语知识运用(共两节,满分45分)第一节:完形填空(共20小题;每小题1.5分,满分30)Mr. Smith gave his wife ten pounds for her birthday. The day after her birthday Mrs. Smith went shopping. She got on 41 and sat down next to an old lady.42 she noticed that the old lady’s handbag was 43 . Inside it, she found a wad(沓)of pound notes 44 the one her husband had given her. She quickly 45 her own bag—the notes were 46 . Mrs. Smith was now sure that the old lady sitting 47 her must have stolen them. She thought 48 not have to call the 49 as she didn’t like getting people 50 .So she decided to take back the money 51 the lady’s handbag and say 52 about it. She looked around the bus to make sure 53 was watching, then she carefully put her hand into 54 handbag, took out the notes and 55 her own handbag.When she got home that evening, she showed 56 the beautiful hat she had bought.“How did you 57 it?” he asked.“58 you gave me for my birthday, of course.”“Oh, 59 then?”he asked, as he 60 a wad of pound notes on the table.41. A.a bus B.a train C.an old ship D.a plane42. A.In a minute B.After a while C.For a second D.On the moment 43. A.good B.old C.open D.shut44. A.the same that B.perhaps was C.probably as D.exactly like 45. A.looked at B.watched carefully C.saw to D.looked into46. A.gone B.bought C.appeared D.found47. A.close B.next to C.before D.behind48. A.she would B.he could C.she must D.he might49. A.driver B.old lady C.police D.husband50. A.to difficulty B.into trouble C.out of work D.seeing her51. A.into B.out C.away D.from52. A.something B.everything C.anything D.nothing53. A.nothing B.somebody C.nobody D.neither54. A.the old lady’s B.her husband’s C.the police’s D.her own55. A.gave it away B.put them into C.brought them out D.took it to 56. A.the driver B.the police C.the old lady D.her husband 57. A.pay for B.spend on C.cost in D.take to58. A.Use them B.With the money C.With that D.Using it59. A.how is it B.what’s that C.where is it D.why is this 60. A.put up B.held out C.pointed to D.handed up第二节:语法填空(共10小题;每小题1.5分,满分15)The Internet is an amazing information resource. Students, teachers, and researchers use it as __61 investigative tool. Journalists use it to find information for stories. Doctors use it to learn more about unfamiliar diseases and the 62 (late) medical development. Ordinary people use it for shopping, banking, bill-paying, and 63 (communicate)with family and friends. People all over the world use it to connect with individuals from _64 countries and cultures. While there are many positive developments __65__ (associate) with the Internet, there are also certain fearsand concerns. One concern relates to a lack __66__control over what appears on the Internet. With television and radio there are__67_(editor) to check the accuracy or appropriateness of the content of programs, and with television there are restrictions on what kinds of programs can __68__ (broadcast) and at what times of the day. With the Internet, parents cannot check a published guide to_69 (determination) what is suitable for__70___ (they) children to see.第四部分:写作(共两节,满分35)第一节:短文改错(共10小题;每小题1分,满分10分)A few days before, I saw an interesting program on TV. It was about the problem of traffic in our cities. It was seemed that a great deal of damage had been done by traffic. But now our cities are still suffered from the problem. Some buildings are actually falling to piece. There is often faster to walk than to go by car or bus, The problem has been getting worse for long time. This is only one problem among thousands of others in our cities. All sorts of terribly things have happened with our cities. And the biggest question is “How can we do about it?” Our roads have ever been designed for such a heavy traffic.第二节:书面表达(满分25分)假如你叫李华,毕业于贵州大学旅游学院。