上海市复旦附中2023高一下学期期末测验(含答案)
2024届上海市复旦中学物理高一第二学期期末质量检测试题含解析
![2024届上海市复旦中学物理高一第二学期期末质量检测试题含解析](https://img.taocdn.com/s3/m/de703453df80d4d8d15abe23482fb4daa58d1d20.png)
2024届上海市复旦中学物理高一第二学期期末质量检测试题注意事项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:本大题共10小题,每小题5分,共50分。
在每小题给出的四个选项中,有的只有一项符合题目要求,有的有多项符合题目要求。
全部选对的得5分,选对但不全的得3分,有选错的得0分。
1、(本题9分)一架飞机以200 m/s的速度在高空沿水平方向做匀速直线运动,每隔1 s 先后从飞机上自由释放A、B、C三个物体,若不计空气阻力,则A.在运动过程中A在B前200 mB.在运动过程中B在C前200 mC.A、B、C在空中排列成一条竖直线D.A、B、C在空中排列成一条抛物线2、(本题9分)如图所示,绷紧的水平传送带始终以恒定速率1v运行。
初速度大小为2v 的小物块从与传送带等高的光滑水平地面上的A处滑上传送带。
若从小物块滑上传送带开始计时,小物块在传送带上运动的v t 图像(以地面为参考系)如图乙所示。
已知2v>1v,则()A.2t时刻,小物块离A处的距离达到最大B.2t时刻,小物块相对传送带滑动的距离达到最大C.0~2t时间内,小物块受到的摩擦力方向先向右后向左D.0~3t时间内,小物块始终受到大小不变的摩擦力作用3、如图所示,A、B为某小区门口自动升降杆上的两点,A在杆的顶端,B在杆的中点处. 杆从水平位置匀速转至竖直位置的过程中,A、B两点A .角速度大小之比2:lB .角速度大小之比1:2C .线速度大小之比2:lD .线速度大小之比1:24、 (本题9分)图中接地金属球A 的半径为R ,球外点电荷的电荷量为Q ,到球心的距离为r.该点电荷的电场在球心的场强大小等于( )A .B .C .0D .5、一个质量为m 的足球,以v 0速度由地面踢起,当它到达离地面高为h 的B 点处(取B 点为重力势能零参考平面,重力加速为g ),下列说法中正确的是A .在B 点重力势能为mghB .在B 点的动能为2012mv mgh + C .在B 点的机械能为2012mv mgh +D .在B 点的机械能为2012mv mgh -6、 (本题9分)如图所示,某同学用绳子拉动木箱,使它从静止开始沿粗糙水平路面运动至具有某一速度(木箱对路面的压力不为0),木箱获得的动能一定( )A .小于拉力所做的功B .等于拉力所做的功C .等于克服摩擦力所做的功D .等于合力所做的功7、质量相等的A 、B 两球在光滑水平面上沿同一直线、同一方向运动,A 球的动量是7 kg·m/s ,B 球的动量是5 kg·m/s ,当A 球追上B 球发生碰撞,则碰撞后A 、B 两球的动量可能值是( )A .p A =6 kg·m/s ,pB =6 kg·m/s B .p A =3 kg·m/s ,p B =9 kg·m/sC .p A =-2 kg·m/s ,p B =14 kg·m/sD .p A =-4 kg·m/s ,p B =17 kg·m/s8、 (本题9分)一重球从高h 处下落,如图所示,到A 点时接触弹簧,压缩弹簧至最低点位置B 。
2023届上海复旦附中物理高一第二学期期末质量跟踪监视试题含解析
![2023届上海复旦附中物理高一第二学期期末质量跟踪监视试题含解析](https://img.taocdn.com/s3/m/192839b9b8d528ea81c758f5f61fb7360b4c2bf0.png)
2022-2023学年高一物理下期末模拟试卷考生须知:1.全卷分选择题和非选择题两部分,全部在答题纸上作答。
选择题必须用2B 铅笔填涂;非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。
2.请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。
3.保持卡面清洁,不要折叠,不要弄破、弄皱,在草稿纸、试题卷上答题无效。
一、选择题:(1-6题为单选题7-12为多选,每题4分,漏选得2分,错选和不选得零分) 1、 (本题9分)两个带等量正电的点电荷,固定在图中P 、Q 两点,MN 为PQ 连线的中垂线,交PQ 于O 点,A 为MN 上的一点.一带负电的试探电荷q ,从A 点由静止释放,只在静电力作用下运动,取无限远处的电势为零,则A .q 由A 向O 的运动是匀加速直线运动B .q 由A 向O 运动的过程电势能逐渐增大C .q 运动到O 点时的动能最大D .q 运动到O 点时电势能为零2、 (本题9分)下列说法正确的是( ) A .匀速圆周运动是匀速运动 B .匀速圆周运动是匀变速运动 C .匀速圆周运动是变加速运动D .做圆周运动的物体其合力一定与速度方向垂直3、 (本题9分)如图,O 、A 、B 三点位于同一直线上,且OA AB 。
现在O 点放置一点电荷,它在A 点产生的电场强度大小为E ,则它在B 点产生的电场强度大小为A .4EB .2E C .2E D .4E4、 (本题9分)如图所示,有两个穿着溜冰鞋的人站在水平冰面上,当其中某人A 从背后轻轻推另一个人B 时,两个人会向相反的方向运动。
不计摩擦力,则下列判断正确的是( )A.A、B的质量一定相等B.推后两人的动能一定相等C.推后两人的总动量一定为零D.推后两人的速度大小一定相等5、3个质量分别为m1、m2、m3的小球,半径相同,并排悬挂在长度相同的3根竖直绳上,彼此恰好相互接触.现把质量为m1的小球拉开一些,如图中虚线所示,然后释放,经球1与球2、球2与球3相碰之后,3个球的动量相等.若各球间碰撞时均为弹性碰撞,且碰撞时间极短,不计空气阻力,则m1:m2:m3为()A.6:3:1 B.2:3:1 C.2:1:1 D.3:2:16、2016年12月11日,风云四号同步气象卫星在西昌成功发射,实现了中国静止轨道气象卫星升级换代和技术跨越,大幅提高天气预报和气候预测能力。
上海市复旦大学附属中学2022-2023学年高一下学期期末数学试题(教师版)
![上海市复旦大学附属中学2022-2023学年高一下学期期末数学试题(教师版)](https://img.taocdn.com/s3/m/467fe448c381e53a580216fc700abb68a982add5.png)
复旦大学附属中学2022学年第二学期高一年级数学期末考试试题时间:120分钟满分:150分一、填空题(本大题共有12题,其中1-6每小题4分,7-12每小题5分,满分54分)1.已知复数1i z =+(i 是虚数单位),则1i z -=________.【答案】【解析】【分析】利用复数的四则运算化简复数1i z -,再利用复数的模长公式可求得1i z -.【详解】因为1i z =+,则()1i 1i 1i 1i 12i z -=-+=-+=-,因此,1i 2i z =--==..2.向量()3,5a =r 在向量()1,1b = 方向上的数量投影为________.【答案】【解析】【分析】利用数量积的几何意义根据题意直接求解即可【详解】向量()3,5a =r 在向量()1,1b = 方向上的数量投影为a b b⋅== ,故答案为:3.如图,是古希腊数学家阿基米德的墓碑文,墓碑上刻着一个圆柱,圆柱内有一个内切球,这个球的直径恰好与圆柱的高相等,相传这个图形表达了阿基米德最引以自豪的发现,在这个伟大发现中,球的体积与圆柱的体积之比为________.【答案】23##2:3【解析】【分析】根据两图形的关系可得圆柱的底面半径与球的半径相等,设半径为r ,计算出两几何体的体积,求出比值即可.【详解】 圆柱内切一个球,∴圆柱的底面半径与球的半径相等,不妨设为r ,则圆柱的高为2r ,23π22πV r r r ∴=⋅=圆柱,34π3V r =球.∴球与圆柱的体积之比为2:3,故答案为:23.4.若ππcos isin 88n ⎛⎫+ ⎪⎝⎭是纯虚数(其中i 是虚数单位),则正整数n 的最小值为________.【答案】4【解析】【分析】求得()ππππcos isin cos isin 8888nn n n *=⎛⎫++∈ ⎪⎝⎭N ,根据复数的概念可得出n 的表达式,即可求得正整数n 的最小值.【详解】因为ππππππππcos isin cos isin cos isin cos isin 88888888n⎛⎫⎛⎫⎛⎫⎛⎫++⋅+⋅⋅+ ⎪ ⎪ ⎪ ⎪=⎝⎭⎝⎭⎝⎭⎝⎭ ()22π2πππππcos isin cos isin cos isin 888888n -⋅=⎛⎫⎛⎫⎛⎫+⋅+⋅+ ⎪ ⎪⎝⎭⎝⎭⎝⎭ 个()()33π3πππππππcos isin cos isin cos isin cos isin 88888888n n n n *-⎛⎫⎛⎫⎛⎫+⋅+⋅⋅=+==+∈ ⎪ ⎪⎝⎭⎝⎭⎝⎭N 个因为ππcos isin 88n ⎛⎫+ ⎪⎝⎭为纯虚数,则πcos 08n =,可得()πππ82n k k =+∈Z ,可得()84n k k =+∈Z ,又因为n *∈N ,当0k =时,正整数n 取最小值4.故答案为:4.5.函数()π2sin cos 2sin cos 12f x x x x x ⎛⎫=-++ ⎪⎝⎭,则()f x 的最大值为________.【答案】【解析】【分析】利用三角恒等变换化简函数()f x 的解析式,利用正弦型函数的基本性质可求得()f x 的最大值.【详解】因为()2π2sin cos 2sin cos 1sin 22cos 1sin 2cos 22f x x x x x x x x x ⎛⎫=-++=-+=- ⎪⎝⎭π24x ⎛⎫=- ⎪⎝⎭,故函数()f x ..6.设方程220x x m -+=的两个根为,αβ,且||2αβ-=,则实数m 的值是________.【答案】0或2【解析】【分析】当,αβ为实数根时,利用根与系数关系即可求出结果;当,αβ为虚数根时,原方程的根是2i 442±,利用复数模的定义即可求出结果.【详解】当,αβ为实数根时,方程220x x m -+=的两个根为,αβ,2m αβαβ∴+=⋅=,,||2αβ-=,()()224444m αβαβαβ∴-=+-=-=,0m ∴=;当,αβ为虚数根时,原方程的根是22±,2αβ∴-===,2m ∴=,2m ∴=或0m =,故答案为:0或2.7.沙漏是古代的一种计时装置,它由两个形状完全相同的容器和一个狭窄的连接管道组成.开始时细沙全部在上部容器中,细沙通过连接管道最后全部流到下部容器,假设在下方也堆积成一个以下底面为底面的圆锥,如图,某沙漏由上下两个圆锥组成,设此圆锥的高为h ,细沙全部在上部时,其高度为圆锥高度的23(细管忽略不计),则细沙全部在下部时堆积成的圆锥的高为________.【答案】827h ##827h 【解析】【分析】求出细沙全在上部时的体积,根据细沙全在上部与全在下部的体积相等,利用圆锥的体积公式可求得答案.【详解】设圆锥的底面半径为r ,细沙在上部时,细沙的底面半径为1r ,则12233h r r h ==,得123r r =,所以细沙的体积为221228π33381V r h r h π⎛⎫=⨯⨯⨯= ⎪⎝⎭,设细沙流入下部的高度为1h ,则22118ππ381r h r h =,得1827h h =,所以细沙全部在下部时堆积成的圆锥的高为827h ,故答案为:827h 8.若异面直线a 、b 所成的角为80 ,P 为空间一定点,则过点P 且与a 、b 所成的角都是50 的直线有且仅有________条.【答案】3【解析】【分析】在空间取一点P ,经过点P 分别作//a a ',//b b ',分析直线PM 满足它的射影PQ 在a '、b '所成角的平分线上时的情况可得出答案.【详解】在空间取一点P ,经过点P 分别作//a a ',//b b ',设直线a '、b '确定平面α,当直线PM 满足它的射影PQ 在a '、b '所成角的平分线上时,PM 与a '所成的角等于PM 与b '所成的角,设直线PM 与a '、b '所成角为θ,因为直线a 、b 所成角为80 ,得a '、b '所成锐角为80 ,①当直线PM 的射影PQ 在a '、b '所成锐角的平分线上时,则PM 与a '、b '所成角的范围是4090θ≤≤ ,这种情况下,过P 点有2条直线与a 、b 所成角都是50 ;②当直线PM 的射影PQ 在a '、b '所成钝角的平分线上时,PM 与a '、b '所成角的范围是5090θ≤≤ ,这种情况下,过P 点有且仅有1条直线(即PM α⊂时)与a 、b 所成角都是50 .综上所述,过P 点且与a 、b 所成角都是50 的直线有3条.故答案为:3.9.在ABC 中,6AC =,8BC =,90C ∠=︒.P 为ABC 所在平面内的动点,且1PC =,则PA PB ⋅ 的取值范围是________.【答案】[]9,11-【解析】【分析】建立直角坐标系表示各点坐标,再利用三角恒等变换即可求出结果.【详解】如图所示建立直角坐标系,设()()6,0,0,8A B ,P 为ABC 所在平面内的动点,且1PC =,∴设()cos ,sin P θθ,则()()6cos ,sin ,cos ,8sin PA PB θθθθ=--=-- ,∴cos (6cos )sin (8sin )18sin 6cos 110sin()PA PB θθθθθθθϕ⋅=----=--=-+ ,其中3tan 4ϕ=,则PA PB ⋅ 的取值范围是[]9,11-,故答案为:[]9,11-.10.已知2023年第57届世界乒乓球锦标赛规定适用的乒乓球直径为4cm.如图,是一个正方形硬纸板,现有同学将阴影部分裁掉,把剩余的扇形部分制作成一个圆锥型的纸筒.若这样的乒乓球能够完全装入该同学所制作的圆锥型的纸筒内,则正方形纸板面积的最小值为________平方厘米.【答案】3203【解析】【分析】根据圆锥的内切球,利用相似即可求解内切球的半径,进而可求解正方形最大的边长,即可求解.【详解】设正方形的边长为a ,设圆锥的底面圆半径为r ,所以扇形的弧长为π2π24a a r r =⇒=,设圆锥的内切球球心为Q ,半径为R ,作出轴截面如图:圆锥的高4OB =,由于~MQP MBO ,所以1515420a R QP QM R R a OB MB r a -=⇒=⇒=,要使乒乓球能放入圆锥容器,则需满足15220R a a ≤=⇒≥所以正方形的面积为2240320315a ⎛⎫≥= ⎪⎝⎭,故最小值为3203,故答案为:320311.在xOy 平面上,将两个函数21y x =-和21y x =+两条直线2y =和0y =围成的封闭图形记为D ,如图所示,记D 绕y 轴旋转一周而成的几何体为Ω,则Ω的体积值________.【答案】10π3【解析】【分析】作直线y x =、y x =-,分析可知,直线y x =、y x =-、1y =围成的区域绕y 轴旋转所形成的几何体为圆锥,确定该圆锥的底面半径与高,可求得这部分几何体的体积;然后利用祖暅原理分别计算出由曲线函数21y x =-以及线段y x =、()02y x y =-≤≤围成的区域绕y 轴旋转所形成的几何体的体积,以及由曲线)2112y x y =+≤≤与线段y x =、()12y x y =-≤≤围成的区域绕y 轴旋转所形成的几何体的体积,进而可得出Ω的体积.【详解】如下图所示,作直线y x =、y x =-,①计算出由直线y x =、y x =-、1y =围成的区域绕y 轴旋转所形成的几何体的体积,由题意可知,该几何体是底面半径为1,高为1的圆锥,其体积为211ππ1133V =⨯⨯=;②计算出由曲线函数y =以及线段y x =、()02y x y =-≤≤围成的区域绕y 轴旋转所形成的几何体的体积,由y =可得221y x =-,即221x y -=,过点()()0,02y y ≤≤作几何体的截面,截面面积为()222ππS x y =-=,由祖暅原理可知,该几何体的体积等于底面积为π,且高为2的柱体的体积,这部分几何体的体积为2π22πV =⨯=;③计算出由曲线)12y y =≤≤与线段y x =、()12y x y =-≤≤围成的区域绕y 轴旋转所形成的几何体的体积,由)12y y =≤≤可得221y x -=,过点()()0,12y y ≤≤作几何体的截面,截面面积为()223ππS y x =-=,所以,该部分几何体的体积等于底面面积为π,高为1的柱体的体积,故该部分几何体的体积为3π1πV =⨯=.综上所述,Ω的体积为123π10π2ππ33V V V V =++=++=.故答案为:10π3.12.如图,棱长为1的正方体12345678A A A A A A A A -的八个顶点分别为128,,,A A A ,记正方体12条棱的中点分别为91020,,,A A A ,6个面的中心为212226,,,A A A ,正方体的中心为27A .记117j j m A A A A =⋅ ,}7{1,22j ∈⋯,,,其中17A A 是正方体的体对角线.则1227m m m ++⋯+=________.【答案】812##40.5【解析】【分析】建立如图所示的空间直角坐标系,利用空间向量数量积的坐标运算,可求1227m m m ++⋯+的值.【详解】建立如图所示的空间直角坐标系,则()10,0,0A ,()21,0,0A ,()31,1,0A ,()40,1,0A ,()50,0,1A ,()61,0,1A ,()71,1,1A ,()80,1,1A ,设向量()1,,j A A x y z = ,而()171,1,1A A = ,故117j j m A A A A x y z =⋅=++ ,故1227m m m ++⋯+表示各点的坐标和的和.现各点的横坐标之和为X ,纵坐标之和为Y ,竖坐标之和为Z ,根据对称性可得1271990922X Y Z ===⨯+⨯+⨯=,故12272781322m m m ++⋯+=⨯=,故答案为:812.【点睛】方法点睛:对于一些较为复杂的计算问题,如果直接算比较麻烦,则可以换一个等价的计算方法,从而使得问题得以简化.二、选择题(本大题共有4题,每题5分,满分20分)13.已知平面α和平面β,直线,m n β⊂,则“//m α且//n α”是“//a β”的().A.充分非必要条件B.必要非充分条件C.充要条件D.既非充分也非必要条件【答案】B【解析】【分析】根据平面平行的性质可判断必要性,根据面面平行的判断可判断充分性.【详解】由于直线,m n 不一定相交,所以由//m α,//n α,不能得到//a β,故充分性不成立,若//a β,则平面β内的任意直线均与平面α平行,故必要性成立,故“//m α且//n α”是“//a β”的必要不充分条件,故选:B14.下面五个命题:(1)两组对边分别相等的四边形是平行四边形;(2)两组对边分别平行的四边形是平行四边形;(3)四个角都是直角的四边形是矩形;(4)有两个侧面是矩形的三棱柱是直三棱柱;(5)有两个侧面是矩形的四棱柱是直四棱柱其中真命题的个数是()A.1B.2C.3D.4【答案】C【解析】【分析】空间四边形也可以满足两组对边分别相等,故(1)错误;四边形一组对边平行可知,则由公理可知四边形共面,故(2)正确;对四边形为平面四边形时和空间四边形进行讨论,可以判断(3)正确;根据直线与平面的判定定理,得到这两个侧面的交线垂直于底面,故(4)正确;若侧棱与底面两条平行的两边垂直,有两个侧面均是矩形,此时的棱柱不一定是直棱柱,故(5)错误.【详解】空间四边形也可以满足两组对边分别相等,故(1)错误;四边形一组对边平行可知,则由公理可知四边形共面,四边形为平面四边形,由平行四边形判定定理可知:两组对边分别平行的平面四边形是平行四边形,故(2)正确;当四边形为平面四边形时,根据矩形的判定可知:四个角都是直角的四边形是矩形;当四边形为空间四边形时,假设存在四个角都是直角的空间四边形A BCD -,则,,AD AB AD CD ⊥⊥AD 为AB ,CD 的公垂线,,,BC AB BC CD ⊥⊥BC 为AB ,CD 的公垂线,这与公垂线的性质矛盾,故不存在四个角都是直角的空间四边形,故命题(3)正确;对有两个侧面是矩形的三棱柱,根据直线与平面的判定定理,得到这两个侧面的交线垂直于底面,故(4)正确;若侧棱与底面两条平行的两边垂直,有两个侧面均是矩形,此时的棱柱不一定是直棱柱,故(5)错误.故选:C.15.过正四面体ABCD 的顶点A 作截面,若满足①截面是等腰三角形;②截面与底面BCD 成75°的二面角,这样的截面个数为()A.6B.12C.18D.24【答案】C【解析】【分析】计算二面角A CD B --的大小,得出与底面BCD 所成的角为75︒的等腰三角形截面与底面三角形交线的位置情况,然后分类求出截面数即可.【详解】作正四面体ABCD 的高AO ,连接BO 并延长交CD 于点E ,连接AE ,则E 为CD 的中点,O 为等边BCD △的中心,所以,BE CD AE CD ⊥⊥,所以AEB ∠为二面角A CD B --的平面角,设2AB =,则BE =,所以12,3333OE BE OB BE ====,所以263AO ===,所以263tan 33AO AEB OE ∠===因为31tan 45tan 303tan 75tan(4530)221tan 45tan 3033+︒+︒︒=︒+︒==+-︒︒所以75AEB ∠<︒,在平面BCD 内,以O 为圆心,以tan 75AO ︒为半径作圆,则圆O 在BCD △内部,所以若截面AMN 与底面BCD 所成角为75︒,则截面AMN 与平面BCD 的交线为圆O 的切线,①若圆O 的切线与BCD △的一边平行,如图1所示,则存在6个符合条件的截面三角形AMN ,②若圆O 的切线过BCD △的顶点,不妨设过点B ,交CD 于M ,如图2所示,则由ACM △≌BCM ,可得AM BM =,所以截面ABM 为符合条件的截面三角形,则这样的三角形有6个,③若圆O 的切线MN 与BCD △的两边相交,不妨设与BC 交于M ,与CD 交于N ,且BM CN =,在等边BCD △中,连接,,OC OM ON ,设MN 与圆O 切于点P ,连接OP ,则OP MN ⊥,因为BM CN =,,30OC OB OBM OCN =∠=∠=︒,所以OBM ≌OCN ,所以BOM CON ∠=∠,OM ON=因为120BOC ∠=︒,所以120MON ∠=︒,所以30OMN ONM ∠=∠=︒,所以2622863242tan 75323AO OM OP ⨯====-︒+,在OBM 中由余弦定理得2222cos30OM OB BM OB BM =+-⋅︒,22286232334223332BM BM ⎛⎫⎛⎫-=+-⨯⨯ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭,解得366938433BM --=或366938433BM +-=如图3,则ABM ≌ACN △,所以AM AN =,所以截面AMN 为符合条件的截面三角形,则这样的三角形有6个,综上,符合条件的截面个数有18个,故选:C16.已知()12,0,πx x ∈且0λ>,0μ>,选项中的命题都正确的是().(1)不等式1212sin sin x x x x -≤-恒成立;(2)设12x x <,()11,sin A x x ,()22,sin B x x ,()11,0A x ,()12,0B x ,如果四边形11A ABB 的面积为s ,那么存在[]012,x x x ∈使()210sin x x x s -=成立;(3)对任意1λμ+=时,不等式12122sin 2sin sin sin x x x x λμ+≥+恒成立;(4)对任意1λμ+=时,不等式()1212sin sin sin x x x x λμλμ+≥+恒成立.A.(1)(2)(3) B.(1)(2)(4) C.(1)(3)(4) D.(2)(3)(4)【答案】B【解析】【分析】可证明0x >时,有sin x x >恒成立,据此可判断(1)的正误,利用零点存在定理可判断(2)的正误,利用反例可判断(3)的正误,利用向量结合正弦函数的性质可判断(4)的正误.【详解】先证明一个不等式:当0x >时,有sin x x >恒成立.证明:若1x >,则1sin x x >≥恒成立.若01x <≤,如图,在单位圆中,弧度为x 的角的终边与单位圆的交点为P ,过P 作PH x ⊥轴,垂足为H ,则sin PH x =,而 PH PS PS <<,故sin x x <,因为此时sin 0x >,故sin sin x x x>=综上,0x >时,有sin x x >.对于(1),若12x x =,则1212sin sin 00x x x x -=≤=-,若12x x ≠,不妨设12x x <,因为1212121212sin sin sin sin 2222x x x x x x x x x x +-+-⎛⎫⎛⎫-=+- ⎪ ⎪⎝⎭⎝⎭12121221212cos sin 2sin 22222x x x x x x x x x x +---=≤<⨯=-,故(1)成立.对于(2),()()12211sin sin 2s x x x x =+-,取()()[]12121sin sin sin ,,2g x x x x x x x =-+∈,则()()1121sin sin 2g x x x =-,()()2211sin sin 2g x x x =-,故()()()212121sin sin 04g x g x x x =--≤,由零点存在定理可得存在[]()0120121,,sin sin sin 2x x x x x x ∈=+,此时()()()21012211sin sin sin 2x x x x x x x s -=+-=,故(2)正确.而210x x ->,故210x x ->,故()0121sin sin sin 2x x x =+,对于(3),取1212ππ,,,3336u x x λ====,则121202sin 2sin sin 3sin 1311132326x x x x λμ+-=-⨯+=-⨯<,故(3)错误.对于(4),设t μλ=,SQ tQT = ,(),Q x y ,且()()1122,sin ,,sin S x x T x x ,则()()1122,sin ,sin x x y x t x x x y --=--,故1212sin sin ,11x tx x t x x y t t++==++,过Q 作x 轴的垂线,垂足为G ,由正弦函数的图象特征可得:1212sin sin sin 11x t x x tx t t ++≤++,故1212sin sin sin 11x x x x μμλλμμλλ++≤++,结合1λμ+=整理得到:()1212sin sin sin x x x x λμλμ+≤+,故(4)成立,故选:B【点睛】思路点睛:正弦函数具有的某些代数性质,可以结合函数的图象来讨论(比如凹凸性等),另外为了研究某些性质,我们可以猜测一些我们需要的结论并给出适当的证明.三、解答题(本大题共有5题,满分78分)17.如图,在四棱锥11A BCC B -中,平面ABC⊥平面11BCC B ,ABC 是正三角形,四边形11BCC B 是正方形,D 是AC 的中点.(1)求证:1AB //平面1BDC ;(2)求直线BC 和平面1BDC 所成角的大小【答案】(1)见解析(2)arcsin 5【解析】【分析】(1)由线面平行的判定定理证明即可;(2)过点C 作1CH C D ⊥,连接BH ,由题意可证得CH ⊥平面1BDC ,所以CBH ∠是直线BC 和平面1BDC 所成角,求解即可.【小问1详解】连接1`,B C BC 交于点O ,连接OD ,在1B AC △中,1//OD AB ,OD ⊂平面1BDC ,1AB ⊄平面1BDC ,所以1AB //平面1BDC ;【小问2详解】因为ABC 是正三角形,D 是AC 的中点,所以BD AC ⊥,设ABC 的边长为2,所以BD ==,因为平面ABC ⊥平面11BCC B ,平面ABC ⋂平面11BCC B BC =,因为四边形11BCC B 是正方形,所以1CC BC ⊥,所以1CC ⊥平面ABC ,AC ⊂平面ABC ,所以1CC AC ⊥,所以1C D ===,1BC ==,所以22211+=BD C D BC ,所以1BD C D ⊥,又因为BD AC ⊥,1AC C D D = ,1,AC C D ⊂平面1DCC ,所以BD ⊥平面1DCC ,过点C 作1CH C D ⊥,连接BH ,CH ⊂平面1DCC ,所以BD CH ⊥,11,,C D BD D C D BD ⋂=⊂平面1BDC ,所以CH ⊥平面1BDC ,所以CBH ∠是直线BC 和平面1BDC 所成角,在1CC D ,11CC CD CH C D ⋅=⋅,所以21CH ⨯=,所以5CH =,所以sin 5CH CBH BC ∠==.所以5arcsin5CBH ∠=,所以直线BC 和平面1BDC 所成角的大小5arcsin 5.18.如图,已知点P 在圆柱1OO 的底面圆O 上,AB 为圆O 的直径,圆柱1OO 的表面积为24π,2OA BP ==.(1)求异面直线1A B 与AP 所成角的大小;(2)求点A 到平面1A PB 的距离.【答案】(1)3arc cos 5(2)677【解析】【分析】(1)过点B 作BQ AP ,连接AQ ,则1QBA ∠为异面直线1A B 与AP 所成的角,进而求出个边长,利用余弦定理即可求出结果;(2)利用等体积转化法即可求解.【小问1详解】过点B 作BQ AP ,连接AQ ,则1QBA ∠为异面直线1A B 与AP 所成的角或其补角,则四边形APBQ 为平行四边形,2OA BP ==,则OPB △为等边三角形,易知90APB ∠=︒,30PAO ∠=︒,故4AB =,则QB PA ==由圆柱1OO 的表面积为212π22π220πBB ⋅+⋅⋅=,可得113AA BB ==,在1A AQ 中,13AA =,2AQ BP ==,1A Q ==在1A AB △中,15A B ==,在1A QB 中,由余弦定理可得:222111123cos 25A B QB A Q QBA A B QB +-∠==⋅,所以1r cos 5QBA a ∠=,异面直线1A B 与AP所成角的大小为arc cos5.【小问2详解】设点A 到平面1A PB 的距离为h ,则11A A PB A APB V V --=,111133A PB APB S h S AA ⋅⋅=⋅⋅,12APB S AP BP =⋅= 因为1AA ⊥平面ABP ,BP AP ⊥,所以BP ⊥平面1A AP ,即1BP A P ⊥,在1Rt A PB中,1A P ==故1112A PB S A P PB =⋅= ,所以7h =,即点A 到平面1A PB的距离为7.19.如果一个正多面体的所有面都是全等的正三角形或正多边形,每个顶点聚集的棱的条数都相等,这个多面体叫做正多面体.有趣的是只有正四面体、正方体、正八面体、正十二面体和正二十面体五种正多面体,现将它们的体积依次记为,4681220,,,,V V V V V .(1)利用金属板分别制作正多面体模型各一个,假设制作每个模型的外壳用料(即表面积)均等于2,分别求出4V 和8V 的值;并猜想12V 与20V 的大小关系(猜想不需证明)(2)多面体的欧拉定理:简单多面体的面数F 、棱数E 与顶点数V 满足:2V F E +-=.已知正多面体都是简单多面体,设某个正多面体每个顶点聚集的棱的条数为m ,每个面的边数为n ,求,,m n E 满足的关系式;并尝试据此说明正多面体仅有五种.【答案】(1)4V =;8V =1220V V <.(2)11112m n E+=+;见解析【解析】【分析】(1)设正四面体的边长为a ,由正四面体的表面积求出a ,再由正四面体的体积公式求出正四面体的体积,同理求出正八面体的体积;在相同的表面积下,正多面体的面越多,其体积越大,所以可得出1220,V V 的大小;(2)由题意可得出2E nF mV ==,代入2V F E +-=即可得出11112m n E +=+,因此1112m n +>,又3,3m n ≥≥,即可说明正多面体仅有五种.【小问1详解】设正四面体的边长为a ,所以正四面体的表面积为:22344S a =⋅==,所以a =ABCD 中,H 是BCD △的中心,则AH 是高,AH DH ⊥,233323DH =⨯⨯=⨯=4AH ==,所以(241434V =⨯⨯⨯=设正八面体的边长为b ,所以正八面体的表面积为:22384S b =⋅==所以b =,如下图所示:在正八面体中,连接AC 交平面EFBH 于点O ,则AO ⊥平面EFBH ,所以212EFBG S b ==,2222AO b ===⨯=,所以正八面体的体积为(3238112222233233EFBG V S AO b b b =⨯⨯⨯=⨯⨯⨯==⨯=,在相同的表面积下,正多面体的面越多,其体积越大,所以1220V V <.【小问2详解】设某个正多面体每个顶点聚集的棱的条数为m ,每个面的边数为n ,则2E nF mV ==,于是22,E E V F m n ==,将其代入2V F E +-=,所以222E E E m n +-=,所以11112m n E +=+,所以,,m n E 满足的关系式为:11112m n E +=+,因此1112m n +>,又3,3m n ≥≥,所以所有可能的正整数解为:()()()()()3,3,3,4,3,5,4,3,5,3,共5种,所以正多面体仅有五种.20.已知直角梯形ABCD ,//AD BC ,π2ABC ADE ∠=∠=,1AB =,扇形圆心角BAE x ∠=,π0,2x ⎛⎫∈ ⎪⎝⎭,如图,将ADC △,ABC 以及扇形BAE 的面积分别记为()()()p x q x s x ,,(1)写出()()()p x q x s x ,,的表达式,并指出其大小关系(不需证明);(2)用tan2x表示梯形ABCD 的面积()t x ;并证明:()()2t x s x >⋅;(3)设()()()p x f x s x =,π02ααϕ<<+<,试用代数计算比较()f α与()f αϕ+的大小.【答案】(1)()()()111sin tan 222p x x q x x s x x ==,=,,()()()p x s x q x <<(2)()42tan21tan 2xt x x=-,证明见解析,(3)()()f f αϕα+<【解析】【分析】(1)根据锐角三角函数,以及扇形的面积公式即可求解,(2)根据二倍角公式即可得()42tan21tan2x t x x=-,利用()22224tantanπ22221tantan ,4221tan 1tan 1tan 1tan 1tan 22222x x x xx x x t x x x x x x =>>∴=+>=+-+--,即可由放缩法求证,或者构造函数()sin tan 2,g x x x x =+-利用导数求解单调性即可求证,(3)利用和差角公式,以及tan sin ,sin cos 0,sin 0ααααααϕϕ>>∴>>>>即可作差比较大小,或者构造函数()cos sin ,m x x x x =-求导判断单调性,即可利用()f x 的单调性求解.【小问1详解】由题意可得sin sin ,tan tan AD AE x x BC AB x x ====,所以()()()2111111sin tan 222222p x AD AB x q x BC AB x s x x x ⋅=⋅==⋅==,=,,如图:在单位圆中,设AOB x ∠=,π0,2x ⎛⎫∈ ⎪⎝⎭,则2111111sin sin ,tan ,222222AOB BOM AOB S OB OA x x S OB BM x S OA x x =⋅==⋅=== 扇形,由于AOB BOM AOB S S S << 扇形,所以sin tan <<x x x ,π0,2x ⎛⎫∈ ⎪⎝⎭,因此()()()p x s x q x <<.【小问2详解】()2242sin cos tan 2tan tan 2tansin tan 2222222221tan 1tan 1tan 222x x x x x x AD BC x x t x AB x x x ⎛⎫+⨯ ⎪++⎝⎭=⋅===+=+--,方法一:由()22224tantanπ22221tantan ,4221tan 1tan 1tan 1tan 1tan 22222x x x x x x xt x x x x x x =>>∴=+>+=+-+--.所以()()44111tan 1tan 222t x x x x x x x s ⎛⎫ ⎪-=- ⎪ ⎪⎝⋅>--⎭-,由于π0,2x ⎛⎫∈ ⎪⎝⎭,则()π0,,tan 0,1242x x ⎛⎫∈∈ ⎪⎝⎭,所以()()41101tan 22x x t x s x ⎛⎫--⋅>⎪->⎪ ⎪⎝⎭故()()2t x s x >⋅,方法二:由于()()sin tan sin tan 2222t x x x x xs x x x ++---⋅==,令()sin tan 2,g x x x x =+-则()21cos 2cos g x x x'=+-,由于π0,2x ⎛⎫∈ ⎪⎝⎭,所以()2cos 0,1,cos cos x x x ∈∴>,故()22211cos 2cos 220cos cos g x x x x x '=+->+->=,因此()g x 在π0,2x ⎛⎫∈ ⎪⎝⎭单调递增,故()(0)0g x g >=,所以sin tan 20,x x x +->因此()()()()sin tan 22202t x s x t x s x xx x +⇒-⋅>-=⋅>.【小问3详解】方法一:由于()sin (),()p x xf x s x x ==所以()()()()()()sin sin sin sin f f αϕααϕααϕαααϕααϕααϕ++-+-+=-=++,()()()()sin sin sin sin sin cos cos sin ααϕααϕααϕαααϕααϕααϕααϕ+-++--==++[]()sin 1cos sin cos sin ααϕϕαααϕααϕ-+-=+由于π02ααϕ<<+<,所以tan sin ,sin cos 0,sin 0ααααααϕϕ>>∴>>>>,故sin cos sin 0ϕαααϕ->,()sin 1cos 0,αααϕ-+>⎡⎤⎣⎦()()[]()sin 1cos sin cos sin 0f f ααϕϕαααϕααϕααϕ-+--+=>+,因此()()f f αϕα+<.方法二::()2()sin cos sin (),()p x x x x xf x f x s x x x -'===,记()cos sin ,m x x x x =-π0,2x ⎛⎫∈ ⎪⎝⎭,()cos sin cos sin 0m x x x x x x x '=--=-<,故()m x 在π0,2x ⎛⎫∈ ⎪⎝⎭单调递减,故()(0)0m x m <=,所以()2cos sin 0x x x f x x -'=<,故()f x 在π0,2x ⎛⎫∈ ⎪⎝⎭单调递减,由于π02ααϕ<<+<,所以()()f f αϕα+<.【点睛】本题考查了三角恒等变换,应用面积关系证明出关键不等式sin tan <<x x x ,π0,2x ⎛⎫∈ ⎪⎝⎭,结合二倍角公式以及弦切互化关系,即可由三角函数的性质求解,而证明不等式时,常采用放缩法或者作差法,将一些基本的不等关系进行适当的放缩,或者利用作差法求解,多注意不等式的变形形式,比如本题的由sin tan <<x x x 得sin ,cos sin x x x x x <<是解决本题第三问的关键.21.在菱形ABCD 中,已知120ABC ∠=︒,2AB =.E 是对角线AC 上一点,沿BE 把菱形折成二面角P BE C --,将折成二面角后的A 点记作P ,设ABE α∠=,点P 在平面BCD 上的射影记为H .(1)当E 是AC 的中点时,如图1,求证BE ⊥平面PEC ;(2)当H 落在菱形的边CD 上时,如图2,求二面角P BE C --的取值范围;(3)设折痕BE 与菱形的边AD 交于点F ,求四棱锥P BCDF -体积的最大值(说明:可以用到必修一探究实践活动中得到的不等式3a b c ++³(),,R a b c +∈).【答案】(1)证明见解析(2))0,arccos1θ⎡⎤∈⎣⎦(3)(32max41V-=+【解析】【分析】(1)由菱形的结构特征,通过线线垂直,证明线面垂直;(2)连接AH 交BF 于Q ,PQH ∠就是二面角P BE C --的平面角,表示出cos PQH ∠,由ππ,63α⎡⎤∈⎢⎥⎣⎦求的取值范围;(3)PQH ∠就是二面角P BE C --的平面角,记PQH θ∠=,用,αθ表示出四棱锥P BCDF -的体积,利用基本不等式和三角函数的性质求最大值.【小问1详解】菱形ABCD 中,当E 是AC 的中点时,BE AC ⊥,所以,BE EP BE EC ⊥⊥,又EP EC E = ,,EP EC ⊂平面PEC ,所以BE ⊥平面PEC .【小问2详解】若折叠后A 与D 点重合,π6α=,A 与C 点重合,π3α=,当H 落在菱形的边CD 上时,ππ,63α⎡⎤∈⎢⎥⎣⎦,连接AH 交BF 于Q ,连接PQ ,于是AH BQ ⊥,PQH ∠就是二面角P BE C --的平面角,记PQH θ∠=,由题设菱形得πsin 2AH α⎛⎫⋅-=⎪⎝⎭,则3cos AH α=,2sin PQ AQ α==,32sin cos HQ AH AQ a a=-=-,所以3cos 1sin 2QH PQ θα==-,由ππ,63α⎡⎤∈⎢⎥⎣⎦1cos 1θ≤≤,则)0,arccos 1θ⎡⎤∈⎣⎦;【小问3详解】连接AH 交BF 于Q ,连接PQ ,于是AH BF ⊥,所以PQH ∠就是二面角P BE C --的平面角,记PQH θ∠=,sin 2sin sin PH PQ θαθ==,π03,α⎛⎤∈ ⎥⎝⎦,在ABF △中,22πsin sin 3AFαα=⎛⎫- ⎪⎝⎭,得AF =四边形BCDF 的面积()11422S DF BC ⎛=+-⋅ ⎝四棱锥P BCDF -的体积1331sin cos V S PH θαα=⋅=≤222313123sin cos sin cos sin cos αααααα⎛⎫+=++ ⎪ ⎪⎝⎭()())2231cot 1tan tan cot a ααα=+++++()()(32243cot tan 41αααααα=++++≥+=,当且仅当223cot tan αααα⎧=⎪⎨=⎪⎩,即tan α=时等号成立,得()32min311sin cos αα⎛⎫+=+ ⎪ ⎪⎝⎭,所以当πtan 2αθ==时,四棱锥P BCDF -体积的最大值(32max41V -=.【点睛】方法点睛:作二面角的平面角可以通过垂线法进行,几何体展开、折叠问题,要抓住前后两个图形间的联系,找出其中的量的关系.。
上海市复旦大学附属中学2022-2023学年高一下学期期末考试英语试题
![上海市复旦大学附属中学2022-2023学年高一下学期期末考试英语试题](https://img.taocdn.com/s3/m/abda08c0951ea76e58fafab069dc5022aaea460a.png)
上海市复旦大学附属中学2022-2023学年高一下学期期末考试英语试题学校:___________姓名:___________班级:___________考号:___________一、单项选择1.It ________ a long time before he died of a terminal cancer.A.would be B.should be C.was D.has been 2.—Do you have clothes ________? I’ll wash them for you.—No, thank you. I’ll wash them myself.A.to be washed B.to wash C.washing D.being washed 3.________ the proposal for some time, I decided to cancel it because I knew it was no use ________ risks.A.Considering; taking B.Considered; to takeC.Having considered; taking D.Having considered; to have taken 4.The expert recommended us ________ the room every day.A.airing B.should air C.having aired D.to air 5.Human beings are superior to animals ________ they can use language as a tool to communicate.A.provided that B.in that C.except that D.for fear that 6.Would you please put the book ________ it belongs?A.to whom B.to which C.to that D.where 7.You see the lightning ________ it happens, but you hear the thunder later.A.the instant B.for an instant C.the instant when D.in an instant 8.Occasions are quite rare ____ I have the time to spend a day with my kids.A.who B.which C.why D.when 9.—Dad, I've finished my assignment.—Good,and ________you play or watch TV, you mustn't disturb me.A.whenever B.whetherC.whatever D.no matter10.Within the core of each of us ________.A.is the child we once were B.we once were is the childC.is we once were the child D.we once were the child is11.George is going to talk about the geography of his country, but I'd rather he ________ more on its culture.A.focus B.focusedC.would focus D.had focused12.As we all know, very loud noise ________ make people sick or drive them mad.A.should B.need C.can D.mustA.shouldn’t eat B.mustn’t have eatenC.shouldn’t have eaten D.mustn’t eat14.The jury ________ him with having committed the crime and he was sentenced to imprisonment.A.assured B.chargedC.confirmed D.accused15.All the documents ________ by the company suggested that its exports ________ last year.A.relieved... decrease B.issued ... decreasedC.exposed ... decreased D.released ... should decrease16.We would like to ________ our sincere gratitude and appreciation to all of our wonderful sponsors for their continued support and cooperation.A.expand B.inform C.display D.extend 17.Everyone has a duty to ________ “cancer of corruption”, says Secretary-General in message, urging all to wrestle with it.A.ride out B.carry on C.stamp out D.dive into 18.Her commitment to redecorating the big house kept her ________ for a whole week.A.refreshed B.dominated C.occupied D.restrictedA.for ... add to B.of... refer toC.for ... turn to D.of... attach to20.The girl who aspires to participate and win in the beauty contest starve herself every day to have a pretty face and a good ________.A.figure B.image C.appearance D.form21.________ is never ________ with the progress he has made will be a success.A.Whoever; content B.Who; identifiedC.Whoever; identified D.Who; content22.Allen ________ himself through rigorous training and learned to live life without any luxuries—physical or psychological.A.concerned B.disciplined C.contained D.drained 23.His lack of cooperation consciousness ________ for his failure in the election last month.A.reasoned B.excused C.explained D.accounted 24.Indeed, by ________ estimates, there may be as many as 40 million stray dogs in China, carrying disease or becoming aggressive and attacking ________.A.tough, passers-by B.rough, passers-byC.tough, passer-bys D.rough, passer-bys25.Tourists often ________ the delicate balance of nature on the island.A.upset B.beat C.offend D.decline二、选用适当的单词或短语补全短文Directions: Fill in each blank with a proper word chosen from the box. Each word can beChip flow interruptedA stable global supply chain of chips had been maintained before disruptive moves by the US.Two of the US’ top chipmakers—NVIDIA and AMD-were ordered to stop exports of two high-end chips to China on Aug 31. The ban 26 sophisticated (精密的) chips for graphics processing units (GPUs); which have been widely used in applications including AI and creative production.This came after US President Joe Biden signed an order to pass the $52.7 billion (about 369.5 billion yuan) semiconductor chip manufacturing subsidy (补贴) and research law on Aug 25.It aims to 27 efforts to “make the United States more competitive with China’sscience and technology efforts”, Reuters noted.Biden also signed the CHIPS and Science Act of 2022 into law on Aug 9. According to the act, chip makers that shift their factories to the US can receive subsidies and tax benefits with 28 conditions that restrict US companies from increasing investments in China for 10 years.“The US and its allies,” Eric Schmidt, former CEO of Google and a financier for the Bill Clinton, Obama and Biden presidential campaigns, said in March, “should utilize targeted export controls on high-end semiconductor manufacturing equipment... to protect 29 technical advantages and slow the advancement of China’s semiconductor industry”.In 30 to the US latest act, Woo Jin-hoon, a guest professor at Beijing Foreign Studies University, wrote for China Daily, this is “a move that can be profitable for the US in the short term, but harmful in the long run”.The design, manufacturing and even raw materials of a complete and complex product like semiconductors (especially chips) are usually 31 across many different countries and regions, forming a huge trade network.No matter how hard countries or regions try to support their own manufacturing bases and 32 their production, a certain degree of interdependence among countries and regions is unavoidable, China Daily commented.Chinese Foreign Ministry spokesperson Wang Wenbin said on Sept 1 at a press 33 that the US move is typical “sci-tech hegemony (霸权)”.“With its technological advantages, the US has abused the concept of national security and its state power to 34 down on the development of 35 economies and developing countries,” said Wang. “The move violates market economy principles, harms international economic and trade orders and disrupts the stability of global industrial and supply chains.”三、完形填空Background noise—like the chatter in a coffee shop or the drone of passing traffic—might slow our reading speed, but according to a study of Russian readers, it doesn’twhile working, the study has some interesting points to make. In particular, it examined how we might change our reading style to compensate for auditory noise and visual distractions such as typos or poor formatting.“Overall, previous studies reported a harmful effect of both auditory and visual noise on reading fluency and 38 , though their results varied,” write linguistics researcher Nina Zdorova and colleagues. “So far, none of the studies exploring the influence of noise 39 it in the framework of the language processing theories.”One of the language processing theories examined was the noisy channel model, which proposes that our brain deals with noise by looking at the meaning of 40 words more and at entire sentences less. We then use a bit of smart guesswork to 41 the overall meaning and relationships between words.The second theory is the good enough model; that’s when our brains aren’t analyzing every single detail of a text but instead only grabbing enough words for a ‘good enough’ understanding. By focusing less on the precise words, our brains can 42 some cognitive resources to deal with noise.To see how reading was affected by noise 43 these models, the researchers ran two experiments: one on auditory noise (71 participants) and one on visual noise (70 participants). When it came to the auditory noise test, background chatter from overlapping podcasts caused people to spend longer looking at the key section of sentences before completing their reading. This extra time could 44 the noise, meaning sentence comprehension isn’t affected by it. In the visual noise test, comprehension remained the same while reading speed 45 . That’s a bit 46 considering previous studies, but the researchers think people just wanted to finish the task, with the visual noise an uncomfortable distraction.“In both experiments, we observed that longer total reading time was 47 with an accuracy increase for incorrect sentences,” write the researchers.There’s a lot going on in this study, but overall it’s a bigger win for the good-enough language processing theory-and an indication that auditory and visual noise doesn’t make us 48 any more or less on any particular comprehension method while we’re reading.With so many variables to measure in terms of what’s being read and what the 49 noise is, further study is required to learn more. 50 potential distractions may not interrupt your reading as much as you think.36.A.reinforce B.estimate C.affect D.interpret 37.A.First of all B.For example C.Above all D.To start with 38.A.context B.efficiency C.comprehension D.device 39.A.evaluated B.identified C.established D.employed 40.A.individual B.different C.new D.unfamiliar 41.A.confirm B.imply C.refer D.infer 42.A.exploit B.spare C.commit D.consume 43.A.on account of B.regardless of C.in regard to D.in contrast to 44.A.make up for B.live up to C.catch up with D.put up with 45.A.declined B.shrank C.expanded D.increased 46.A.embarrassing B.depressing C.puzzling D.annoying 47.A.associated B.compared C.replaced D.mixed 48.A.take B.set C.rely D.base 49.A.accompanying B.strange C.deafening D.distant 50.A.Therefore B.However C.Instead D.Otherwise四、阅读理解Like expensive watches that never break, the world’s best airports can be boring. You land, move through passport control and check into a hotel within minutes. The experience is pleasant, but not memorable. The worst airports have more characters. To adapt Tolstoy, lovely airports are all alike, but every wretched airport is wretched in its own way.To work out which is the world’s worst airport, we conducted a survey of our correspondents who travelled a lot. It attracted more, and more passionate, responses than nearly any other internal survey we have done.Although each awful airport is unique, four themes occur again and again: danger, bullying by officials, theft and delay. Sometimes, all these enhance each other. For example, it takes ages to get through Lubumbashi airport (in the Democratic Republic of Congo) because security officials slow things down in the hope that passengers will give them “un Cadeau” to hurry up. If you hand over $1, they let you board without your bags getting checked at all. Such deals make air travel in places like Congo slower, riskier, costlier and much more unpleasant.Air travellers make tempting targets for thieves. They are rich enough to afford an air ticket, which in many places makes them rich indeed. They carry luggage, some of it valuable. They are often far from home and unfamiliar with local rules. And airports are full of choke points through which travellers must pass if they are to board their planes, creating opportunities for dishonest officials to charge them. The ones in Manila are especially creative. Some have been known to plant bullets in luggage so they can “find” them and demand money not to have the owners arrested.Rules change at borders, and some airport officials enforce them mindlessly. One correspondent recalls that in Santiago, Chile: “I once got detained for two hours for failing to declare an unopened, sealed bag of almonds. I then had to write a declaration expressing my regret for bringing the nuts. When I failed to do so without cracking up I was threatened with arrest. The lady next to me was being interrogated for carrying a lone banana.”Poor countries have an excuse for poor airports. Rich countries do not, which is perhaps why travellers are particularly annoyed to find grottiness (恶心) in, say, Brussels, the heart of the European Union. Our Charlemagne columnist writes of Charleroi, its second airport: “It is dirty and crowded, and has terrible food. The planes leave and land at unreasonble hours. And the only real way into town is a coach that runs every 30 minutes and is frequently overbooked: more than once I’ve queued in the rain only to see it drive off as I reach the front.”51.The last sentence of the first paragraph implies that _______.A.each bad airport is uniqueB.good airports are hard to findC.awful airports have a lot in commonD.the world’s best airports are not that good52.Lubumbashi airport is mentioned in paragraph 3 in order to _______.A.explain how delay occurs in African airportsB.illustrate how the four themes are interrelatedC.argue against the necessity of airport security officialsD.give an example of what $1 means to people in Congo53.The phrase “choke points” (paragraph 4) is closet in meaning to “_______”.A.agents B.passengers C.stores D.barriers 54.What can be learned about Charleroi?A.It is located in a rich country.B.It used to be dirty and crowded.C.It used to be close to the city center.D.It is the country’s second largest airport.Yvonne Morones, who has a pet dog named Scamp the Tramp, is talking with the Pet Journal about Scamp.76When I saw Scamp on Petfinder, all of a sudden I understood what love was like. I suddenly found I loved him because I fell in love with his face.What did you know about him when you adopted him?He’d been living on the street in Compton, California, and people were feeding him McDonald’s. And his name was Muffin Man, which didn’t seem to fit him at all.What do you do to give him that bed head look?It’s au naturel! He does get a mango shampoo and a coconut conditioner, but these gray dreads just appear on his head, back, and tail. The pet hairdresser says his hair is uncontrollable.Scamp works with you as a social therapy dog too. How do people react to his unusual looks?The first time he went to the senior center with me, the seniors just laughed and said, “Yvonne, what have you got there?” Then they fell in love with him. They’ll even write little adventure stories about Scamp. He just inspires people.Has being awarded the world’s ugliest dog changed him?He’s no longer Scamp the Tramp. He’s now Scamp the Champ. Now I have to get him a new dog tag.55.What can be learned about Scamp the Tramp?A.He loves fast food.B.He looks unattractive.C.He dislikes his new tag.D.He used to live with seniors.A.Why did you go to Petfinder so often?B.What did you often find on Petfinder?C.Who brought you Scamp the Tramp?D.How did you first meet Scamp the Tramp?57.By “It’s au naturel!”, Yvonne means that ______.A.Scamp turned gray after being adopted by herB.Scamp is fond of being washed with shampooC.nothing has been done to change Scamp’s appearanceD.it is the pet hairdresser who has given Scamp his new lookOn January 15th, the Guardian showed off its new, smaller look, shifting from its distinctive “Berliner” format to a tabloid(娱乐小报) shape with a redesigned logo in black type. But the more dramatic makeover is of the financial books of Guardian Media Group (GMG), publisher of the Sunday Observer and the daily Guardian, which may find its new operation in the black next financial year. A newspaper business that two years ago was threatened with existentially worrying losses appears on the edge of breaking even.The turnaround is partly due to steep cost-cutting, which is a dog-bites-man story in journalism. But the Guardian would manage the achievement while still giving away news free online, and that is a story worth telling.In January 2016 David Pemsel, the new chief executive of GMG, and Katharine Viner, the new editor-in-chief of the Guardian, informed staff that GMG’s endowment fund, meant to ensure the financial security of the paper in the long run, had lost £100m ($140m) in just half a year, taking it to £740m. Mr Pemsel was advised by industry peers to cut costs and put online news behind a paywall. He and Ms Viner cut costs by 20%, or more than £50m. Alan Rusbridger, Ms Viner’s predecessor(前任), had led the newspaper to global relevance with a large online readership. But he spent without thinking of the consequences. In two years GMG has reduced its employees by 400, to about 1,500.Yet unlike a growing number of newspapers, the Guardian has not put up a paywall. Instead it has pursued a membership model, asking online readers to contribute whatever they like. About 600,000 now do, with annual payments or one-off amounts. American readers tend to choose the latter option, Ms Viner says. GMG says the total figure amounts to tens of millions of pounds per year. Ms Viner says revenue from readers (including 200,000 print subscribers) is now greater than revenue from advertisers.The result is steadily declining operating losses: from £69m two years ago to £45m last financial year and, Mr Pemsel says, less than £25m in the year that ends on April 1st. He predicts breaking even next year. Giving up its own printing presses and going tabloid willhelp, saving several million pounds a year. The Guardian may now physically look more like its peers, but its turnaround story remains distinctive.58.The phrase “in the black”(Paragraph 1) most probably means ______.A.making profit B.taking on a new lookC.losing support D.enjoying great popularity59.What does the writer imply about Alan Rusbridger?A.He advised GMG to cost costs.B.He got the Guardian into trouble.C.He was the founder of GMG’s endowment fund.D.He was fired due to his failure to bring the Guardian online.60.What does the Guardian allow its online readers to do?A.To pay as they like.B.To skip the advertisements.C.To join its membership club.D.To connect to other newspapers. 61.What conclusion can be drawn from the passage?A.The Guardian has been reduced to a tabloid.B.The Guardian succeeds by giving away news free online.C.The Guardian turns around by looking more likes its peers.D.The Guardian has broken even by cutting its operation costs.五、六选四What a chatbot can teach us about the art of conversation After lunchtime on May 2, 1989, a student at Drake University in Iowa started an online text chat with a user at University College Dublin. The UCD user’s handle was “MGonz”.Over the next hour and 20 minutes, the two exchanged insults (谩骂). When the student logged off, he wrote off MGonz as an abusive troll (山怪). But while MGonz was abusive, it was not a troll—it was a simple chatbot programmed by UCD undergrad Mark Humphrys. The next day, Humphrys reviewed the chat logs in astonishment. His MGonz chatbot had passed the Turing test.The Turing test was invented by the mathematician, codebreaker and computing pioneer Alan Turing in 1950. 62 Turing’s test is a benchmark for artificial intelligence—but I am less interested in the test itself than in the moral of the story of MGonz’s success.Faced with the difficult task of convincing a human that a chatbot is human, the obvious strategy is to increase the sophistication (复杂) of the chatbot, Humphrys try an alternative: reduce the sophistication of the human. MGonz had passed the Turing test, but is it not also fair to say that the student had failed it? 63 These are all things that any chatbot finds hard.But MGonz generates dialogue because insults need neither context nor memory. And it is impossible to read the MGonz transcript without thinking of ugly parallels on social media.We are at our best when our conversation explores complex issues and is sensitive to context. 64 The qualities that distinguish us from MGonz are the qualities that get driven out by a fast-moving, soundbite-driven world.Brian Christian’s book The Most Human Human explores the history of chatbots, while reflecting on the nature of good conversation. Christian argues that chatbots tend to pass for human because we humans set the benchmark so low. 65 No wonder the chatbots find us easy to imitate.Conversation is not easy. But the best conversations are delightful. So let’s start by promising to do better than MGonz and see what we can build from there.A.So many of our interactions are predictable or downright rude.B.A good conversation involves give and take, builds over time and exists in a context.C.That is understandable response to the limited range of modern communication.D.But complexity and context do not play well on social media.E.The test is simply for a computer to successfully pretend to be a human in a text-based conversation with another human.F.Turing had something more uplifting in mind than MGonz’s exchange with the student.六、用单词的适当形式完成短文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 word; for the other blanks, use one word that best fits each blank.In the 19th century, thousands of poor people struggled to survive in the slums of London, where diseases spread 66 the dense population. At one time, 600 people diedof cholera (霍乱) in a week, a fatal disease then 67 (suppose) to be spread by airborne germs.A young doctor, John Snow, anxious to help, obtained a map of the slums, on 68 he marked the buildings where each person died. He soon noticed that most of the deaths occurred around the center of the circle. The number of deaths 69 (be) greatest around the center of the circle and then decreased 70 the distance from the center of the circle increased.Snow concluded that at the center there 71 be somebody or something that was causing or spreading cholera. When he went to the district, he saw a pump bringing water up from an open well, which was the sole source of water for people there. He examined a sample of water to see what it was. Then he suspected the water was contaminated, so he took the handle of the pump away, thus 72 (stop) people from drinking water from the well. Snow urged that the city authorities 73 (investigate) the water in the well. At first they were reluctant to spend money on 74 had not yet proved to be a danger, but finally they found that the wall of the well cracked in several places and that the raw sewage was seeping(渗漏) into it. Then the men filled the cracks and dug another well 75 clean water could be obtained. The cholera slackened and then disappeared.七、汉译英(整句)76.这个孩子不太会因为考试的好坏而承受巨大压力。
2023-2024学年上海市复旦大学附属中学高一下学期期末考试化学试卷
![2023-2024学年上海市复旦大学附属中学高一下学期期末考试化学试卷](https://img.taocdn.com/s3/m/e725f147cd7931b765ce0508763231126edb77e2.png)
2023-2024学年上海市复旦大学附属中学高一下学期期末考试化学试卷1.冬奥会上短道速滑运动员使用的速滑冰刀的材质是合金钢。
下列关于合金钢的说法正确的是A.熔点一般比纯铁低B.硬度一般比纯铁小C.含碳量比生铁高D.延展性比生铁差2.吴老师用打磨过的铝片进行如图实验,下列分析不合理的是A.①中铝片发生了钝化B.②中产生的气泡是氮氧化合物,不是氢气C.③中沉淀溶解的离子方程式是D.将铝片换成铁片,实验现象类似3.铝热反应释放大量的热,常用于冶炼难熔金属。
用铝粉和 Fe2O3做铝热反应实验,需要的试剂还有A.KCl B.KClO 3C.MnO 2D.Mg4.小韩同学进行上述铝热反应实验得到了一块黑灰色物质。
为证明其中含有金属铝,她设计了一个简单实验。
该实验所用试剂是___________(填化学式),该试剂与金属铝反应的离子方程式为___________。
5.小金同学推测灰黑色物质中还含有Fe2O3,为此他设计如下实验:取一小块样品投入到稍过量稀硫酸中,向反应后的混合液中滴加物质甲的溶液,观察到溶液颜色未变红,证明黑灰色物质中不含Fe2O3。
则物质甲是___________(填化学式)。
小金的实验方案是否合理?___________(填“合理”或“不合理”)。
理由:___________(用离子方程式说明)。
6.因在航空、核能、电池等高技术领域的重要作用,锂被称为“21世纪的能源金属”。
(1)氢化锂(LiH)中负离子半径大于正离子半径,其原因是___________。
(2)下列关于碱金属元素及其单质的叙述中,正确的是___________。
A.金属钠着火,要用泡沫灭火剂扑灭B.水溶液中正离子氧化性:C.单质熔点: Li<Na<K<Rb D.与水反应的能力: Na<K(3)Li2O2与Na2O2的组成与性质相似,小郭同学将Li2O2投入水中,产生大量气体,请写出化学方程式___________。
上海复旦大学附属中学2024届物理高一第二学期期末综合测试试题含解析
![上海复旦大学附属中学2024届物理高一第二学期期末综合测试试题含解析](https://img.taocdn.com/s3/m/a8da126766ec102de2bd960590c69ec3d5bbdb89.png)
上海复旦大学附属中学2024届物理高一第二学期期末综合测试试题注意事项:1.答题前,考生先将自己的姓名、准考证号码填写清楚,将条形码准确粘贴在条形码区域内。
2.答题时请按要求用笔。
3.请按照题号顺序在答题卡各题目的答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试卷上答题无效。
4.作图可先使用铅笔画出,确定后必须用黑色字迹的签字笔描黑。
5.保持卡面清洁,不要折暴、不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。
一、选择题:(1-6题为单选题7-12为多选,每题4分,漏选得2分,错选和不选得零分)1、在奥运比赛项目中,高台跳水是我国运动员的强项。
质量为m的跳水运动员进入水中后受到水的阻力而做减速运动,设水对他的阻力大小恒为F,那么在他减速下降高度为h的过程中,下列说法正确的是(g为当地的重力加速度)A.他的动能减少了FhB.他的机械能减少了FhC.他的机械能减少了(F-mg)hD.他的重力势能增加了mgh2、下列有关静电学的说法正确的是A.摩擦起电和感应起电的本质是产生了新的电子B.电荷量是不能连续变化的物理量C.电场力做功与移动电荷的路径有关D.沿着电场线方向电势逐渐升高3、(本题9分)小船匀速过河,己知船在静水中的速度为,水流速度为,河宽为,以下说法不正确...的是()A.当过河时间为时,小船的船头恰好垂直于河岸B.当小船船头斜向上游方向时,可能使小船的过河位移为C.当1v变为5m/s时,小船不能到达正对岸D.若小船船头始终指向正对岸,则小船在过河过程中2v突然变大,过河时间将变长4、下列关于电场强度、电势、电场线的说法中,正确的是A.电场强度为零的地方,电势一定为零B.沿着电场线方向,电势逐渐降低C.电场强度较大的地方,电势一定较高D.电势为零的地方,电场强度一定为零5、一轻杆一端固定质量为m的小球,以另一端O为圆心,使小球在竖直平面内做半径为R的圆周运动,如图所示,则下列说法正确的是A.小球过最高点的最小速度是gRB.小球过最高点时,杆所受到的弹力可以等于零C.小球过最高点时,杆对球的作用力一定随速度增大而增大D.小球过最高点时,杆对球的作用力一定随速度增大而减小6、(本题9分)汽车在水平公路上转弯,沿曲线由M向N行驶. 下图中分别画出了汽车转弯时所受合力F的四种方向,你认为正确的是A.B.C.D.7、(本题9分)在光滑水平面上A、B两小车中间有一弹簧,如图所示,用手抓住小车并将弹簧压缩后使小车处于静止状态.将两小车及弹簧看做一个系统,下面说法正确的是A.两手同时放开后,系统总动量始终为零B.先放开左手,再放开右手后,在弹簧伸缩的过程中,动量不守恒C.先放开左手,再放开右手后,在弹簧伸缩的过程中,动量守恒且总动量向右D.无论何时放手,两手放开以后,在弹簧伸缩的过程中,系统总动量都保持不变,但系统的总动量不一定为零8、如图所示,将物体从一定高度水平抛出(不计空气阻力),物体运动过程中离地面高度为h时,物体水平位移为x、物体的机械能为E、物体的动能为E k、物体运动的速度大小为v.以水平地面为零势能面.下列图像中,能正确反映各物理量与h的关系的是().A.B.C.D.9、(本题9分)如图为质量相等的两个质点A、B在同一直线上运动的v-t图像.由图可知( )A.在t时刻两个质点在同一位置B.在t时刻两个质点速度相等C.在0-t时间内质点B比质点A位移大D.在0-t时间内合外力对两个质点做功相等10、(本题9分)如图是一汽车在平直路面上启动的速度-时间图像,t1时刻起汽车的功率保持不变.由图像可知()A.0~t1时间内,汽车的牵引力增大,加速度增大,功率不变B.0~t1时间内,汽车的牵引力不变,加速度不变,功率增大C.t1~t2时间内,汽车的牵引力减小,加速度减小D.t1~t2时间内,汽车的牵引力不变,加速度不变11、(本题9分)图中a、b是两个点电荷,它们的电荷量分别为Q1、Q2,MN是ab连线的中垂线,P是中垂线上的一点。
2024届上海市复旦大学附属中学浦东分校数学高一第二学期期末综合测试试题含解析
![2024届上海市复旦大学附属中学浦东分校数学高一第二学期期末综合测试试题含解析](https://img.taocdn.com/s3/m/968316d7bdeb19e8b8f67c1cfad6195f312be8f7.png)
2024届上海市复旦大学附属中学浦东分校数学高一第二学期期末综合测试试题注意事项:1.答题前,考生先将自己的姓名、准考证号码填写清楚,将条形码准确粘贴在条形码区域内。
2.答题时请按要求用笔。
3.请按照题号顺序在答题卡各题目的答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试卷上答题无效。
4.作图可先使用铅笔画出,确定后必须用黑色字迹的签字笔描黑。
5.保持卡面清洁,不要折暴、不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。
一、选择题:本大题共10小题,每小题5分,共50分。
在每个小题给出的四个选项中,恰有一项是符合题目要求的1.已知{}n a 为递增等比数列47565,6a a a a +==,则110a a +=() A .152B .5C .6D .3562.数列{}n a 的通项公式为1(21)(21)n a n n =-⋅+,则数列{}n a 的前100项和100S =( ). A .200201B .200401C .100201D .1004013. 数列{a n }的通项公式是a n =(n +2)910n⎛⎫⎪⎝⎭,那么在此数列中( )A .a 7=a 8最大B .a 8=a 9最大C .有唯一项a 8最大 D .有唯一项a 7最大4.过△ABC 的重心任作一直线分别交边AB ,AC 于点D 、E .若AD xAB =,AE y AC =,0xy ≠,则4x y +的最小值为( )A .4B .3C .2D .15.如果角θ的终边经过点21⎛⎫⎪ ⎪⎝⎭,那么tan θ的值是( )A .12B .CD .6.已知数列{}n a 是首项为2,公差为4的等差数列,若2022n a =,则n = ( ) A .504B .505C .506D .5077.设的内角A ,B ,C 所对的边分别为a ,b ,c ,且6C π=,12a b +=,面积的最大值为() A .6B .8C .7D .98.设集合{}0,1,2,3A =,集合{}1,1B =-,则A B =( )A .{}1,1-B .{}1C .{}1,0-D .{}1,01-,9.在边长为1的等边三角形ABC 中,D 是AB 的中点,E 为线段AC 上一动点,则EB ED ⋅的取值范围为( ) A .233,162⎡⎤⎢⎥⎣⎦ B .233,644⎡⎤⎢⎥⎣⎦ C .23,316⎡⎤⎢⎥⎣⎦D .233,642⎡⎤⎢⎥⎣⎦ 10.为了研究某药品的疗效,选取若干名志愿者进行临床试验,所有志愿者的舒张压数据(单位:kPa )的分组区间为[12,13),[13,14),[14,15),[15,16),[16,17],将其按从左到右的顺序分别编号为第一组,第二组,,第五组,如图是根据试验数据制成的频率分布直方图,已知第一组与第二组共有20人,第三组中没有疗效的有6人,则第三组中有疗效的人数为( )A .6B .8C .12D .18二、填空题:本大题共6小题,每小题5分,共30分。
上海市复旦大学附属中学2023-2024学年高一下学期期末考试数学试卷(A)
![上海市复旦大学附属中学2023-2024学年高一下学期期末考试数学试卷(A)](https://img.taocdn.com/s3/m/df2eaa8da48da0116c175f0e7cd184254b351b09.png)
上海市复旦大学附属中学2023-2024学年高一下学期期末考
试数学试卷(A)
学校:___________姓名:___________班级:___________考号:___________
20.(1)1,2,3,4,6{}
5,A =(2)证明见解析
(3)9
【分析】(1)根据题意列举即可求解;
(2)根据数学归纳法即可证明;
(3)设项数最小值为m ,有2||C 27m
A =³,求出m 的取值范围,扩大数列的d 的最快方法:取除数列中最大值n a 外的k a ,使[0,]k a ÇZ 为数列中的最长连续自然数子列,给数列{}n a 末尾加入一个11n k n a a a +=++,可使d 张至21k n a a ++,据此即可求解.
【详解】(1)1,2,3,4,6{}5,A =;
(2)因为2201S a =+<,所以323123
22S a a a -=++>=,由数学归纳法可设22n n
S ->,因为111122k k k k k S a S S +-++=+>=,
所以原假设正确,所以对任意不小于3的正整数n ,不等式22n n
S ->都成立;(3)设项数最小值为m ,有2||C 27m
A =³,。
上海市复旦大学附属中学2022-2023学年高一下学期期末数学试题
![上海市复旦大学附属中学2022-2023学年高一下学期期末数学试题](https://img.taocdn.com/s3/m/d0a53966302b3169a45177232f60ddccda38e6fd.png)
上海市复旦大学附属中学2022-2023学年高一下学期期末数学试题学校:___________姓名:___________班级:___________考号:___________二、单选题13.已知平面a 和平面b ,直线,m n b Ì,则“//m a 且//n a ”是“//a b ”的( ).A .充分非必要条件B .必要非充分条件C .充要条件D .既非充分也非必要条件14.下面五个命题:(1)两组对边分别相等的四边形是平行四边形;(2)两组对边分别平行的四边形是平行四边形;(3)四个角都是直角的四边形是矩形;(4)有两个侧面是矩形的三棱柱是直三棱柱;(5)有两个侧面是矩形的四棱柱是直四棱柱其中真命题的个数是( )A .1B .2C .3D .415.过正四面体ABCD 的顶点A 作截面,若满足①截面是等腰三角形;②截面与底面BCD 成75°的二面角,这样的截面个数为( )A .6B .12C .18D .24三、解答题17.如图,在四棱锥11A BCCB -中,平面ABC ^平面11BCC B ,ABC V 是正三角形,四边形11BCC B 是正方形,D 是AC 的中点.(1)求证:1AB //平面1BDC ;(2)求直线BC 和平面1BDC 所成角的大小18.如图,已知点P 在圆柱1OO 的底面圆O 上,AB 为圆O 的直径,圆柱1OO 的表面积为24π,2OA BP ==.当直线PM满足它的射影PQ在a¢、b¢所成角的平分线上时,PM与a¢所成的角等于PM与b¢所成的角,设直线PM与a¢、b¢所成角为q,因为直线a、b所成角为80o,得a¢、b¢所成锐角为80o,①当直线PM的射影PQ在a¢、b¢所成锐角的平分线上时,则PM与a¢、b¢所成角的范围是4090qo o,££这种情况下,过P点有2条直线与a、b所成角都是50o;②当直线PM的射影PQ在a¢、b¢所成钝角的平分线上时,PM与a¢、b¢所成角的范围是5090qo o,££这种情况下,过P点有且仅有1条直线(即PM aÌ时)与a、b所成角都是50o.综上所述,过P点且与a、b所成角都是50o的直线有3条.故答案为:3.9.[]-9,11【分析】建立直角坐标系表示各点坐标,再利用三角恒等变换即可求出结果.如图3,则ABMV≌ACN△,所以AM所以截面AMN为符合条件的截面三角形,则这样的三角形有综上,符合条件的截面个数有故选:C16.B分析】可证明0x>时,有理可判断(2)的正误,利用(4)的正误.【详解】先证明一个不等式:证明:若1x>,则1sinx>³,如图,在单位圆中,的联系,找出其中的量的关系.。
2022-2022年上海市复旦附中高一第二学期期末数学试卷〔精品解析
![2022-2022年上海市复旦附中高一第二学期期末数学试卷〔精品解析](https://img.taocdn.com/s3/m/91af59c40408763231126edb6f1aff00bed57073.png)
2022-2022年上海市复旦附中高一第二学期期末数学试卷〔精品解析第1页(共15页)2022-2022学年上海市复旦附中高一第二学期期末数学试卷一.填空题1.(3分)在等差数列{an}中,若a4=0,a6+a7=10,则a7=2.(3分)在数列1、3、7、15、…中,按此规律,127是该数列的第项3.(3分)已知数列{an}的前n项和Sn=n2﹣1,那么数列{an}的通项公式为4.(3分)若在等比数列{an}中,a1a2…a9=512,则a5=5.(3分)方程(3co某﹣1)(co某+in某)=0的解集是6.(3分)若数列{a7.(3分)若数列{a是等差数列,则数列(8.(3分)观察下列式子:,,,…,你可.9.(3分)在我国古代数学著作《孙子算经》中,卷下第二十六题是:今有物,不知其数,三三数之剩二,五五数之剩三,七七数之剩二,问物几何?满足题意的答案可以用数列表示,该数列的通项公式可以表示为an=10.(3分)对于下列数排成的数阵:它的第10行所有数的和为11.(3分)对于数列{an}满足:a1=1,an+1﹣an∈{a1,a2,…,an}(n∈N某),其前n项和为Sn,记满足条件的所有数列{an}中,S12的最大值为a,最小值为b,则a﹣b=12.(3分)设n∈N某,用An表示所有形如++…+的正整数集合,其中0≤r1<r2<…<rn≤n,且ri∈N(i∈N 某),bn为集合An中的所有元素之和.则{bn}的通项公式为bn =.二.选择题13.(3分)“b是与的等差中项”是“b是与的等比中项”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14.(3分)在数列{an}中,a1=1,a2=64,且数列是等比数列,其公比,则数列{an}的最大项等于()A.a7B.a8C.a6或a9D.a1015.(3分)若数列,若k∈N某,则在下列数列中,可取遍数列{an}前6项值的数列为()A.{a2k+1}B.{a3k+1}C.{a4k+1}D.{a5k+1}16.(3分)数列{an}中,若a1=a,,n∈N某,则下列命题中真命题个数是()(1)若数列{an}为常数数列,则a=±1;(2)若a∈(0,1),数列{an}都是单调递增数列;(3)若aZ,任取{an}中的9项,,,…,(1<k1<k2<…<k9)构成数列{an}的子数列{},n=1,2,…,9,则{}都是单调数列.A.0个B.1个C.2个D.3个三.解答题17.已知{an}是一个公差大于0的等差数列,且满足a4a6=96,a3+a7=20,数列{bn}满足等式:(n∈N某).(1)求数列{an}的通项公式;(2)求数列的前n项和Sn.18.已知b、c为常数且均不为零,数列{an}的通项公式为an=,并且a1、a3、a2成等差数列,a1、a2、a4成等比数列.(1)求b、c的值;(2)设Sn是数列{an}前n项的和,求使得不等式S2n>20222成立的最小正整数n.19.王某2022年12月31日向银行贷款100000元,银行贷款年利率为5%,若此贷款分十第2页(共15页)。
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
上海市复旦附中2023高一下学期期末测
验(含答案)
本次测验共分为四个部分,分别是语文、数学、英语和综合素养。
语文部分
一、选择题
1. 下列四句诗句中,表达了抒情感受的是()。
A. 屭(xì)日炎炎走江潭
B. 岁暮阴阳催短景
C. 君问归期未有期
D. 春心莫共花争发
答案:D
2. 识别下列成语中,正确的一项是:
A. 人善被人欺,马善被人骑
B. 朝令夕改,夕令朝改
C. 落花无言,流水无情
D. 金玉其外,败絮其中
答案:D
二、填空题
1. 我们难以想象在那个遥远的山区,他们不仅缺少医疗卫生设备以及__空气__,而且连基本的文化生活也无从谈起。
2. 康庄大道的盛况未必有多少人会知道,只有在黄金岁月中留下的__风土人情__才令人过目难忘。
三、解答题
请根据你的实际文学阅读情况回答下列问题。
1. 你认为《老人与海》中最为感人的情节是什么?
答案略。
数学部分
一、选择题
1. 已知$\log_3\frac{15}{2}=a,\log_5\frac{2}{3}=b$,求$ab$的值。
A. $\frac{1}{2}$
B. $\frac{2}{3}$
C. $\frac{5}{6}$
D. $\frac{2}{5}$
答案:B
2. 返回天津市出租车起步价为14元,每公里2.5元的信息,若小刚乘坐的出租车行驶4千米,则小刚需要付多少钱?
A. 16.50元
B. 17.50元
C. 18.50元
D. 19.50元
答案:D
二、填空题
1. 已知甲、乙、丙三人,若甲比乙多$200$元,丙比甲少$100$元,则三人的财产总和为$\underline{\hspace{1cm}}$元。
答案:$400$。
2. 已知$a-b=3-bc,b^2+c^2=1$,则$a^2-2ab+2b^2$的值为$\underline{\hspace{1cm}}$。
答案:$2$。
三、解答题
已知集合$A=\{x|x=\frac{m}{2n+1},m,n\in
\mathbb{N^{*}},m+n\leq 4\}$,试回答下列问题。
1. 求集合$A$中元素的个数。
答案略。
英语部分
一、选择题
1. ---It's terribly hot in the room.Could you let me have the window________?
---With pleasure.
A. opened
B. be opened
C. to open
D. open
答案:A
2. If you work harder, you___________ promoted.
A. were
B. will be
C. would be
D. had been
答案:B
二、填空题
1. She failed in the examination, ________ made her parents very disappointed.
答案:which。
2. Listening to music ________ one of my favorite ways to kill time.
答案:is。
三、解答题
请根据你所掌握的阅读情况回答下列问题。
1. Can science grow more selfish?
答案略。
综合素养部分
一、选择题
1. 在南极,一年四季差不多都是__白天__。
A. 黑夜
B. 白天
C. 晚上
D. 早晨
答案:B
2. 中国五千年的文明中,与日本的文化一起流传到世界各地的,包括了__墨汁__、__绘画__、__茶道__等等。
A. 墨汁、手工艺品、茶道
B. 音乐、书法、美术
C. 墨汁、绘画、茶道
D. 茶道、武术、细胞学
答案:C
二、填空题
1. __红十字会__是国际红十字运动的中心机构,由国际委员会、联合会和国际基金会组成。
答案:红十字会。
2. 一切____源于____,人生本因爱而精彩。
答案:美好,爱。
三、解答题
请回答“自己的收入能否代表一个人的价值”这个问题,并简述理由。
答案略。
以上为答案,仅供参考。
希望同学们能够取得好成绩。