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2024年浙江省五校联盟高三3月联考英语试题(含答案)

2024年浙江省五校联盟高三3月联考英语试题(含答案)

2024年浙江省五校联盟高三3月联考五校:杭州二中、温州中学、金华一中、绍兴一中、衢州二中命题:浙江省温州中学第一部分听力 ( 共两节,满分 3 0 分 )做题时,先将答案标在试卷上。

录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A 、B 、C 三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.What part of maths is the woman bad at?A.ShapesB.Numbers.C.Angles.2.What is the probable relationship between the speakers?A.Friends.B.Brother and sister.C.Doctor and patient.3.What industry does the woman hope to work in?A.Travel.B.Finance.C.Medicine.4.Where are the speakers probably?A.In a classroomB.In the wild.C.In a hospital.5.When will the woman's mother probably arrive?A.At about 12:00 p.m.B.At about 3:00 p.m.C.At about 6:00 p.m.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的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。

广东省五校2024-2025学年高二10月联考(一)数学试卷(解析版)

广东省五校2024-2025学年高二10月联考(一)数学试卷(解析版)

2024-2025学年第一学期珠海市实验中学、河源高级中学、中山市实验中学、惠州市博罗中学、珠海市鸿鹤中学联考(一)试卷高二数学满分:150分 考试时间:120分钟1.说明:注意事项:1.答卷前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名和考生号、试室号、座位号填写在答题卡上.用2B 铅笔将试卷类型(A )填涂在答题卡相应位置上.2.选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目选项的答案信息点涂黑.如需改动,用橡皮擦干净后,再选涂其他答案,答案不能答在试卷上.一、选择题:本题共8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.310y −−=的倾斜角为() A. 30° B. 135°C. 60°D. 150° 【答案】A 【解析】【分析】根据直线倾斜角与斜率之间的关系即可得倾斜角. 【详解】设直线的倾斜角为α, tan 180αα=°≤<°,所以30α=°, 故选:A2. 设()()(),,1,1,1,1,,,,4,2x y a b y z c x ∈===−R ,且,//a c b c ⊥,则2a b +=( ) A. B. 0C. 3D. 【答案】D 【解析】【分析】由向量的共线与垂直条件求解,b c的坐标,再由向量坐标运算及求模公式可得.【详解】2,,,,,,,11114,a b y z c x ===−,由a c ⊥,则有420a c x ⋅=−+= ,解得2x =,则()2,4,2c =− .由//b c ,则有1242y z==−,解得2y =−,1z =, 所以()1,2,1b =−,故()23,0,3a b += ,则2a b + .故选:D.3. 下列命题中正确的是( )A. 点()3,2,1M 关于平面yOz 对称点的坐标是()3,2,1−−B. 若直线l 的方向向量为()1,1,2e=−,平面α的法向量为()6,4,1m =−,则l α⊥ C. 若直线l 方向向量与平面α的法向量的夹角为120 ,则直线l 与平面α所成的角为30D. 已知O 为空间任意一点,A ,B ,C ,P 四点共面,且任意三点不共线,若12OP mOA OB OC =−+,则12m =−【答案】C 【解析】【分析】由空间点关于平面的对称点的特点可判断A ;由向量的数量积的性质可判断B ;由线面角的定义可判断C ;由共面向量定理可判断D.【详解】对于A ,点()3,2,1M 关于平面yOz 对称的点的坐标是()3,2,1−,A 选项错误;对于B ,若直线l 的方向向量为()1,1,2e=−,平面α的法向量为()6,4,1m =−, ()()1614210e m ⋅=×+−×+×−=,有e m ⊥ ,则//l α或l α⊂,B 选项错误;对于C ,若直线l 的方向向量与平面α的法向量的夹角为120 , 则直线l 与平面α所成的角为()9018012030−−=,C 选项正确; 对于D ,已知O 为空间任意一点,A ,B ,C ,P 四点共面,且任意三点不共线,若12OP mOA OB OC =−+ ,则1112m −+=,解得12m =,D 选项错误. 故选:C.4. 如图,从光源P 发出的一束光,遇到平面镜(y 轴)上的点B 后,反射光线BC 交x轴于点)C,若光线PB 满足的函数关系式为:1y kx =+,则k 的值为( ) 的的A.B.C. 1D. -1【答案】A 【解析】【分析】根据题意,求得(0,1)B 和点C 关于y 轴的对称点()C ′,求得BC k ′,结合,,P B C ′三点共线,即可求解.【详解】为光线PB 满足的函数关系式为1y kx =+, 令0x =,可得1y =,即点(0,1)B ,又因为)C,则点C 关于y 轴的对称点为()C ′,可得BC ′的斜率为BC k ′=,因为,,P B C ′三点共线,可得BC k k ′=,所以k =. 故选:A.5. 过点1,13作直线l ,则满足在两坐标轴上截距之积为2的直线l 的条数为( ) A. 1 B. 2C. 3D. 4【答案】B【分析】设直线l 的方程为()102x ay a a +=≠,将点1,13 代入直线l 的方程,然后由判别式判断即可. 【详解】设直线l 的方程为()102x ay a a +=≠, 将点1,13代入,可得()11032aa a +=≠, 即23620a a −+=,由于Δ36432120=−××=>, 所以方程23620a a −+=有两个根, 故满足题意的直线l 的条数为2. 故选:B.6. 如图,在三棱锥O ABC −中,点D 是棱AC 的中点,若OA a = ,OB b = ,OC c = ,则BD等于( )A 1122a b c −+B. a b c +−C. a b c −+D. 1122a b c −+−【答案】A 【解析】【分析】根据空间向量的基本定理结合线性运算的坐标表示求解. 【详解】点D 是棱AC 的中点,则有()()()11211122222BD BA BC OA OB OC OB a b c a b c =+=−+−=−+=−+.故选:A7. 已知长方体1111ABCD A B C D −,下列向量的数量积一定不为0的是( ).A. 11AD B C ⋅B. 1BD AC ⋅C. 1AB AD ⋅D. 1BD BC ⋅【答案】D 【解析】【分析】当四边形ADD 1A 1为正方形时,可证AD 1⊥B 1C 可判断A ;当四边形ABCD 为正方形时,可证AC ⊥BD 1可判断B ;由长方体的性质可证AB ⊥AD 1,分别可得数量积为0,可判断C ;可推在△BCD 1中,∠BCD 1为直角,可判BC 与BD 1不可能垂直,可得结论可判断D.【详解】选项A ,当四边形ADD 1A 1为正方形时,可得AD 1⊥A 1D ,而A 1D ∥B 1C ,可得AD 1⊥B 1C ,此时有110⋅=AD B C ,故正确;选项B ,当四边形ABCD 为正方形时,可得AC ⊥BD ,1AC BB ⊥,1BD BB B ∩=, 1,BD BB ⊂平面BB 1D 1D ,可得AC ⊥平面BB 1D 1D ,故有AC ⊥BD 1,此时有10⋅=BD AC ,故正确;选项C ,由长方体的性质可得AB ⊥平面ADD 1A 1,1AD ⊂平面ADD 1A 1,可得AB ⊥AD 1,此时必有1AB AD ⋅=0,故正确; 选项D ,由长方体的性质可得BC ⊥平面CDD 1C 1,1CD ⊂平面CDD 1C 1,可得BC ⊥CD 1,△BCD 1为直角三角形,∠BCD 1为直角,故BC 与BD 1不可能垂直,即10⋅≠BD BC ,故错误.故选:D.8. 如图已知矩形,1,ABCD AB BC==AC 将ABC 折起,当二面角B AC D −−的余弦值为13−时,则B 与D 之间距离为( )A. 1B.C.D.【答案】C 【解析】【分析】过B 和D 分别作BE AC ⊥,DF AC ⊥,根据向量垂直的性质,利用向量数量积进行转化求解即可.【详解】解:过B 和D 分别作BE AC ⊥,DF AC ⊥,在矩形,1,ABCD AB BC ==2AC ∴=, ABC ADC S S =△△,1122AB BC AC BE ∴⋅=⋅BE DF ∴==, 则12AECF ==,即211EF =−=, 平面ABC 与平面ACD 所成角的余弦值为13−,cos EB∴< ,13FD >=− , BD BE EF FD =++ ,∴2222233()22212cos 44BD BE EF FD BE EF FD BE EF FD BE EF FD EB FD EB =++=+++⋅+⋅+⋅=++−⋅<,51512()32322FD >=−−=+= ,则||BD =即B 与D , 故选:C .二、选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求.全部选对的得6分,部分选对的得部分分,有选错的得0分.9. 已知直线l 过点()2,3M −,且与x 轴、y 轴分别交于A ,B 点,则( ) A. 若直线l 的斜率为1,则直线l 的方程为5y x =+B. 若直线l 在两坐标轴上的截距相等,则直线l 的方程为1x y +=C. 若M 为AB 的中点,则l 的方程为32120x y −+=D. 直线l 的方程可能为3y = 【答案】AC 【解析】【分析】根据直线点斜式判断A ,由过原点直线满足题意判断B ,由中点求出A ,B 坐标得直线方程判断C ,由直线与坐标轴有交点判断D.【详解】对于A ,直线l 的斜率为1,则直线l 的方程为32y x ,即5y x =+,故A 正确; 对于B ,当直线l 在两坐标轴上的截距都为0时,l 的方程为32y x =−,故B 错误; 对于C ,因为中点()2,3M −,且A ,B 在x 轴、y 轴上,所以()4,0A −,()0,6B ,故AB 的方程为146x y−+=,即32120x y −+=,故C 正确; 对于D ,直线3y =与x 轴无交点,与题意不符,故D 错误. 故选:AC .10. 如图,在平行六面体1111ABCD A B C D −中,以顶点A 为端点的三条棱长均为6,且它们彼此的夹角都是60°,下列说法中正确的是( )A. CC 1⊥BDB. 1136AA BD ⋅=C. 11B C AA与夹角是60°D. 直线AC 与直线11A C 的距离是【解析】【分析】设1,,AB a AD b AA c ===,依题得||||||6,18,a b c a b b c c a ===⋅=⋅=⋅= 运用向量数量积的运算律计算即可判断A,B 两项;利用向量夹角的公式计算排除C 项;利用空间向量关于点到直线的距离公式计算即可验证D 项.【详解】如图,设1,,AB a AD b AA c ===, 则||||||6,66cos 6018,a b c a b b c c a ===⋅=⋅=⋅=××=对于A ,因1,CC c BD b a ==−,则1()0CC BD c b a c b c a ⋅=⋅−=⋅−⋅=,故A 正确; 对于B ,因1AA c = ,1BD b a c =−+,则211()||18183636AA BD c b a c c b c a c ⋅=⋅−+=⋅−⋅+=−+= ,故B 正确; 对于C ,11,B C b c AA c =−= 211()||183618B C AA b c c b c c ⋅=−⋅=⋅−=−=− ,且11||6,||6,B C AA ==设11B C AA 与夹角为θ,则1111181cos 662||||B C AA B C AA θ⋅==−=−×⋅,因[0,π]θ∈,则2π3θ=,即C 错误;对于D,在平行六面体1111ABCD A B C D −中,易得111111////,AA BB CC AA BB CC ==, 则得11ACC A ,故11//AC A C ,故点1A 到直线AC 的距离d 即直线AC 与直线11A C 的距离.因,AC a b =+ 1()36AA AC c a b ⋅=⋅+=,且1||6,||AA AC==则d ===,故D 正确.11. 如图,已知正方体1111ABCD A B C D −的棱长为2,E ,F ,G 分别为AD ,AB ,11B C 的中点,以下说法正确的是( )A. 三棱锥1C EFG −的体积为13B. 1A C ⊥平面EFGC. 1BC ∥平面EFGD. 二面角G EF C −−【答案】ABC 【解析】【分析】建立如图所示的空间直角坐标系,由向量法证明1//BC 面EFG ,1A C ⊥平面EFG ,转换后求棱锥的体积,由空间向量法求二面角,从而判断各选项.【详解】如图,分别以1,,DA DC DD 为,,x y z 轴建立空间直角坐标系,则(0,0,0)D ,(2,0,0)A ,(0,2,0)C ,1(0,0,2)D ,(2,2,0)B ,1(0,2,2)C ,1(2,2,2)B ,1(2,0,2)A ,E ,F ,G 分别为AD ,AB ,11B C 的中点,则(1,0,0)E ,(2,1,0)F ,(1,2,2)G ,(1,1,0),(0,2,2)EF EG ==,1(2,0,2)BC − ,易知12BC EG EF =−,所以1,,BC EF EG 共面, 又1BC ⊄平面EFG ,所以1//BC 面EFG ,C 正确;1111111123323C EFG B EFG G BEF BEF V V V S BB −−−===⋅=××××= ,A 正确; 1(2,2,2)A C =−− ,12200AC EF ⋅=−++= ,同理10A C EG ⋅=, 所以1AC是平面EFG 的一个法向量,即1A C ⊥平面EFG ,B 正确; 平面CEF 的一个法向量是(0,0,1)n =,111cos ,A C n A C n A C n ⋅===G EF C −−D 错误.三、填空题:本题共3小题,每小题5分,共15分.12. 若直线1l :10x ay +−=与直线2l :420ax y ++=平行,则a =___________. 【答案】2 【解析】【分析】结合已知条件,利用直线间的平行关系求出参数a ,然后对参数a 进行检验即可求解.【详解】因为直线1l :10x ay +−=与直线2l :420ax y ++=平行, 所以2140a ×−=,解得,2a =±,当2a =时,直线1l :210x y +−=,直线2l :2420x y ++=,即210x y ++=,满足题意; 当2a =−时,直线1l :210x y −−=,直线2l :2420x y −++=,即210x y −−=, . 综上所述,2a =. 故答案为:2.13. 已知()()2312A B −,,,,若点(),P x y 在线段AAAA 上,则3yx −的取值范围是_______. 【答案】13,2−−【解析】【分析】设(3,0)Q ,利用斜率计算公式可得:QA k ,QB k .再利用斜率与倾斜角的关系即可得出. 【详解】设(3,0)Q ,则30323AQ k −==−−,201132BQ k −==−−−, 点(,)P x y 是线段AB 上的任意一点, ∴3y x −的取值范围是[3−,1]2−,故答案为:[3−,1]−14. 《九章算术》中的“商功”篇主要讲述了以立体几何为主的各种形体体积的计算,其中堑堵是指底面为直角三角形的直棱柱.如图,在堑堵111ABC A B C −,中,M 是11A C 的中点,122AB AA AC ==,113BN BB = ,3MG GN =,若1AG xAA y AB z AC =++ ,则x y z ++=_________.【答案】118【解析】【分析】建立空间直角坐标系,利用空间向量可以解决问题.【详解】设2AB =,如下图所示,建立空间直角坐标系, ()000A ,, ,()200B ,,,()001C ,,,()1010A ,,1012M ,,,1203N,,,则1121200123232MN=−=−,,,,,-, 所以13213110122432228AG AM MG++−,,,-,,, 又因为()131122,,228AG xAA y AB z AC y x z y x z ++⇒,, 所以131112488x y z ++=++= 故答案为:118四、解答题:本题共5小题.解答应写出文字说明、证明过程或演算步骤.15. 已知ABC 的两顶点坐标为()1,1A −,()3,0C ,()10,1B 是边AB 的中点,AD 是BC 边上的高. (1)求BC 所在直线的方程; (2)求高AD 所在直线的方程.【答案】(1)3490x y +−=; (2)4370x y −−=. 【解析】【分析】(1)由条件结合中点坐标公式求B 的坐标,利用点斜式求直线BC 方程,再化为一般式即可; (2)根据垂直直线的斜率关系求直线AD 的斜率,利用点斜式求直线AD 方程,再化为一般式即可. 【小问1详解】因为1()0,1B 是边AB 的中点,所以()1,3B −, 所以直线BC 的斜率34BC k =−, 所以BC 所在直线的方程为:()334y x =−−,即3490x y +−=, 【小问2详解】因为1()0,1B 是边AB 的中点,所以()1,3B −, 因为AD 是BC 边上的高,所以1BC AD k k ⋅=−,所以30113AD k −⋅=−−−, 所以43AD k =, 因此高AD 所在直线的方程为:41(1)3y x +=−,即4370x y −−=.16. 已知直线()()1231:−=−+a y a x l . (1)求证:直线l 过定点;(2)若直线l 不经过第二象限,求实数a 的取值范围;(3)若直线l 与两坐标轴的正半轴围成的三角形面积最小,求l 的方程. 【答案】(1)证明见解析 (2)1a ≤(3)240x y +−=【解析】【分析】(1)由方程变形可得()2310a x y x y −−++=,列方程组,解方程即可; (2)数形结合,结合直线图像可得解;(3)求得直线与坐标轴的交点,可得面积,进而利用二次函数的性质可得最值. 【小问1详解】由()():1231l a y a x −=−+,即()2310a x y x y −−++=, 则20310x y x y −= −++=,解得12x y = = ,所以直线过定点()1,2; 【小问2详解】如图所示,结合图像可知,当1a =时,直线斜率不存在,方程为1x =,不经过第二象限,成立; 当1a ≠时,直线斜率存在,方程为11213ya a a x +−−−, 又直线不经过第二象限,则2301101a a a − > −≤ − ,解得1a <; 综上所述1a ≤; 【小问3详解】已知直线()():1231l a y a x −=−+,且由题意知1a ≠,令0x =,得101=>−y a ,得1a >, 令0y =,得1032>−xa ,得32a <,则22111112132410651444S a a a a a =××==−−−+−−−+, 所以当54a =时,S 取最小值, 此时直线l 的方程为55123144y x−=×−+,即240x y +−=. 17 已知()()()0,0,0,2,5,0,1,3,5A B C .(1)求AC 在AB上的投影向量;(2)若四边形ABCD 是平行四边形,求顶点D 的坐标; (3)若点(0,3,0)P ,求点P 到平面ABC 的距离.【答案】(1)3485,,02929(2)()1,2,5−−(3【解析】【分析】(1)利用投影向量公式可求投影向量;.(2)根据AD BC =可求D 的坐标;(3)根据点面距公式可求点P 到平面ABC 的距离. 【小问1详解】()1,3,5AC = ,()2,5,0AB = ,故AC 在AB上的投影向量为AC AB AB ABAB⋅, 而()21534852,5,0,,0292929AC AB AB AB AB⋅+ ==.【小问2详解】设(),,D x y z ,则AD BC =,故()(),,1,2,5x y z =−−, 故D 的坐标为()1,2,5−−. 【小问3详解】()0,3,0AP =,设平面ABC 的法向量为mm ��⃗=(xx ,yy ,zz ),则00m AB m AC ⋅= ⋅=即250350x y x y z += ++= ,取5x =−,则2y =,15z =−, 故15,2,5m=−−,故点P 到平面ABC18. 如图,在长方体1111ABCD A B G D −中,11,2AD AA AB ===,点E 在棱AB 上移动.(1)求证:11D E A D ⊥.(2)当点E 为棱AB 的中点时,求CE 与平面1ACD 所成角的正弦值. (3)在棱AB 上是否存在点M ,使平面1D MC 与平面AMC 所成的角为π6?若存在,求出AM 的值;若不存在,请说明理由.【答案】(1)证明见解析(2(3)存在,2AM =. 【解析】【分析】(1)依题意建立空间直角坐标系,利用空间向量数量积为0即可证得垂直; (2)先求得平面1ACD 的法向量,再利用空间向量法求线面角即可得解;(3)先求得平面1D MC 与平面AMC 法向量,再利用空间向量法求线面角即可得解. 【小问1详解】以D 为坐标原点,直线DA ,DC ,1DD 分别为x ,y ,z 轴,建立空间直角坐标系,设AE x =,02x <<,则()11,0,1A ,()10,0,1D ,()1,,0E x ,AA (1,0,0),()0,2,0C ,所以()()111,0,11,,10DA D E x ⋅=⋅−=,则11DA D E ⊥, 所以11D E A D ⊥. 【小问2详解】因为E 为AB 的中点,所以()1,1,0E ,从而()1,1,0CE=−,()1,2,0AC =− ,()11,0,1AD =−,设平面1ACD 的法向量为(),,n a b c = ,则100n AC n AD ⋅=⋅= , 即200a b a c −+=−+= ,得2a b a c= = ,令2a =,则()2,1,2n =, 设CE 与平面1ACD 所成角为π02θθ<<,的则sin cos ,CE θ=〈 所以CE 与平面1ACD. 【小问3详解】设这样的点M 存在,且AM x =,02x <<,平面1D MC 与平面AMC 所成的角为π6, 则()1,,0M x ,()10,0,1D ,()0,2,0C ,()1,2,0CM x =− ,()10,2,1CD =−,设平面1D MC 的法向量为(),,m a b c ′′=′ ,则()12020m CM a x b m CD b c ⋅=+−= ⋅′=−′+=′′, 取1b ′=,得()2,1,2mx =−, 易知平面AMC 的一个法向量()0,0,1p =,所以πcos 6m p m p⋅== ,由02x <<,解得2x =,所以满足题意的点M 存在,此时2AM =. 19. 已知111(,,)a x y z = ,222(,,)b x y z = ,333(,,)c x y z =,定义一种运算:123231312132213321()a b c x y z x y z x y z x y z x y z x y z ×⋅=++−−−,已知四棱锥P ABCD −中,底面ABCD是一个平行四边形,(2,1,4)AB =− ,(4,2,0)AD = ,(1,2,1)AP −(1)试计算()AB AD AP ×⋅的绝对值的值,并求证PA ⊥面ABCD ;(2)求四棱锥P ABCD −的体积,说明()AB AD AP ×⋅的绝对值的值与四棱锥P ABCD −体积的关系,并由此猜想向量这一运算()AB AD AP ×⋅的绝对值的几何意义.【答案】(1)48,证明见解析;(2)体积为16,()3P ABCD AB AD AP V −×⋅=,()AB AD AP ×⋅的绝对值表示以,,AB AD AP 为邻边的平行六面体的体积. 【解析】【分析】(1)根据新定义直接计算,由向量法证明线线垂直,得线面垂直;(2)计算出棱锥体积后,根据数据确定关系.【详解】(1)由题意()AB AD AP ×⋅221424(1)(1)0=××+××+−×−×202−××4(1)1−×−×(1)24−−××=48.122(1)140AP AB ⋅=−×+×−+×= ,1422100AP AD ⋅=−×+×+×=,∴,AP AB AP AD ⊥⊥,即,AP AB AP AD ⊥⊥.,AB AD 是平面ABCD 内两相交直线,∴AP ⊥平面ABCD .(2)由题意2221,20AB AD == ,24(1)2406AB AD ⋅=×+−×+×=,sin ABCDS AB AD BAD=∠==,AP =∴111633P ABCD ABCD V S PA −==×=. ∴()3P ABCD AB AD AP V −×⋅=, 猜想:()AB AD AP ×⋅的绝对值表示以,,AB AD AP 为邻边的平行六面体的体积.【点睛】本题考查向量的新定义运算,解题时根据新定义的规则运算即可.考查学生的创新意识,同时考查学生的归纳推理能力.。

2024-2025学年福建省福宁古五校教学联合体高三上学期期中联考物理试题

2024-2025学年福建省福宁古五校教学联合体高三上学期期中联考物理试题

2024-2025学年福建省福宁古五校教学联合体高三上学期期中联考物理试题1.物理学在长期的发展进程中,形成了一整套系统的研究问题的思想方法。

如微元法、比值定义法、极限法、类比法等。

这些思想方法极大地丰富了人们对物质世界的认识,拓展了人们的思维方式。

下列说法中不正确的是()A.动摩擦因数的定义用的是比值定义法B.卡文迪许巧妙地采用了放大法,运用扭秤测出万有引力常量C.瞬时速度的定义用到极限的思想方法,且瞬时速度方向和平均速度方向总是一致的D.推导匀变速直线运动位移公式时,把整个运动过程划分成很多小段,每一小段近似看做匀速直线运动,然后把各小段的位移相加,其和代表物体的位移,采用的是微元法,且位移方向与平均速度的方向总是一致的2.关于曲线运动,下列说法正确的是()A.在变力作用下,物体不可能做曲线运动B.做曲线运动的物体,相等时间内速度的变化量可能相同C.做曲线运动的物体,受到的合外力一定在不断改变D.只要物体做圆周运动,它所受的合外力一定指向圆心3.如图所示,轻杆的一端固定在通过O点的水平转轴上,另一端固定一小球,轻杆绕O点在竖直平面内沿逆时针方向做匀速圆周运动,轨迹经A,B,C,D四点,其中A点为最高点、C点为最低点,B点与O点等高,下列说法正确的是()A.小球经过B点时,所受杆的作用力方向沿着BO方向B.从A点到C点的过程,杆对小球的作用力做正功C.从A点到C点的过程,小球重力的瞬时功率保持不变D.小球经过D点时,所受杆的作用力方向可能沿切线方向4.在跳台滑雪比赛中,运动员在空中滑翔时身体的姿态会影响其下落的速度和滑翔的距离,如图(a)所示;某运动员先后两次从同一跳台起跳,每次都从离开跳台开始计时,用表示他在竖直方向的速度,其图像如图(b)所示,和是他落在倾斜雪道上时刻,则()A.第一次滑翔过程中在竖直方向上的位移比第二次的大B.第一次滑翔过程中在水平方向上的位移比第二次的大C.第二次滑翔过程中在竖直方向上的平均加速度比第一次的大D.竖直方向速度大小为时,第二次滑翔在竖直方向上所受阻力比第一次的大5.人造地球卫星失效后一般有两种处理方案,即“火葬”与“冰冻”。

2024届浙江省五校联盟高三下学期3月联考语文试题

2024届浙江省五校联盟高三下学期3月联考语文试题

2024年浙江省五校联盟高三3月联考语文试题卷命题:浙江省衢州第二中学考生须知:1.本卷满分150分,考试时间150分钟;2.答题前,在答题卷指定区域填写班级、姓名、试场号、座位号及准考证号。

3.所有答案必须写在答题纸上,写在试卷上无效;4.考试结束后,只需上交答题纸。

一、现代文阅读(35分)(一)现代文阅读I(本题共5小题,19分)阅读下面的文字,完成1—5题。

材料一:经典是“恒久之至道,不刊之鸿教”。

中华经典承载了古圣先贤的志向、智慧与才情,是中华优秀传统文化之渊薮。

而经典的产生有其特定的历史文化语境,亦有其超越时空的传世性和普适性。

诞生于齐梁之际的《文心雕龙》是中国文论元典,中国文章学巨著,中华文化宝典。

这条精雕细刻的“文龙”距今已一千五百多年,依然优美耐看,“灵动多姿”。

究其原因,主要是由于“古典诚然是过去的东西,但是我们的兴趣和研究是现代的,不但承认过去东西的存在并且认识到过去东西里的现实意义。

”(钱钟书语)《文心雕龙》为新文论建设树立“经典范式”。

海通以来,“西学东渐”。

传统的“诗文评”被现代学科意义上的“文学理论”所替代,范畴、术语、命题以及表述方式都发生了质的转换。

这种转换更新了研究视角与研究方法,催生了“文学理论”学科的独立,具有正面意义。

但伴随而来的是“以西律中”的“强制阐释”,文学与文论的民族特点被遮蔽,以至于某些研究者对中国文论产生了隔膜,一味地“竞新逐奇”,自觉或不自觉地切割与中国传统文论的联系。

尽管通行的文学理论教材也吸纳了“意境”等个别中国文论范畴,并引述“诗文评”的只言片语;其实不过是给西式文论做注脚,“虽轩翥出辙,而终入笼内”。

建设新文论,固然要“别求新声于异邦”,望今以制奇;亦须“资于故实”,参古以定法。

而《文心雕龙》为新文论建设树立了“经典范式”。

《文心雕龙》由“文之枢纽”“论文叙笔”“剖情析采”和《序志》等四个部分组成。

其中“文之枢纽”本乎道,师乎圣,体乎经,酌乎纬,变乎骚,这五篇可视为“文原论”;“论文叙笔”自《明诗》至《书记》,先“文”后“笔”,这二十篇可视为“文体论”;“剖情析采”从《神思》至《程器》,先“情”后“采”,这二十四篇可视为“文术论”。

2024吉林省长春市五校联考高三数学试卷(含答案)

2024吉林省长春市五校联考高三数学试卷(含答案)

2024届高三联合模拟考试数学试题东北师大附中 长春十一高中 吉林一中 四平一中 松原实验中学注意事项:1.答卷前,考生务必将自已的考生号、姓名、考场号填写在答题卡上,2.回答选择时,选出每小题答案后,用铅笔把答题卡上对应题目的答案标号涂黑.如需要改动,用橡皮擦干净后,再选涂其他答案标号.回答非选择题时,将答案写在答题卡上,写在本试卷上无效.一、选择题:本题共8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.已知集合(){}{}22log 2,2x A xy x B y y −==−==∣∣,则A B ⋂=( )A.()0,2B.[]0,2C.()0,∞+D.(],2∞− 2.已知复数iz 1i=−,则z 的虚部为( ) A.12−B.1i 2− C.12 D.1i 2 3.将一枚质地均匀的骰子连续抛掷6次,得到的点数分别为1,2,4,5,6,x ,则这6个点数的中位数为4的概率为( ) A.16 B.13 C.12 D.234.刍薨是《九章算术》中出现的一种几何体,如图所示,其底面ABCD 为矩形,顶棱PQ 和底面平行,书中描述了刍薨的体积计算方法:求积术曰,倍下袤,上袤从之,以广乘之,又以高乘之,六而一,即()126V AB PQ BC h =+⋅(其中h 是刍薨的高,即顶棱PQ 到底面ABCD 的距离),已知28,AB BC PAD ==和QBC 均为等边三角形,若二面角P AD B −−和Q BC A −−的大小均为120︒,则该刍薨的体积为( )A.303B.203 9932D.4843+ 5.中国空间站的主体结构包括天和核心舱、问天实验舱和梦天实验舱.假设中国空间站要安排甲,乙,丙,丁4名航天员开展实验,其中天和核心舱安排2人,问天实验舱与梦天实验舱各安排1人.若甲、乙两人不能同时在一个舱内做实验,则不同的安排方案共有( )种 A.8 B.10 C.16 D.20 6.已知π3cos sin 6αα⎛⎫−+= ⎪⎝⎭,则5πsin 6α⎛⎫− ⎪⎝⎭的值是( ) A.3 B.14− C.14 37.已知点F 为地物线2:4C y x =的焦点,过F 的直线l 与C 交于,A B 两点,则2AF BF +的最小值为( )A.22B.4C.322+D.6 8.已的1113sin ,cos ,ln 3332a b c ===,则( ) A.c a b << B.c b a << C.b c a << D.b a c <<二、多选题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求.全部选对的得6分,部分选对的得部分分,有选错的得0分.9.已知数列{}n a 满足*1121,,N 1n n a na n a n +==∈+,则下列结论成立的有( ) A.42a =B.数列{}n na 是等比数列C.数列{}n a 为递增数列D.数列{}6n a −的前n 项和n S 的最小值为6S10.已知正方体1111ABCD A B C D −的棱长为2,M 为空间中动点,N 为CD 中点,则下列结论中正确的是( )A.若M 为线段AN 上的动点,则1D M 与11B C 所成为的范围为ππ,62⎡⎤⎢⎥⎣⎦B.若M 为侧面11ADD A 上的动点,且满足MN ∥平面1AD C ,则点M 2C.若M 为侧面11DCC D 上的动点,且2213MB =,则点M 的轨迹的长度为23π9D.若M 为侧面11ADD A 上的动点,则存在点M 满足23MB MN +=11.已知()()()()1ln ,e 1xf x x xg x x =+=+(其中e 2.71828=为自然对数的底数),则下列结论正确的是( )A.()f x '为函数()f x 的导函数,则方程()()2560f x f x ⎡⎤−'+=⎣⎦'有3个不等的实数解 B.()()()0,,x f x g x ∞∃∈+=C.若对任意0x >,不等式()()2ln ex g a x g x x −+≤−恒成立,则实数a 的最大值为-1D.若()()12(0)f x g x t t ==>,则()21ln 21t x x +的最大值为1e三、填空题:本题共3小题,每小题5分,共15分.12.622x x ⎛⎫− ⎪⎝⎭展开式的常数项为__________.13.已知向量a ,b 为单位向量,且12a b ⋅=−,向量c 与3a b +共线,则||b c +的最小值为__________. 14.已知双曲线2222:1(0,0)x y C a b a b−=>>的左,右焦点分别为12,,F F P 为C 右支上一点,21122π,3PF F PF F ∠=的内切圆圆心为M ,直线PM 交x 轴于点,3N PM MN =,则双曲线的离心率为__________.四、解答题:本题共5小题,共77分,解答应写出文字说明、证明过程或演算步骤.15.(本小题13分)为了更好地推广冰雪体育运动项目,某中学要求每位同学必须在高中三年的每个冬季学期选修滑冰、滑雪、冰壶三类体育课程之一,且不可连续选修同一类课程若某生在选修滑冰后,下一次选修滑雪的概率为13:在选修滑雪后,下一次选修冰壶的概率为34,在选修冰壶后,下一次选修滑冰的概率为25. (1)若某生在高一冬季学期选修了滑雪,求他在高三冬季学期选修滑冰的概率:(2)苦某生在高一冬季学期选修了滑冰,设该生在高中三个冬季学期中选修滑冰课程的次数为随机变量X ,求X 的分布列及期望, 16.(本小题15分)在ABC 中,角,,A B C 的对边分别为,,a b c ,已知1,cos cos 2cos 0a C c A b B =+−=. (1)求B ;(2)若2AC CD =,且3BD =c . 17.(本小题15分)如图,在四棱锥P ABCD −中,底面是边长为2的正方形,且6PB BC =,点,O Q 分别为棱,CD PB 的中点,且DQ ⊥平面PBC .(1)证明:OQ ∥平面PAD ; (2)求二面角P AD Q −−的大小. 18.(本小题17分)已知椭圆2222:1(0)x y C a b a b +=>>的两焦点()()121,0,1,0F F −,且椭圆C 过33,P ⎛ ⎝⎭. (1)求椭圆C 的标准方程;(2)设椭圆C 的左、右顶点分别为,A B ,直线l 交椭圆C 于,M N 两点(,M N 与,A B 均不重合),记直线AM 的斜率为1k ,直线BN 的斜率为2k ,且1220k k −=,设AMN ,BMN 的面积分别为12,S S ,求12S S −的取值范围18.(本小题17分) 已知()2e2e xx f x a x =−(其中e 2.71828=为自然对数的底数).(1)当0a =时,求曲线()y f x =在点()()1,1f 处的切线方程, (2)当12a =时,判断()f x 是否存在极值,并说明理由; (3)()1R,0x f x a∀∈+≤,求实数a 的取值范围.五校联合考试数学答案一、单选题1-8ACADB BCD二、多选题9.ABD 10.BC 11.AC三、填空题12.60 13.211414.75四、解答题15.解:(1)若高一选修滑雪,设高三冬季学期选修滑冰为随机事件A , 则()3234510P A =⨯=. (2)随机变量X 的可能取值为1,2.()()323113221171,2.534320534320P X P X ==⨯+⨯===⨯+⨯=所以X 的分布列为:X 1 2P1320 720()137272.202020E X =+⨯= 16.解:(1)1,cos cos 2cos cos cos 2cos 0a C c A b B a C c A b B =∴+−=+−=.()sin cos sin cos 2sin cos sin 2sin cos 0.A C C A B B A C B B ∴+−=+−=又()1ππ,sin sin 0,cos 23A B C A C B B B ++=∴+=≠∴=∴=.(2)2AC CD =,设CD x =,则2AC x =,在ABC 中2222141cos ,1422c x B c x c c +−==∴+−=.在ABC 与BCD 中,22222142cos ,cos ,63042x c x BCA BCD x c x x∠∠+−−==∴−−=.2321321330,0c c c c c ±+∴−−=∴=>∴=. 17.解:(1)取PA 中点G ,连接,GQ GD ∴点Q 为PB 中点,GQ ∴∥1,2AB GQ AB =. 底面是边长为2的正方形,O 为CD 中点,DO ∴∥1,2AB DO AB =. GQ ∴∥,OD GQ OD =∴四边形GQOD 是平行四边形.OQ ∴∥DG . OQ ⊄平面,PAD GD ⊂平面,PAD OQ ∴∥平面PAD .(2)DQ ⊥平面,PBC BC ⊂平面PBC DQ BC ∴⊥.又底面是边长为2的正方形,,,DC BC DQ DC D BC ∴⊥⋂=∴⊥平面DCQ .OQ ⊂平面,DCQ BC OQ ∴⊥.又CQ ⊂平面,DCQ BC CQ ∴⊥. 26,6,2,2PB QB BC QC =∴==∴=底面是边长为2的正方形,22,2DB DQ DQ CQ ∴=∴==,O 为CD 中点,OQ DC ∴⊥.又,,BC OQ DC BC C OQ ⊥⋂=∴⊥平面ABCD .取AB 中点E ,以,,OE OC OQ 所在直线分别为,,x y z 轴建立如图所示的空间直角坐标系O xyz −, 则()()()()()()0,0,0,0,0,1,2,1,0,2,1,0,0,1,0,2,1,2O Q A B D P −−−−所以()()()4,0,2,2,0,0,2,1,1AP AD AQ =−=−=−, 设平面PAD 法向量为(),,m x y z =,则()4200,1,020m AP x z m m AD x ⎧⋅=−+=⎪∴=⎨⋅=−=⎪⎩ 设平面QAD 法向量为(),,n x y z =,则()200,1,120n AQ x y z n n AD x ⎧⋅=−++=⎪∴=−⎨⋅=−=⎪⎩ 2cos ,2m n m n m n⋅>==⋅ 又二面角P AD Q −−范围为()0,π,所以二面角P AD Q −−的大小为π4. 18.解:(1)由题意可得:2222213314c a b c ab ⎧⎪=⎪−=⎨⎪⎪+=⎩,解得2,31a b c =⎧⎪=⎨⎪=⎩22143x y +=;(2)依题意,()()2,0,2,0A B −,设()()1122,,,M x y N x y ,直线BM 斜率为BM k .若直线MN 的斜率为0,则点,M N 关于y 轴对称,必有120k k +=,不合题意.所以直线MN 的斜率必不为0,设其方程为()2x ty m m =+≠±,与椭圆C 的方程联立223412,,x y x ty m ⎧+=⎨=+⎩得()2223463120t y tmy m +++−=,所以()22Δ48340t m=+−>,且12221226,34312.34tm y y t m y y t ⎧+=−⎪⎪+⎨−⎪=⎪+⎩因为()11,M x y 是椭圆上一点,满足 2211143x y +=,所以2121111221111314322444BM x y y y k k x x x x ⎛⎫− ⎪⎝⎭⋅=⋅===−+−−−, 则12324BM k k k =−=,即238BM k k −⋅=.因为()()1221222BM y y k k x x ⋅=−−()()()()121222121212222(2)y y y y ty m ty m t y y t m y y m ==+−+−+−++−()()()()()22222222223123432334,4(2)42831262(2)3434m m m t m m t m t m m m t t −−++====−−−−−−+−++ 所以23m =−,此时22432Δ4834483099t t ⎛⎫⎛⎫=+−=+> ⎪ ⎪⎝⎭⎝⎭,故直线MN 恒过x 轴上一定点2,03D ⎛⎫−⎪⎝⎭. 因此()12222122264,343431232.34334tm t y y t t m y y t t ⎧+=−=⎪++⎪⎨−⎪==−++⎪⎩,所以12S S −=12121212222323y y y y ⎛⎫⎛⎫−−−−−−−− ⎪ ⎪⎝⎭⎝⎭.()()()22212121222833243342283399433334t t y y y y y y t ++−=−=+−==+()2228314334934t t =−++令2122118340,,34439x S S x x t ⎛⎤=∈−=−+ ⎥+⎝⎦ 当211344t =+即0t =时,12S S −86212834860,399S S x x ⎛∴−=−+ ⎝⎦19.解:(1)当0a =时,()()()2,21x x f x xe f x x e =−=+'−.()14.f e =−∴'曲线()y f x =在点()()1,1f 处的切线方程为 ()41242.y e x e ex e =−−−=−+(2)当12a =时,()2122x xf x e xe =−,定义域为(),∞∞−+ ()()()22122,x x x x f x e x e e e x '=−+=−−令()e 22xF x x =−−,则()2xF x e '=−,当()(),ln2,0x F x ∞∈−'<;当()()ln2,,0x F x ∞∈+'>; 所以()F x 在(),ln2∞−递减,在()ln2,∞+上递增,()min ()ln222ln222ln20F x F ==−−=−< ()()2110,260F F e e−=>=−> 存在()11,ln2x ∈−使得()10F x =,存在()2ln2,2x ∈使得()20F x =,()1,x x ∞∈−时,()()()0,0,F x f x f x >'>单调递增; ()12,x x x ∈时,()()()0,0,F x f x f x <'<单调递减; ()1,x x ∞∈+时,()()()0,0,F x f x f x >'>单调递增;所以12a =时,()f x 有一个极大值,一个极小值. (3)()()()222121xx x x f x ae x e e ae x '=−+=−−,由()()21111,0,00a x f x f a aa a a+∀∈+≤+=+=≤R ,得0a <,令()e 1xg x a x =−−,则()g x 在R 上递减,0x <时,()()()e 0,1,e ,0,e 11x x xa a g x a x a x ∈∈∴=−−>−−,则()()1110g a a a ∴−>−−−=又()110g ae −−=<,()01,1x a ∃∈−−使得()00g x =,即()000e 10x g x a x =−−=且当()0,x x ∞∈−时,()0g x >即()0f x '>; 当()00,x x ∞∈+时,()0g x <即()0f x '<,()f x ∴在()0,x ∞−递增,在()0,x ∞+递减,()002max 00()2x x f x f x ae x e ∴==−,由()000001e 10,exx x g x a x a +=−−==, 由max 1()0f x a+≤得()000000e 1e 201x x x x x e x +−+≤+即()()00011101x x x −++≤+, 由010x +<得20011,21x x −≤∴−<−,001,e x x a +=∴设()1(21)e x x h x x +=−≤<−,则()0xxh x e −=>', 可知()h x 在)2,1⎡−⎣上递增,()((()()221221210h x h e h x h e −−≥−==<−=实数a 的取值范围是()212e ⎡⎣.。

江苏省盐都区2024年秋学期九年级五校期中联考2024-2025学年九年级上学期11月期中物理试题

江苏省盐都区2024年秋学期九年级五校期中联考2024-2025学年九年级上学期11月期中物理试题

2024-2025学年秋学期九年级物理期中考试试卷一、选择题(每小题2分,共26分)1.如图所示的杠杆中,使用时利用了其省距离特点的是( )2.丹丹同学从地上拿起一个鸡蛋,并用大约2s 的时间将它缓缓举过头顶,在这2s 时间内,丹丹做功的功率约为( )A .0.25WB .0.5WC .1WD .2W3.如图所示生活实例中,力对物体做功的有( )A .甲和乙B .甲和丙C .乙和丙D .丙和丁4.关于比热容,下列说法中错误的是( )A .比热容可用来鉴别物质B .水的比热容较大,可用作汽车发动机的冷却剂C .沙的比热容较小,所以沙漠地区昼夜温差较大D .一桶水的比热容比一杯水的比热容大5.在如图所示电路中,下列说法正确的是( )A .只闭合开关S 1、S 4时,只有L 1、L 3亮,且两灯并联B .只闭合开关S 3、S 4时,只有L 2、L 3亮,且两灯串联C .只闭合开关S 1、S 2、S 3时,三个灯均亮D .只闭合开关S 2、S 3、S 4时,三个灯均亮6.某汽车集团研发了一款汽车。

该汽车发动机工作时,效率更高,动力更强劲。

如图为其发动机某一冲程的示意图,下列说法正确的是( )A .该款汽车发动机使用的汽油的热值更大B .如图所示为压缩冲程,内能转化为机械能,需要靠飞轮的惯性来完成C .该款汽油机的吸气冲程,只用吸入汽油D .若该汽油机飞轮的转速为1 200/min ,则在1 s 内汽油机对外做了10次功7.用四只完全相同的滑轮和两根相同绳子组成如图所示的甲、乙两个滑轮组,现用它们来提升相同的重物,不计绳重及摩擦,则( )A .甲较省力且机械效率较高 B.乙较省力且机械效率较高甲:小车在推力的作用乙:提着滑板在 丙:物体在绳子 丁:用尽全力搬下前进了一段距离 水平路面上前行 拉力作用下升高 石头,搬而未起D .食品夹A .开瓶器B .扳手C.核桃钳C .两个滑轮组省力程度不同,机械效率相同D .两个滑轮组省力程度相同,机械效率不同8.根据你对生活中物理量的认识,下列数据最符合实际情况的是( )A .重庆南滨路的大气压约为B .中学生跳绳一分钟的功率约为6000WC .普通中学生浮在水中时的浮力约为50ND .冰箱保鲜室中矿泉水的温度约为9.关于温度、比热容、热量、内能,以下说法正确的是( )A .一个物体吸收了热量,它的温度一定会升高B .寒冬房檐下的“冰喇叭”,它既具有内能,也具有机械能C .一个物体温度升高了,一定是吸收了热量D .比热容大的物质升温时吸收的热量一定较多10.如图所示,在光滑的水平台面上,一轻质弹簧左端固定不动,右端连接一金属小球,O 点是弹簧保持原长时小球的位置。

2024年浙江省五校(杭二、金一、绍一、衢二、温中)联盟高三3月联考试卷及答案

2024年浙江省五校(杭二、金一、绍一、衢二、温中)联盟高三3月联考试卷及答案

2024年浙江省五校联盟高三3月联考数学试题卷命题:浙江省杭州第二中学一、选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项符合题目要求.1.若全集U ,集合,A B 及其关系如图所示,则图中阴影部分表示的集合是()A.()U A B ðB.()U A B ðC.()U BA ð D.()U A B ð2.已知(1,2)a =r,2b =r ,且a b ⊥r r ,则a b -r r 与a 的夹角的余弦值为()A.B.C.D.3.设,b c 表示两条直线,,αβ表示两个平面,则下列说法中正确的是()A.若,b c αα⊂∥,则b c ∥B.若,b c b α⊂∥,则c α∥C.若,c αβα⊥∥,则c β⊥ D.若,c c αβ⊥∥,则αβ⊥4.已知角α的终边过点(3,2cos )P α-,则cos α=()A.2B.2-C.2± D.12-5.设等比数列{}n a 的公比为q ,前n 项和为n S ,则“2q =”是“{}1n S a +为等比数列”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.已知实数,x y 满足3x >,且2312xy x y +-=,则x y +的最小值为()A.1+ B.8C. D.1+7.已知双曲线2222:1(0,0)x y C a b a b-=>>的左、右焦点分别为12,F F ,点A 为双曲线的左顶点,以12F F 为直径的圆交双曲线的一条渐近线于,P Q 两点,且23PAQ π∠=,则该双曲线的离心率为()A.B.C.213D.8.在等边三角形ABC 的三边上各取一点,,D E F ,满足3,90DE DF DEF ==∠=︒,则三角形ABC 的面积的最大值是()A. B. C.D.二、选择题:本题共3小题,每小题6分,共18分.在每小题给出的选项中,有多项符合题目要求.全部选对的得6分,部分选对的得部分分,有选错的得0分.9.在学校组织的《青春如火,初心如炬》主题演讲比赛中,有8位评委对每位选手进行评分(评分互不相同),将选手的得分去掉一个最低评分和一个最高评分,则下列说法中正确的是()A.剩下评分的平均值变大B.剩下评分的极差变小C.剩下评分的方差变小D.剩下评分的中位数变大10.在三棱锥A BCD -中,已知3,2AB AC BD CD AD BC ======,点,M N 分别是,AD BC 的中点,则()A.MN AD⊥B.异面直线,AN CM 所成的角的余弦值是78C.三棱锥A BCD -的体积为3D.三棱锥A BCD -的外接球的表面积为11π11.已知函数()(sin cos )x f x e x x =⋅+,(浦江高中数学)则()A.()f x 的零点为,4x k k Z ππ=-∈B.()f x 的单调递增区间为32,2,22k k k Z ππππ⎡⎤++∈⎢⎥⎣⎦C.当0,2x π⎡⎤∈⎢⎥⎣⎦时,若()f x kx ≥恒成立,则22k e ππ≤⋅D.当10031005,22x ππ⎡⎤∈-⎢⎥⎣⎦时,过点1,02π-⎛⎫⎪⎝⎭作()f x 的图象的所有切线,则所有切点的横坐标之和为502π三、填空题:本题共3小题,每小题5分,共15分.12.直线3430x y -+=的一个方向向量是________.13.甲、乙两人争夺一场羽毛球比赛的冠军,比赛为“三局两胜”制.如果每局比赛中甲获胜的概率为23,乙获胜的概率为13,则在甲获得冠军的情况下,比赛进行了三局的概率为________.14.已知函数()f x 及其导函数()f x '的定义域均为R ,记()()g x f x =',若(21),(2)f x g x --均为偶函数,且当[1,2]x ∈时,3()2f x mx x =-,则(2024)g =________.四、解答题:本题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤.15.(本小题满分13分)如图,斜三棱柱111ABC A B C -的底面是直角三角形,90ACB ∠=︒,点1B 在底面ABC 内的射影恰好是BC 的中点,且2BC CA ==.(1)求证:平面11ACC A ⊥平面11B C CB ;(2,求平面1ABB 与平面11AB C 夹角的余弦值.16.(本小题满分15分)已知函数()ln f x x ax =-,其中a R ∈.(1)若曲线()y f x =在1x =处的切线在两坐标轴上的截距相等,求a 的值;(2)是否存在实数a ,使得()f x 在(0,]x e ∈上的最大值是3-?若存在,求出a 的值;若不存在,说明理由.17.(本小题满分15分)记复数的一个构造:从数集中随机取出2个不同的数作为复数的实部和虚部.重复n 次这样的构造,可得到n 个复数,将它们的乘积记为n z .已知复数具有运算性质:()()()()a bi c di a bi c di +⋅+=+⋅+,其中,,,a b c d R ∈.(1)当2n =时,记2z 的取值为X ,求X 的分布列;(2)当3n =时,求满足32z ≤的概率;(3)求5n z <的概率n P .18.(本小题满分17分)在平面直角坐标系xOy 中,我们把点*(,),,x y x y N ∈称为自然点.按如图所示的规则,将每个自然点(,)x y 进行赋值记为(,)P x y ,例如(2,3)8P =,(4,2)14,(2,5)17P P ==.(1)求(,1)P x ;(2)求证:2(,)(1,)(,1)P x y P x y P x y =-++;(3)如果(,)P x y 满足方程(1,1)(,1)(1,)(1,1)2024P x y P x y P x y P x y +-+++++++=,求(,)P x y 的值.19.(本小题满分17分)在平面直角坐标系xOy 中,过点(1,0)F 的直线l 与抛物线2:4C y x =交于,M N 两点(M 在第一象限).(1)当||3||MF NF =时,求直线l 的方程;(2)若三角形OMN 的外接圆与曲线C 交于点D (浦江高中数学)(异于点,,O M N ),(i )证明:MND ∆的重心的纵坐标为定值,并求出此定值;(ii )求凸四边形OMDN 的面积的取值范围.参考答案一、选择题:本题共8小题,每小题5分,共40分. 在每小题给出的四个选项中,只有一项是符合要求的.题号 1 2 3 4 5 6 7 8 答案CBDBCACA选对的得6分,部分选对的得部分分,有选错的得0分.题号 9 10 11 答案BCABDACD12. 3(1,)4 (答案不唯一) 13.2514. 6− 四、解答题:本大题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤. 15.(本小题满分13分)(第Ⅰ问,6分;第Ⅱ问,7分)解:(Ⅰ)取BC 中点为M ,连接1B M ,∵1B 在底面内的射影恰好是BC 中点, ∴1B M ⊥平面ABC ,又∵AC ⊂平面ABC ,∴1B M AC ⊥, 又∵90ACB ∠=,∴AC BC ⊥, ∵1,B M BC ⊂平面11B C CB ,1B MBC M =,∴AC ⊥平面11B C CB ,又∵AC ⊂平面11ACC A ,∴平面11ACC A ⊥平面11B C CB .(Ⅱ)以C 为坐标原点,建立如图所示空间直角坐标系,∵2BC CA ==, ∴11(2,0,0),(0,2,0),(0,1,0),(0,1,3),(0,1,3),A B M B C − 111(2,1,3),(2,2,0),(0,2,0)AB AB B C =−=−=−,设平面1BAB 的法向量为(,,)n x y z =,∴100n AB n AB ⎧⋅=⎪⎨⋅=⎪⎩则有230220x y z x y ⎧−++=⎪⎨−+=⎪⎩,令3,z =则3x y ==,∴(3,3,3)n =,设平面1BAB 的法向量为(,,)m a b c =,∴1110m AB m B C ⎧⋅=⎪⎨⋅=⎪⎩则有23020a b c b ⎧−++=⎪⎨−=⎪⎩,令3a =则0,2b c ==,∴(3,0,2)n =,∴||535|cos ,|||||7993304n m n m n m ⋅<>===++⨯++,平面1ABB 与平面11AB C 夹角的余弦值为57.16.(本小题满分15分)(第Ⅰ问,6分;第Ⅱ问,9分)∴f (x )的最大值是f (e)=1-a e =-3,解得a =4e >0,舍去;②当a >0时,由f ′(x )=1x -a =1-ax x =0,得x =1a,当0<1a <e ,即a >1e 时,∴x ∈⎝⎛⎭⎫0,1a 时,f ′(x )>0;x ∈⎝⎛⎭⎫1a ,e 时,f ′(x )<0, ∴f (x )的单调递增区间是⎝⎛⎭⎫0,1a ,单调递减区间是⎝⎛⎭⎫1a ,e , 又f(x )在(0,e]上的最大值为-3,∴f (x )max =f ⎝⎛⎭⎫1a =-1-ln a =-3,∴a =e 2; 当e≤1a ,即0<a ≤1e 时,f (x )在(0,e]上单调递增,∴f (x )max =f (e)=1-a e =-3,解得a =4e >1e,舍去.综上,存在a 符合题意,此时a =e 217.(本小题满分15分) (第Ⅰ问,6分;第Ⅱ问,4分;第Ⅲ问,5分) (Ⅰ)由题意可知,可构成的复数为{}11i +, 且1112i i ====+=+=.X 的可能取值为1234,,,()11221166119C C P X C C ⋅===⋅,(1142116629C C P X C C ⋅===⋅,()11421166229C C P X C C ⋅===⋅,()11221166139C C P X C C ⋅===⋅,(1142116629C C P X C C ⋅===⋅,()11221166149C C P X C C ⋅===⋅,所以分布列为:(Ⅱ)共有666216C C C ⋅⋅=种, 满足32z ≤的情况有:①3个复数的模长均为1,共有1112228C C C ⋅⋅=种;②3个复数中,2个模长均为1,12,共有2111322448C C C C ⋅⋅⋅=种; 所以()38487221627P z +≤==. (Ⅲ)当1n =或2时,显然都满足,此时1n P =; 当3n ≥时,满足5n z <共有三种情况: ①n 个复数的模长均为1,则共有()122nn C =;②1n −个复数的模长为1,剩余12,则共有()11111242n n n n C C C n −−+⋅⋅=⋅;③2n −个复数的模长为1,剩余2或者2,则共有()()22111124412n n n n C C C C n n −−+⋅⋅⋅=−⋅.故()()()()211216212*********n n n n n nnnn n n n n P z C ++++⋅+−⋅+<===,此时当12n ,=均成立.所以()21253n nn P z +<=.18. (本小题满分17分)(第Ⅰ问,4分;第Ⅱ问,7分;第Ⅲ问,6分) 解:(Ⅰ)根据图形可知()()1,11232x x P x x +=++++=, (Ⅱ)固定x ,则(),P x y 为一个高阶等差数列,且满足()(),1,1P x y P x y x y +−=+−,()()1,,P x y P x y x y +−=+,所以()()()()()1,1,112112y y P x y P x y y x y x ++−=++++−=+−,()()()()11,1122y y x x P x y y x +++=+−+,所以()()()()()11,1122x x y y P x y x y +−=++−−,()()()()()111,2122x x y y P x y x y −−−=++−−,所以()()()()()()()()()()221111,11,21122222322,x x y y y y x x P x y P x y x y y x x y xy y x P x y −−++++−=++−−++−+=++−−+=(Ⅲ)()()()()1,1,11,1,12024P x y P x y P x y P x y +−+++++++=,等价于()()()(),,11,1,12023P x y P x y P x y P x y +++++++=,等价于()(),131,2023P x y P x y +++=,即()()()()()()131211212202322x x y y x x x y y x +++−++++−+=⎡⎤⎡⎤⎣⎦⎣⎦,化简得()()2221010121010y xy x y x x y x y x ++−+=⇔+−++=,由于x y +增大,()()1x y x y +−+也增大,当31x y +=时,()()129921010x y x y x +−++<<,当33x y +=时,()()1210561010x y x y x +−++>>,故当32x y +=时,()()1210109,23x y x y x x y +−++=⇒==, 即()91023229,2382247422P ⨯⨯=++⨯=.19. (本小题满分17分)(第Ⅰ问,4分;第Ⅱ问,5分;第Ⅲ问,8分) 解:(Ⅰ)设直线MN :1x my =+,1122(,),(,)M x y N x y联立241x xy y m =+=⎧⎨⎩,消去x ,得2440y my −−=,所以12124,4y y m y y +=⋅=−,3MF NF =,则123y y =−∴122212224,34y y y m y y y +=−=⋅=−=−,则213m=,又由题意0,m >∴3m =,直线的方程是y =(Ⅱ)(ⅰ)方法1:设112233(,),(,),(,)M x y N x y D x y因为,,,O M D N 四点共圆,设该圆的方程为220x y dx ey +++=,联立22204x y dx ey y x⎧+++=⎨=⎩,消去x ,得()42416160y d y ey +++=,即()()3416160y y d y e +++=,所以123,,y y y 即为关于y 的方程()3416160y d y e +++=的3个根,则()()()()312341616y d y e y y y y y y +++=−−−,因为()()()()()32123123122313123y y y y y y y y y y y y y y y y y y y y y −−−=−+++++−,由2y 的系数对应相等得,1230y y y ++=,所以MND ∆的重心的纵坐标为0.方法2:设112233(,),(,),(,)M x y N x y D x y ,则1213234444,,,OM ON MD ND k k k k y y y y y y ====++, 因为,,,O M C N 四点共圆,所以MON MDN π∠+∠=,即tan tan 0MON MDN ∠+∠=,21124()tan 116OM ON OM ON k k y y MON k k y y −−∠==+⋅+,1213234()tan 1()()16ND MD ND MD k k y y MDN k k y y y y −−∠==+⋅+++,化简可得:312y y y =−−, 所以MND ∆的重心的纵坐标为0.(ⅱ)记,OMN MND △△的面积分别为12,S S ,由已知得直线MN 的斜率不为0 设直线MN :1x my =+,联立241x xy y m =+=⎧⎨⎩,消去x ,得2440ymy −−=,所以12124,4y y m y y +=⋅=−,所以1121122S OF y y =⋅⋅−==, 由(i )得,()3124y y y m =−+=−, 所以()22233114444x y m m ==⨯−=,即()24,4D m m −, 因为()212122444MN x x m y y m =++=++=+,点D 到直线MN的距离d =,所以()22211448122S MN d m m =⋅⋅=⋅+=−,所以)221281181S S S m m =+=+−=+− M 在第一象限,即120,0y y ><,340y m =−<,依次连接O ,M ,D ,N 构成凸四边形OMDN ,所以()3122y y y y =−+< ,即122y y −<,又因为124y y ⋅=−,2242y y <,即222y <,即20y <<,所以122244m y y y y =+=−>=,即4m >,即218m >,所以)218116S m m =+−=设t =4t >, 令()()2161f t t t =−,则()()()2221611614816f t t t t t '='=−+−−,因为4t >,所以()248160f t t −'=>,所以()f t在区间,4∞⎛⎫+ ⎪ ⎪⎝⎭上单调递增, 所以()42f t f ⎛⎫>= ⎪⎪⎝⎭, 所以S的取值范围为,2∞⎛⎫+ ⎪ ⎪⎝⎭.。

湖北省鄂东南五校一体联盟联考2025届高考数学一模试卷含解析

湖北省鄂东南五校一体联盟联考2025届高考数学一模试卷含解析

湖北省鄂东南五校一体联盟联考2025届高考数学一模试卷注意事项1.考生要认真填写考场号和座位序号。

2.试题所有答案必须填涂或书写在答题卡上,在试卷上作答无效。

第一部分必须用2B 铅笔作答;第二部分必须用黑色字迹的签字笔作答。

3.考试结束后,考生须将试卷和答题卡放在桌面上,待监考员收回。

一、选择题:本题共12小题,每小题5分,共60分。

在每小题给出的四个选项中,只有一项是符合题目要求的。

1.已知单位向量a ,b 的夹角为34π,若向量2m a =,4n a b λ=-,且m n ⊥,则n =( ) A .2B .2C .4D .62.已知函数()f x 的导函数为()f x ',记()()1f x f x '=,()()21f x f x '=,…,()()1n n f x f x +'=(n ∈N *). 若()sin f x x x =,则()()20192021f x f x += ( )A .2cos x -B .2sin x -C .2cos xD .2sin x3.已知0,2πα⎛⎫∈ ⎪⎝⎭,0,2πβ⎛⎫∈ ⎪⎝⎭,cos2tan 1sin 2βαβ=-,则( )A .22παβ+=B .4παβ+=C .4αβ-=π D .22παβ+=4.已知n S 是等差数列{}n a 的前n 项和,若312S a S +=,46a =,则5S =( )A .5B .10C .15D .205.已知抛物线24x y =上一点A 的纵坐标为4,则点A 到抛物线焦点的距离为( ) A .2B .3C .4D .56.执行如图的程序框图,若输出的结果2y =,则输入的x 值为( )A .3B .2-C .3或3-D .3或2-7.执行如图所示的程序框图,若输入ln10a =,lg b e =,则输出的值为( )A .0B .1C .2lg eD .2lg108.已知直线l 320x y ++=与圆O :224x y +=交于A ,B 两点,与l 平行的直线1l 与圆O 交于M ,N 两点,且OAB 与OMN 的面积相等,给出下列直线1l 330x y +-=320x y +-=,③320x -+=,④3230x y ++=.其中满足条件的所有直线1l 的编号有( ) A .①②B .①④C .②③D .①②④9.已知函数f (x )=sin 2x +sin 2(x 3π+),则f (x )的最小值为( ) A .12B .14C .34D .2210.已知点()25,310A 在双曲线()2221010x y b b-=>上,则该双曲线的离心率为( )A .103B .102C .10D .21011.阅读下侧程序框图,为使输出的数据为,则①处应填的数字为A .B .C .D .12.如图,圆O 是边长为23的等边三角形ABC 的内切圆,其与BC 边相切于点D ,点M 为圆上任意一点,BM xBA yBD =+(,)x y ∈R ,则2x y +的最大值为( )A 2B 3C .2D .22二、填空题:本题共4小题,每小题5分,共20分。

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2014学年第二学期高三英语五校联合质量调研试卷考生注意:1.考试时间120分钟,试卷满分150分。

2.本考试设试卷和答题纸两部分。

试卷分为第Ⅰ卷(第1-9页)和第Ⅱ卷(第10页),全卷共10页。

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第I卷I. Listening ComprehensionPart A Short ConversationsDirections: In Part A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. The newsstand. B. The hair salon. C. The grocery store. D. The bookstore.2. A. A sportsman. B. A doctor. C. A news reporter. D. A game designer.3. A. She didn’t t each class today. B. She didn’t collect the homework.C. She usually talks quietly.D. She usually assigns homework.4. A. Chocolate pudding is his favorite food. B. He has tasted the chocolate pudding.C. There is no more chocolate pudding left.D. He doesn’t want any chocolate pudding.5. A. She disagrees with the man. B. She doesn’t enjoy boring speeches.C. She wonders how long the speech will be.D. She doesn’t think highly of the speaker.6. A. She has just sold all the texts. B. She has kept some science texts.C. She’s no longer interested in science.D. She gave away the latest texts.7. A. She’d like to talk to Larry about the problem.B. Larry should get along well with his brother.C. It’s necessary for Larry to apologize to his brother.D. Larry’s brother may be partly responsible for the problem.8. A. Still live in the dormitory. B. Find out the cost of living in the dormitory.C. Ask for a reduction in her rent.D. Move into an apartment with a roommate.9. A. She has no time to work in a garden. B. She’ll consider the man’s invitation.C. She doesn’t want to join the club.D. She is willing to accept the invitation.10. A. The woman will have to take more than seven courses.B. The woman shouldn’t have attended so many courses.C. His current schedule is also very demanding.D. The woman is likely to graduate early with this schedule.Section BDirections: In Section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. They used to be shorter. B. They used to be made by machine.C. They are lighter now.D. They are made of wood now.12. A. It can stop the board going far when they fall off.B. It helps to keep them warm in cold water.C. It enables their feet to stick to the board.D. It can increase the pleasure of surfing.13. A. A brief introduction to a sport. B. Essential objects for surfing.C. The origin of a professional sportD. How to surf skillfully in waves.Questions 14 through 16 are based on the following passage.14. A. It is produced in small quantities. B. It is sold at a lower price.C. It is served mainly in McDonald’s.D. It is grown from cows alone.15. A. The land and the water system have been polluted seriously.B. Not enough meat has been produced to meet people’s needs.C. Much land has been used up for animals and their food.D. It has consumed fewer and fewer natural resources.16. A. Steaks and hamburgers. B. Animal rights.C. The food crisis in the future.D. Lab-grown meat.Section CDirections: In Section C, you will hear two longer conversations. The conversations will be read twice. After you hear each conversation, you are required to fill in the numbered blanks with the information you have heard. Write your answers on your answer sheet.Blanks 17 through 20 are based on the following conversation.Complete the form. Write ONE WORD for each answer.Blanks 21 through 24 are based on the following conversation.Complete the form. Write NO MORE THAN THREE WORDS for each answer.II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)Lots of people love buying clothes and Gucci is one of the most famous fashion houses in the world. It was started in Italy in 1921 by a man named Guccio Gucci. He was a designer who had a small shop in Florence 25.__________ (specialize) in leather bags and suitcases. The shop was 26.__________ beginning of the family business and by 1953 Guccio’s four sons, Aldo, Ugo, Vasco and Rodolfo were all working for the company.When Guccio died in 1953, his 27.__________ (old) son Aldo became the head of Gucci and took the Gucci label to America, while Rodolfo managed the Italian side of the business. But Aldo’s son, Paolo, made plans to start his own company called Paolo Gucci. When Aldo discovered this, he sacked Paolo and made 28.__________ impossible for his son to start his own fashion business. So angry was Paolo 29.__________ he told the Italian police his father wasn’t paying enough tax. Aldo 30. __________ (send) to prison for a year and a day.After Aldo died in 1990, his nephew, Maurizio, became the head of Gucci. Unfortunately, Maurizio wasn’t a very good businessman and the company lost $60 million in 1991, 31.__________ became the worst year in Gucci’s history.32.__________ (sell) all over the world, Gucci products are still popular with fashion hunters, but there are no members of the Gucci family in the successful company we know today.(B)Having a good speech prepared is only part of being a successful public speaker—you33.__________ also get your message across in a clear and interesting way.●Alter the volume of your voice. It can help to keep the audience alert and stress a point,but 34.__________ (make) sure everyone can hear you comfortably at all times.●Vary your pace. It will add interest to your speech: speaking quickly can make yourwords more exciting, and speaking slowly or pausing can emphasise an idea. But don’tspeak too quickly or too slowly 35. __________ __________ the audience can’tfollow you or will get bored.●Use stress to make spoken words and phrases clearer or give them importance. If youspeak 36.__________ stress, your audience will go to sleep! Too much stress, on theother hand, 37.__________ (sound) unnatural. Just stress the most important words inyour speech.●Speak with clarity. If you mumble (嘟嘟囔囔), your audience won’t understand you.Pronounce the words and phrases in your speech clearly.●Avoid 38.__________ (use) fillers as much as possible, such as “er”, “um” and “youknow”. If you need time to think, just pause.Why not 39.__________ (record) yourself speaking? Listening to the recording is a great way to find out how well you perform and 40.__________ you need to improve on. Remember—practice makes perfect!Section BDirections: Complete the following passage by using the words in the box. Each word can only be used once. Note that there is one word more than you need.Big data could soon be stored in a very small package: DNA. A team of scientists has demonstrated that storing information in synthetic DNA could represent a(n) 41 approach to managing data in the long term, bumping aside the magnetic tape 42 by archivists (档案管理员) today.The approach, published online January 23 in Nature, relies on 43 likely to become faster and cheaper, says biologist and engineer Drew Endy of Stanford University, who was not44 in the work.Unlike record players, which are good only for playing music encoded on now-out-of-date vinyl discs (塑料唱片), machines that make and read DNA find 45 throughout science and always will. “Human beings are never going to stop caring about DNA,” says Endy. DNA is also tiny, lightweight, and can potentially remain undamaged for thousands of years if 46 in a dark, cool environment.This new report comes on the heels of 47 research published last August in Science. The new research projects that, if the costs of making DNA continue to drop, the approach might be 48 for long-term storage in as little as 10 years. “It’s 49 exciting,” Endy says.In the next decade, the approach could store information that needs to last for at least 50 years, such as government records or library texts. And who knows where it will go, 50 Goldman. Perhaps, he s ays, “when the cloud sucks things off your computer, it will be to store it as DNA.”III. Reading ComprehensionSection ADirections: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.Is love really blind? Yes, it is, at least when it comes to 51 others, US researchers reported. College students who reported they were in love were less likely to 52 other attractive men or women, the team at the University of California Los Angeles and dating Web site eHarmony found.“Feeling love for your 53 partner appears to make everybody else less attractive, and the emotion appears to enable you to push thoughts of that 54 other out of your mind,”said Gian Gonzaga of eHarmony, whose study is published in the journal Evolution and Human Behavior.“It’s almost like 55 puts blinders(眼罩)on people,”added Martie Haselton, an associate professor of psychology and communication studies at UCLA.Gonzaga and Haselton asked 120 undergraduates in committed relationships to 56 photographs of attractive members of the opposite sex from an eHarmony Web site. The 57 were asked to choose the most attractive photos, and write an essay either about their current lover, or the 58 of their choice.Those who wrote about their lovers were six times less likely to 59 that they thought of the attractive others than volunteers who wrote about the people on 60 photos.And later asked to 61 the good-looking people in the pictures, the students who wrote about their lovers remembered fewer details about the physical appearance of the attractive 62 .“These people could remember the color of a shirt or whether the photo was taken in NewYork 63 anything attractive about the person,” Gonzaga said.“It’s not like their overall 64 was hurt; it’s as if they had 65 screened out things that would make them think about how attractive the alternative was.”51. A. talking with B. looking at C. working with D. smiling at52. A. take notice of B. be jealous of C. be ignorant of D. catch up with53. A. loving B. handsome C. romantic D. considerate54. A. thrilling B. exciting C. tempting D. puzzling55. A. relationship B. love C. mood D. attraction56. A. arrange B. examine C. develop D. deliver57. A. members B. writers C. lovers D. volunteers58. A. subject B. reason C. desire D. motivation59. A. delay B. stop C. continue D. admit60. A. recent B. casual C. special D. random61. A. report B. repeat C. receive D. recall62. A. strangers B. partners C. friends D. researchers63. A. as well as B. but for C. in spite of D. other than64. A. structure B. memory C. function D. emotion65. A. traditionally B. physically C. selectively D. equallySection BDirections: 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)Because of the politics and history of Africa, wild animals there, which are interested in finding food and water not in politics, are in trouble. In the past, there were no borders between African countries, and the animals could travel freely according to the season or the weather. However, in the 19th and 20th centuries, the continent was divided up into colonies and then into nations. Fences were put up along the borders, so the animals could no longer move about freely.Some countries decided to protect their animals by creating national parks. Kruger National Park, created in South Africa in 1926, was one of the first. By the end of the twentieth century, it had become an important tourist attraction and a home for many kinds of animals. Among these, there were about 9,000 elephants, too many for the space in the park. It was not possible to let any elephants leave the park, however. They would be killed by hunters, or they might damage property or hurt people. South African park officials began to look for other solutions to the elephant problem.As early as 1990, the governments of South Africa and Mozambique had begun talking about forming a new park together. In 1997, Zimbabwe agreed to add some of its land to the park. A new park would combine the Kruger National Park with parks in Mozambique and Zimbabwe. There would be no national border fences within the park, so that elephants and other animals from the crowded Kruger Park could move to areas of Mozambique and Zimbabwe. This new “transfrontier”park would cover 13,150 square miles (35,000 square kilometers). The idea of a transfrontier park interested several international agencies, which gave money and technical assistance to Mozambique to help build its part of the park.In April 2001, the new park was opened, with new borders and a new name: The Great Limpopo Transfrontier Park. A border gate was opened between Kruger National Park and Mozambique, and seven elephants were allowed through. They were the first of 1,000 elephants that would be transferred to the world’s greatest animal park.66. The passage begins with _____.A. a common senseB. a factC. a mysteries eventD. a theory67. Which of the following was a problem facing Kruger National Park?A. It was not big enough to hold all its elephants.B. A lot of hunters slipped in to hunt animals.C. As the first national park in Africa, it was not well designed.D. Too much tourism did great damage to it.68. Which of the following can be inferred from the passage about the new park?A. It is divided into three parts by fences along borders.B. It is built mainly for elephants rather than other animals.C It is located across the border of South Africa and Mozambique.D. It is the result of a talk between Mozambique and some international agencies.69. The passage talks mainly about _____.A. how international aid has functioned in AfricaB. how the Kruger National Park will save its elephantsC. how three African countries cooperated to make a new parkD. how many African animals have suffered because of natural disasters70. If you _____, you may find this advertisement very useful.A. lack exerciseB. eat unhealthilyC. are overweightD. have an irregular way of life71. The slimming technique advertised is different from others in that ______.A. it offers a final goal instead of a daily oneB. it requires you to eat much less than beforeC. it suggests no change to your lifestyle at allD. it tells exactly the amount of food you can take72. If you are not healthy enough, it’s better to _____ before applying the slimming technique.A. get the doctor’s permissionB. take some medicine offered by doctorsC. make sure of the possible resultD. have a thorough physical examination73. You can’t get the £4.50 back unless you _____.A. have paid the money by chequeB. have made great changes to your weightC. have returned the material before the deadlineD. have finished the course in no more than ten days(C)As students are discussing their favorite colleges, there’s one characteristic they can’t co ntrol: their race. That’s one reason voters, courts and politicians in six states have outlawed racial preferences in college admissions, while other colleges, fearful of lawsuits, play down their affirmative-action efforts these days. But make no mistake: race still matters. How much depends on the school and the state.In Texas, public universities have managed to reduce the effect of racial-preference bans by automatically admitting the top 10% of the graduating class of every high school, including those schools where most students are minorities. But Rice University in Houston, private and highly selective, has had to reinvent its admissions strategies to maintain the school’s minority enrollment. Each February, 80 to 90 black, Hispanic (西班牙裔的) and Native American kids visit Rice on an expenses-paid trip. Rice urges headmasters from high schools with large minority populations to recommend qualified students. And in the fall, Rice sends two recruiters (招生人员) on the road to find minority applicants; each recruiter visits about 80 mainly black or Hispanic high schools. Two weeks ago, Rice recruiter Tamara Siler dropped in on Westlake High in Atlanta, where 99% of the 1,296 students are black. Siler went hearing literature and advice, and thou gh only two kids showed up, she said, “I’m pleased I got two.”Rice has also turned to some almost comical end-runs around the spirit of the law. The university used to award a yearly scholarship to a Mexican-American student; now it goes to a student who speaks Spanish really well. Admissions officers no longer know an applicant’s race. But a new essay question asks about each student’s “background” and“cultural traditions.” When Rice officials read applications, they look for “diverse life experiences” and what they awkwardly call “overcome students,” who have triumphed over hardship.Last spring, admissions readers came across a student whose SAT score was lower than 1,200 and who did not rank in the top 10% of her class. Numerically speaking, she was far behind most accepted applicants. But her essay and recommendations indicated a strong interest in civil rights and personal experience with racial discrimination (歧视). She was admitted. “All the newspapers say affirmative action is done,” says a n experienced adviser at a large New York City high school. “But nothing has changed. I have a (minority) kid at Yale with an SAT score in the high 900s.”74. What does the word “outlaw” (in Para. 1) most probably mean?A. supportB. considerC. banD. hate75. What can we infer from the passage about affirmative action?A. It guarantees students of different races to be admitted equally.B. It discriminates against minority students in college admissions.C. It gives preference to minority students in college admissions.D. It is popular with American colleges but not with the American public.76. Why does Rice University send two recruiters to find minority applicants?A. Rice has a large minority population.B. Rice wants to maintain its minority enrollment.C. Minority students do not favor Rice very much.D. Minority students have better school performance.77. Which of the following might be the best title of the passage?A. Here Comes Equality at LastB. Yes, Your Race Still MattersC. Well Done, Affirmative ActionD. Minorities Are Still a Minority in UniversitiesSection DDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.How can a company improve its sales? One of the keys to more effective selling is for a company to first decide on its “sales strategy”. In other words, what is the role of the sales person? Is the salesperson’s job narrative, suggestive, or consultive?The “narrative” sales strategy depends on the salesperson moving quickly into a standard sales presentation. His or her pitch highlights the benefit for the customer of a particular product or service. This approach is most effective for customers whose buying motives are basically the same and is also well suited to companies who have a large number of prospects on which to call.The “suggestive” approach is tailored more for the individual customer. The salesperson must be in a position to offer alternative recommendations that meet a partic ular customer’s needs. One key aspect of the suggestive approach is the need for the salesperson to engage the buyer in some sort of discussion. The salesperson can then use the information from the customer to suggest an appropriate product or service.“We tell our salespeople to be like wine stewards,” says Mindy Sahlawannee, a corporate sales trainer, “the wine steward first checks to see what food the customer has ordered and then opens by suggesting the wine that best complements the dish. Most companies who use a narrative strategy should be using a suggestive strategy. Just like you can’t drink red wine with every dish, you can’t have one sales recommendation to suit all consumers”The final strategy demands that a company’s sales staff act as “consultants” for the buyer. In this role, the salesperson must acquire a great deal of information about the customer. They do this through market research, surveys, and face-to-face discussions. Using this information, the salesperson makes a detailed presentation tailored specifically to a consumer’s needs.“Good sales consultant”, says Alan Goldfarb, president of which publishes the following weekly newspapers, seasonal magazines and specialty publications, “are the people who use a wide range of skills including probing, listening, analysis, and persuasiveness. The best sales consultant, however, are the ones who can think outside the box and use their creativity to present a product and close a sale. The other skills you can teach. Creativity is what we can’t. It’s something we look for in every employee we hire. ”More and more sales teams are switching from a narrative or a suggestive approach to a more consultative strategy. As a result, corporations are looking more at intangibles (无形资产) such as creativity and analytical skills and less at educational background and technical skills.(Note:Answer the questions or complete the statements in NO MORE THAN TWELVE WORDS.)78. The main difference between narrative sales strategy and suggestive sales strategy is that theformer involves __________ while the latter doesn’t.79. Mindy compares salespeople to wine stewards because they both _______________.80. What is the biggest challenge for a consultative salesperson?81. Corporations that favor a consultative strategy may prefer those who __________ whenchoosing employees.第Ⅱ卷I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1. 方便起见,你最好租一辆自行车。

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