复旦附中高一下期中(2019.4)

复旦大学附属中学第二学期 高一年级数学期中考试试卷考试时间120分钟满分150分所有答案均写在答题纸上一、填空题（本大题共有12题，满分54分，第1-6题每题4分，第7-12题每题5分）1．函数()3sin 26f x x π⎛⎫=-⎪⎝⎭的最小正周期为__________．2．α为第三象限角，且coscos22αα=-，则2α在第_______象限． 3．已知扇形的周长为20cm ，面积为216cm ，则扇形的圆心角α的弧度数为_________．4．函数2()log (tan f x x =的定义域为_________． 5．方程1sin 3x =在3,22ππ⎡⎤⎢⎥⎣⎦内的解为x =_________（用反三角函数表示） 6．已知奇函数()f x 的一个周期为2，当(0,1)x ∈时，()cos 3xf x π=，则(7.5)f =_______．7．已知角α满足1sin 63πα⎛⎫-= ⎪⎝⎭，则sin 26πα⎛⎫+= ⎪⎝⎭_______． 8．函数()cos 23f x x π⎛⎫=-⎪⎝⎭的单调递增区间为_________． 9．函数2()cos sin 1f x x x =++在7,46ππ⎛⎤⎥⎝⎦上的值域是_______． 10．已知()4sin 3cos ,()f x x x f x =+向右平移(0)ααπ<<个单位后为奇函数，则tan α=_______．11．我们知道函数的性质中，以下两个结论是正确的：①偶函数()f x 在区间[,]()a b a b <上的取值范围与在区间[,]b a --上的取值范围是相同的；②周期函数()f x 在一个周期内的取值范围也就是()f x 在定义域上的值域，由此可求函数()|sin |cos |g x x x =的值域为_________．12．已知定义在R 上的奇函数，满足(2)()0f x f x -+=，当(0,1]x ∈时，2()log f x x =-，若函数()()sin F x f x x π=-，在区间[2,]m -上有2021个零点，则m 的取值范围是___________．二、选择题（本大题共有4题，满分20分，每题5分）每题有且只有一个正确选项．13．已知函数()2sin(2)f x x ϕ=+，则“2πϕ=”是“()f x 为偶函数”的（ ）条件．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．在ABC 中，2sin 22A c bc-=，则ABC 的形状为（ ） A ．正三角形 B ．直角三角形 C ．等腰直角三角形 D ．等腰三角形 15．设定义在R 上的函数()sin 3f x x π⎛⎫=+⎪⎝⎭．则()f x （ ） A ．在区间27,36ππ⎡⎤⎢⎥⎣⎦上是增函数 B ．在区间,2ππ⎡⎤--⎢⎥⎣⎦上是减函数C ．在区间,84ππ⎡⎤⎢⎥⎣⎦上是增函数 D ．在区间5,36ππ⎡⎤⎢⎥⎣⎦上是减函数 16．设函数()2sin()1(0)f x x ωϕω=+->，若对于任意实数ϕ，()f x 在区间3,44ππ⎡⎤⎢⎥⎣⎦上至少有2个零点，至多有3个零点，则ω的取值范围是（ ） A ．816,33⎡⎫⎪⎢⎣⎭ B ．164,3⎡⎫⎪⎢⎣⎭ C ．204,3⎡⎫⎪⎢⎣⎭ D ．820,33⎡⎫⎪⎢⎣⎭三、解答题（本大题共有5题，满分76分）解答下列各题必须在答题纸的相应位置写出必要的步骤．17．（本题满分14分，第1小题满分7分，第2小题满分7分）在平面直角坐标系xOy 中，角θ的始边为x 轴正半轴，终边在第二象限且与单位圆交于点P ． （1）若点P 的横坐标为35-，求cos 3sin 3sin cos θθθθ+-的值．（2）若将射线OP 绕点O 逆时针旋转4π，得到角α，若1tan 3α=，求tan θ的值． 18．（本题满分14分，第1小题满分6分，第2小题满分8分）已知函数()sin()(0,0,02)f x A x A ωϕωϕπ=+>><<的部分图像如图所示．（1）求函数()f x 的解析式；（2）若()(),0,64h x f x f x x ππ⎛⎫⎡⎤=⋅-∈ ⎪⎢⎥⎝⎭⎣⎦，求()h x 的取值范围．19．（本题满分14分，第1小题满分6分，第2小题满分8分） 如图某公园有一块直角三角形ABC 的空地，其中,,26ACB ABC AC ππ∠=∠=长a 千米，现要在空地上围出一块正三角形区域DEF 建文化景观区，其中D E F 、、分别在BC AC AB 、、上．设DEC θ∠=．（1）若3πθ=，求DEF 的边长；（2）当θ多大时，DEF 的边长最小？并求出最小值．20．（本题满分16分，第1小题满分4分，第2小题满分6分，第3小题满分6分） 已知数2()32sin 1(0)6212x f x x πωπωω⎛⎫⎛⎫=+++-> ⎪ ⎪⎝⎭⎝⎭的相邻两对称轴间的距离为2π．（1）求()f x 的解析式； （2）将函数()f x 的图像向右平移6π个单位长度，再把横坐标缩小为原来的12（纵坐标不变），得到函数()y g x =的图像，当,126x ππ⎡⎤∈-⎢⎥⎣⎦时，求函数()g x 的值域． （3）对于第（2）问中的函数()g x ，记方程4()3g x =在4,63x ππ⎡⎤∈⎢⎥⎣⎦，上的根从小到依次为12,,n x x x ，试确定n 的值，并求1231222n n x x x x x -+++++的值．21．（本题满分18分，第1小题满分4分，第2小题第（ⅰ）问满分6分，第2小题第（ⅱ）问满分8分） 在非直角三角形ABC 中，角,,A B C 的对边分别为,,a b c ． （1）若2a c b +=，且3B π=，判断三角形ABC 的形状；（2）若(1)a c mb m +=>，（ⅰ）证明：1tantan 221A C m m -=+； （可能运用的公式有sin sin 2sincos22αβαβαβ+-+=）（ⅱ）是否存在函数()m ϕ，使得对于一切满足条件的m ，代数式cos cos ()()cos cos A C m m A Cϕϕ++恒为定值？若存在，请给出一个满足条件的()m ϕ，并证明之；若不存在，请给出一个理由．复旦大学附属中学第二学期 高一年级数学期中考试试卷考试时间120分钟 满分150分 所有答案均写在答题纸上一、填空题（本大题共有12题，满分54分，第1-6题每题4分，第7-12题每题5分） 1．【答案】：π【解析】：函数()3sin 26f x x π⎛⎫=-⎪⎝⎭的最小正周期22T ππ==． 2．【解析】：因为α以为第三象限角，所以322,2k k k Z πππαπ+<<+∈，所以3,224k k k Z παπππ+<<+∈又因为cos cos 22αα=-，所以cos 02α<，所以在第二象限．3．【答案】：12【解析】：设扇形半径为r ，弧长为l ，所以2201162l r lr +=⎧⎪⎨=⎪⎩，解得162l r =⎧⎨=⎩或48l r =⎧⎨=⎩，所以8l r α==（舍）或12l r α==所以扇形的圆心角α的弧度数为124．【答案】：,32xk x k k ππππ⎧⎫+<<+∈⎨⎬⎩⎭Z【解析】：根据题意得，tan 0x ->，即tan x >22,32k x k k Z ππππ+<<+∈所以函数2()log (tan f x x =的定义域,32x k x k k ππππ⎧⎫+<<+∈⎨⎬⎩⎭Z5．【答案】：1arcsin3π- 6．【答案】：2-【解析】：根据题意得，(7.5)(0.58)(0.5)(0.5)cos 6f f f f π=-+=-=-=-=7．【答案】：79【解析】：227sin 2cos 2cos 212sin 16263699πππππαααα⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫+=-+=-=--=-= ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭8．【答案】：,,36k k k ππππ⎡⎤-++∈⎢⎥⎣⎦Z【解析】：()cos 2cos 233f x x x ππ⎛⎫⎛⎫=-=-⎪ ⎪⎝⎭⎝⎭，所以222,3k x k k Z ππππ-≤-≤∈，解得,36k x k k Z ππππ-+≤≤+∈，所以单调递增区间为,,36k k k ππππ⎡⎤-++∈⎢⎥⎣⎦Z9．【解析】：222()cos sin 21sin sin 1sin sin 2f x x x x x x x =++=-++=-++ 令7sin ,,46t x x ππ⎛⎤=∈⎥⎝⎦，则1,12t ⎡⎤∈-⎢⎥⎣⎦，2()2f t t t =-++ 所以22minmin 115119()2,()2224224f t f t ⎛⎫⎛⎫=---+==-++= ⎪ ⎪⎝⎭⎝⎭所以59(),44f t ⎡⎤∈⎢⎥⎣⎦，所以59(),44f x ⎡⎤∈⎢⎥⎣⎦，所以函数2()cos sin 1f x x x =++在7,46ππ⎛⎤ ⎥⎝⎦上的值域是59,44⎡⎤⎢⎥⎣⎦10．【答案】：34【解析】：由根据题意可得，函数()4sin 3cos 5sin()f x x x x ϕ=+=+，其中3tan 4ϕ= 因为()f x 向右平移α个单位后，可得()5sin[()]5sin()g x x x αϕαϕ=-+=-+，又由()5sin()g x x αϕ=-+为奇函数，所以,k k Z ϕαπ-=∈，即,k k Z αϕπ=-∈， 又因为0απ<<，所以αϕ=，所以3tan tan 4αϕ==．11．【答案】：[1,2]【解析】：()|sin |cos |g x x x =是偶函数，周期为π．当[0,]x π∈时，sin ,0,2()|sin ||cos |sin ,,2x x x g x x x x x x πππ⎧⎛⎫⎡⎤+∈⎪ ⎪⎢⎥⎣⎦⎪⎝⎭==⎨⎛⎫⎡⎤⎪-∈ ⎪⎢⎥⎪⎣⎦⎝⎭⎩所以2sin 0,32()2sin ,32x x g x x x πππππ⎧⎛⎫⎛⎫⎡⎤+∈ ⎪⎪ ⎪⎢⎥⎝⎭⎣⎦⎪⎝⎭=⎨⎛⎫⎛⎫⎡⎤⎪-∈ ⎪ ⎪⎢⎥⎪⎝⎭⎣⎦⎝⎭⎩，所以()g x 值域为[1,2]12．【解析】：由题意，函数()f x 为R 上奇函数，所以(0)0f =，且()()f x f x -=-， 又(2)()0f x f x -+=，可得(2)()f x f x -=-，可得函数()f x 的图像关于点(1,0)对称， 联立可得(2)()f x f x -=-，所以()f x 是以2为周期的周期函数， 又由函数sin y x π=的周期为2，且关于点(,0)()k k Z ∈对称， 因为当(0,1]x ∈时，2()log f x x =-，由图像可知，函数2()log f x x =-和sin y x π=的图像在[1,1)-上存在1234111,,0,22x x x x =-=-==四个零点，即一个周期内有4个零点，要使得函数()()sin F x f x x π=-，在区间[2,]m -上有2021个零点，其中1234312,,1,22x x x x =-=-=-=-都是函数的零点，即函数()()sin F x f x x π=-在[0,]m 上有2017个零点，如果m 是第207个零点，则20171210084m -=⨯=，如果m 是第2018个零点，则12017100822m =+=，即20171008,2m ⎡⎫∈⎪⎢⎣⎭．二、选择题（本大题共有4题，满分20分，每题5分）每题有且只有一个正确选项．13．【答案】：A 【解析】：①当2πϕ=时，()2sin 22cos22f x x x π⎛⎫=+= ⎪⎝⎭，∵()2cos(2)2cos2()f x x x f x -=-==，∴()f x 为偶函数， ②当()f x 为偶函数时，,Z 2k k πϕπ=+∈，综上所述，2πϕ=是()f x 为偶函数的充分不必要条件，故选：A14．【答案】：B【解析】：21cos sin 222A c b A c --==，即cos b A c =，所以sin cos sin BA C=，所以cos cos sin A C B =cos sin sin sin()sin cos cos sin A C B A C A C A C ==+=+，所以sin cos 0A C =，因为sin 0A ≠，所以cos 0C =，所以cos 0C =是直角数学卷，故选B 15．【答案】：4【解析】：|sin |y x =图像如图，可知|sin |y x =的周期为π，严格单调递增区间是,,2k k k Z πππ⎡⎤+∈⎢⎥⎣⎦所以()sin 3f x x π⎛⎫=+⎪⎝⎭的严格单调递增区间是,,36k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦，令1k =，易知A 正确16．【答案】：B【解析】：令()0f x =，则1sin()2x ωϕ+=，令t x ωϕ=+，则1sin 2t = 则问题转化为sin y t =在区间3,44ππωϕωϕ⎡⎤++⎢⎥⎣⎦上至少有两个，至多有三个t ，使得1sin 2t =， 作出sin y t =和12y =的图像，观察交点个数，可知使得1sin 2t =的最短区间长度为2π，最长长度为223ππ+， 由题意列不等式的：3222443πππωϕωϕππ⎛⎫⎛⎫≤+-+<+ ⎪ ⎪⎝⎭⎝⎭解得：1643ω≤<，故选B三、解答题（本大题共有5题，满分76分）解答下列各题必须在答题纸的相应位置写出必要的步骤．17．【答案】：见解析【解析】：（1）因为P 在单位圆上，且点P 的横坐标为35-，可求得纵坐标为45，所以4tan 3θ=-， 所以cos 3sin 13tan 33sin cos 3tan 15θθθθθθ++==--．（2）由题知4παθ=+，则4πθα=-则11tan 113tan tan 141tan 213παθαα--⎛⎫=-===- ⎪+⎝⎭+18．【答案】：见解析 【解析】：（1）由图可得37341264A T πππ⎛⎫==--= ⎪⎝⎭，∴T π=， ∴22πωπ==，则())f x x ϕ=+，又7721212f ππϕ⎛⎫⎛⎫=⨯+=⎪ ⎪⎝⎭⎝⎭2,3k k Z πϕπ=+∈， ∵02ϕπ<<，∴3πϕ=，∴()23f x x π⎛⎫=+ ⎪⎝⎭；（3）()()226363h x f x f x x x ππππ⎡⎤⎛⎫⎛⎫⎛⎫=⋅-=+-+ ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦233sin 2sin 23sin 2cos cos2sin sin 2sin 22cos233322x x x x x x x x πππ⎛⎫⎛⎫=+=+=+ ⎪ ⎪⎝⎭⎝⎭3333sin 4cos4sin 4444264x x x π⎛⎫=-+=-+ ⎪⎝⎭． ∵0,4x π⎡⎤∈⎢⎥⎣⎦，∴54,666x πππ⎡⎤-∈-⎢⎥⎣⎦，则当462x ππ-=时，()h x 取得最小值为0， 当462x ππ-=时，()h x 取得最大值为94，()h x 的取值范围为90,4⎡⎤⎢⎥⎣⎦19．【答案】：见解析【解析】：（1）设DEF 的边长为x 千米 由3πθ=，得11,22CE x AE a x ==-．在AEF 中，,333FEA A ππππθ∠=--=∠=，∴AEF 为等边三角，得12AE x a x ==-，解得23x a =．所以DEF 的边长等于23a 千米．（2）设DEF 的边长为x 千米．则cos CE x θ=，cos AE a x θ=- 在AEF 中，2,,33FEA A EFA ππθθ∠=-∠=∠=，∴cos sin sin 3x a x θπθ-=，解得x ===，当2πθ+=，arctan 2πθ=-min 7ax ==所以当2πθ=-时，DEF的边长取得最小值为7千米 20．【答案】：见解析【解析】：（1）由题意，函数2())2sin 12x f x x ωϕωϕ+⎛⎫=++- ⎪⎝⎭)cos()2sin 6x x x πωϕωϕωϕ⎛⎫=+-+=+- ⎪⎝⎭因为函数()f x 图像的相邻两对称轴间的距离为2π，所以T π=，可得2w =．（2）将函数()f x 的图像向右平移6π个单位长度，可得2sin 23y x π⎛⎫=-⎪⎝⎭的图像． 再把橫坐标缩小为原来的12，得到函数()2sin 43y g x x π⎛⎫==- ⎪⎝⎭的图像．当,126x ππ⎡⎤∈-⎢⎥⎣⎦时，24,333x πππ⎡⎤-∈-⎢⎥⎣⎦， 当432x ππ-=-时，函数()g x 取得最小值，最小值为2-，当433x ππ-=时，函数()g x()g x的值域[-．（3）由方程4()3g x =，即42sin 433x π⎛⎫-= ⎪⎝⎭，即2sin 433x π⎛⎫-= ⎪⎝⎭， 因为4,63x ππ⎡⎤∈⎢⎥⎣⎦，可得4,533x πππ⎡⎤-∈⎢⎥⎣⎦，设43x πθ=-，其中,53πθπ⎡⎤∈⎢⎥⎣⎦，即2sin 3θ=， 结合正弦函数sin y θ=的图像，可得方程2sin 3θ=在区间,53ππ⎡⎤⎢⎥⎣⎦有5个解，即5n =， 其中122334453,5,7,9θθπθθπθθπθθπ+=+=+=+=， 即12233445443,445,447,44933333333x x x x x x x x ππππππππππππ-+-=-+-=-+-=-+-= 解得1223344511172329,,,12121212x x x x x x x x ππππ+=+=+=+= 所以()()()()1234512233445202223x x x x x x x x x x x x x π+++++=+++++++=． 21．【答案】：见解析 【解析】：（1）由余弦定理得222b ac ac =+-，将2a cb +=代入得到ac =，所以ABC 为等边三角形． （2）（ⅰ）由a c mb +=及正弦定理sin sin sin a b c A B C==得sin sin sin A C m B +=， 所以2sin cos 2sin cos 2222A C A C B B m +-=，因为222A C B π+=-， 所以()2sin cos 2sin cos 2sin cos 22222222B A C A C B A C B m m πππ--++⎛⎫⎛⎫-==-⎪ ⎪⎝⎭⎝⎭， 有cos cos 22A C A C m -+=，由两角和、差的余弦公式可得 cos cos sin sin cos cos sin sin cos cos sin sin 222222222222A C A C A C A C A C A C m m m ⎛⎫+=-=- ⎪⎝⎭，整理得(1)sinsin (1)cos cos 2222A C A C m m +=-， 故1tan tan 221A C m m -=+． （ⅱ）由1tan tan 221A C m m -=+及半角正切公式1cos sin tan 2sin 1cos ααααα-==+可得 21cos sin 1cos sin 1cos 1cos tan tan 22sin 1cos sin 1cos 1cos 1cos A C A A C C A C A A C C A C ----⎛⎫=⋅⋅⋅=⋅ ⎪++++⎝⎭22(1)(1)m m -=+， 展开整理得()2421(cos cos )4cos cos m m A C m A C -++=-， 即()2421(cos cos )4cos cos m m A C m A C-++=-， 即222cos cos 21cos cos 1mA C m m A C m +-+=+， 即222cos cos 112cos cos 1m A C m m A C m +-+=--+，与原三角式作比较可知()m ϕ存在且22()1m m m ϕ=-+．。

2019-2020学年上海市杨浦区复旦大学附中高一下学期期中物理试卷一、单选题（本大题共7小题，共28.0分）1.如图所示，洗衣机的脱水筒采用带动衣物旋转的方式脱水，脱水筒旋转稳定前，转动越来越快，则在这过程中，下列物理量变小的是()A. 线速度B. 角速度C. 转速D. 周期2.利用盛沙的漏斗演示简谐振动，如果考虑漏斗里砂子逐渐减少，则沙摆的频率将()A. 逐渐增大B. 逐渐减少C. 先增大后减少D. 先减小后增大3.在光滑水平面上，a、b两球沿水平面相向运动．当两球间距小于或等于L时，受到大小相等、相互排斥的水平恒力作用；当两球间距大于L时，则相互作用力为零．两球在相互作用区间运动时始终未接触，两球运动的v−t图象如图所示，则()A. a球质量小于b球质量B. t1时刻两球间距最小C. 0−t2时间内，两球间距逐渐减小D. 0−t3时间内，b球所受排斥力方向始终与运动方向相反4.2017年10月24日，在地球观测组织(GEO)全会期间举办的“中国日”活动上，我国正式向国际社会免费开放共享我国新一代地球同步静止轨道气象卫星“风云四号”(如图所示)和全球第一颗二氧化碳监测科学实验卫星(简称“碳卫星”)的数据。

“碳卫星”是绕地球极地运行的卫星，在离地球表面700公里的圆轨道对地球进行扫描，汇集约140天的数据可制作一张无缝隙全球覆盖的二氧化碳监测图。

下列关于这两颗卫星的说法正确的是()A. “风云四号”卫星的线速度大于第一宇宙速度B. “风云四号”卫星的线速度小于“碳卫星”的线速度C. “风云四号”卫星的运行周期小于“碳卫星”的运行周期D. “风云四号’’卫星的向心加速度大于“碳卫星”的向心加速度5.一弹簧振子做简谐运动，它所受的回复力F随时间t变化的图线为正弦曲线，如图所示，下列说法不正确的是()A. 在t从0到2 s时间内，弹簧振子做减速运动B. 在t1=3s和t2=5s时，弹簧振子的速度大小相等，方向相反C. 在t1=5s和t2=7s时，弹簧振子的位移大小相等，方向相同D. 在t从0到4 s时间内，t=2s时刻弹簧振子所受回复力做功功率最小6.如图为一质点作简谐运动的图象，则在图中t1和t2两个时刻，振子具有相同的物理量是()A. 加速度B. 位移C. 速度D. 回复力7.人在沼泽地上行走时会陷下去，当人加速下陷时()A. 人处于超重状态B. 人对沼泽地的压力大小等于沼泽地对人的支持力大小C. 人对沼泽地的压力大小大于沼泽地对人的支持力大小D. 沼泽地对人的支持力等于人的重力二、多选题（本大题共5小题，共20.0分）8.下列选项与多普勒效应有关的是()A. 科学家通过比较星球与地球上同种元素发出光的频率来计算星球远离地球的速度B. 医生利用超声波探测病人血管中血液的流速C. 技术人员用超声波探测金属、陶瓷、混凝土中是否有气泡D. 交通警察向车辆发射超声波并通过测量反射波的频率确定车辆行进的速度9.一质点做简谐运动的图象如图所示，下列说法不正确的是()A. 质点振动频率是4HzB. 在10s内质点经过的路程是20cmC. 第4s末质点的速度是零D. 在t=1s和t=3s两时刻，质点位移大小相等、方向相同E. 在t=2s和t=6s两时刻，质点速度相同10.关于波的认识，下列说法正确的是()A. 潜艇利用声呐探测周围物体的分布情况，利用的是波的反射原理B. 发生多普勒效应时，波源的频率发生了变化C. 机械波在介质中的传播速度是由介质本身的性质决定的D. 波在传播过程中绕过障碍物向前传播的现象，是波的折射现象E. 医生利用超声波探测病人血管中血流的速度，利用的是多普勒效应11.下列说法正确的是()A. 入射波的波长等于反射波的波长B. 在纵波中，两个密部或两个疏部之间的距离等于一个波长C. 介质中的质点的沿水平方向振动，波沿水平方向方向传播，此类波为纵波D. 做简谐运动的物体在振动方向上所受的合外力，可以使是恒力，也可以是变力E. 在敲响古刹里的大钟时，有的同学发现，停止对大钟的敲击后，大钟仍“余音未绝”，究其原因是：大钟停止了振动，但空气仍在振动12.如图所示，在一根长为L的细线下面系一质量为m的小球，将小球拉离竖直位置，使悬线与竖直方向成α角，给小球一个初速度，使小球在水平面内做匀速圆周运动，悬线旋转形成一个圆锥面，这就是常见的圆锥摆模型。

2018-2019学年上海市复旦大学附属中学高一下学期期中考试数学试题一、单选题1．在ABC ∆中，“1sin 2A =”是“6A π=”的（ ） A ．充分非必要条件 B ．必要非充分条件 C ．充要条件 D ．既非充分又非必要条件 【答案】B【解析】试题分析：ABC ∆中，若1sin 2A =，则6A π=或56π，反之，若6A π=，则一定有1sin 2A =，所以在ABC ∆中，“1sin 2A =”是“6A π=”的必要非充分条件，故选B ．【考点】1、已知三角函数求角；2、充分条件与必要条件. 2．设函数的图象为，下面结论中正确的是A ．函数的最小正周期是B ．图象关于点对称C ．图象可由函数的图象向右平移个单位得到D ．函数在区间上是增函数【答案】B 【解析】根据的周期计算公式，对称中心，单调区间及图形变换规律依次判断即可。

【详解】 函数的最小正周期为，故A 错误； ∵，∴图象关于点对称，故B 正确；易知图象可由函数的图象向右平移个单位得到，故C 错误；易得函数的单调递增区间是，当时，，∴函数在区间上是先增后减，故D 错误.故选B.【点睛】本题考查的相关性质，关键要对此部分知识加强记忆，深刻理解其处理思想和方法，此类问题是各类考试的热点，应给予足够的重视。

二、填空题3．已知,，若角与的终边相同，则____________【答案】【解析】利用终边相同的角的特点可知，再将其化为弧度制的角得到结果. 【详解】与的终边相同本题正确结果：【点睛】本题考查终边相同的角的问题，弧度制与角度制的互化，属于基础题.4．已知函数的最小正周期为，则____________【答案】【解析】根据正切型函数最小正周期为构造方程求得结果.【详解】的最小正周期本题正确结果：【点睛】本题考查的最小正周期问题，属于基础题.5．一个半径为r的扇形，若它的周长等于弧所在的半圆的长，那么该扇形的圆心角是____________弧度【答案】【解析】根据扇形弧长公式表示出扇形的周长，从而建立起方程，求解得到圆心角. 【详解】设扇形的圆心角为：则扇形的周长本题正确结果：【点睛】本题考查扇形弧长公式的应用问题，属于基础题.6．已知是第三象限的角，则的符号是____________号（填正或负）【答案】负【解析】根据角的范围可得和的范围，进而可确定和的符号，从而得到结果.【详解】为第三象限角，；本题正确结果：负【点睛】本题考查三角函数在各个象限内的符号问题，属于基础题.7．角终边上有点，且，则____________【答案】【解析】根据构造方程，求出，根据的定义求得结果.【详解】由题意得：本题正确结果：【点睛】本题考查三角函数的定义问题，属于基础题.8．若，则____________【答案】【解析】根据二倍角公式可得，进而得到，代入得到结果.【详解】本题正确结果： 【点睛】本题考查函数解析式的求解以及利用解析式求解函数值的问题，关键是能够通过二倍角公式构造出关于的形式，根据整体法得到函数解析式.9．已知函数，且是其单调区间，则的取值范围是____________ 【答案】【解析】根据的范围得到的范围；根据函数单调可知，解不等式得到结果. 【详解】 当时，，即本题正确结果：【点睛】 本题考查利用的单调性求解参数范围的问题，关键是能够通过的范围得到整体所处的范围，放入的单调区间中构造不等式.10．已知，，____________【答案】【解析】根据诱导公式和二倍角公式可求得，再根据角的范围求得，利用两角和差公式求解得到结果.【详解】即：本题正确结果：【点睛】本题考查三角函数中诱导公式、二倍角公式、两角和差公式的应用以及同角三角函数的求解问题，关键是能够通过配凑角的方式通过已知角将所求角表示出来，从而利用公式求解得到结果.11．张老师整理旧资料时发现一题部分字迹模糊不清，只能看到：在中，分别是角是的对边，已知，，求边，显然缺少条件，若他打算补充的大小，并使得有两解，那么的取值范围是____________【答案】【解析】问题为三角形有两个解，根据画圆法可确定，从而得到所求范围. 【详解】由题意可知三角形有两个解由上图可知：若有两解，可知以为圆心，为半径的圆弧与有两个交点则，即【点睛】本题考查三角形解的个数的问题，关键是能够将问题转化为与之间的大小关系的比较.12．函数的值域____________【答案】【解析】首先确定定义域，根据二倍角公式将整理为，从而根据定义域可知，进而得到函数值域.【详解】定义域为：当时，值域为本题正确结果：【点睛】本题考查正切函数值域的求解问题，忽略原函数的定义域是本题的易错点.13．为了竖一块广告牌，要制造三角形支架，如图，要求，的长度大于1米，且比长0.5米，为了稳固广告牌，要求越短越好，则最短为____________米【答案】【解析】根据余弦定理构造出，利用换元法可将右侧式子凑成符合基本不等式的形式，根据基本不等式求得最小值.【详解】设，则由余弦定理得令，则当且仅当，即时，即时，取得最小值本题正确结果：【点睛】本题考查利用基本不等式解决实际问题，关键是能够通过余弦定理将所求长度化为关于变量的和的形式，根据基本不等式求解出和的最小值.14．设是定义在上的周期为4的函数，且，记，若函数在区间上零点的个数是8个，则的取值范围是____________【答案】【解析】将问题转化为与的图象在区间之间有个交点的问题，根据解析式和周期画出函数图象，通过数形结合得到结果.【详解】由题意可转化为与的图象在区间之间有个交点由解析式及周期，可得函数的图象如下图：若与在有个交点，则位置如图所示数形结合可知：本题正确结果：【点睛】本题考查利用函数区间内的零点个数求解参数范围问题，关键是能够将问题转化为曲线与直线的交点个数问题，从而通过数形结合的方式求得结果.15．设函数，其中，若、、是的三条边长，则下列结论：①对于一切都有；②存在使、、不能构成一个三角形的三边长；③为钝角三角形，存在，使，其中正确的个数为______个A．3 B．2 C．1 D．0【答案】A【解析】构造函数，根据函数单调性可知，根据三角形三边关系可知，可推导出，从而可得，可知①正确；通过取值可知存在取值使得取值不满足三边关系，可知②正确；根据余弦定理可知，可得，再结合，可知，由零点存在性定理可知③正确；由此可得选项. 【详解】①令在上单调递减在上单调递减当时，根据三角形三边关系可知：又时，都有，可知①正确；②取，，，则，不满足三角形三边关系，可知②正确；③为钝角三角形，从而又，由零点存在性定理，可知③正确本题正确选项：【点睛】本题考查函数与解三角形知识的综合应用问题，其中涉及到零点存在定理的应用、余弦定理及三角形三边关系的应用、函数单调性问题，关键是能够构造出合适的函数来对问题进行求解.16．若函数的最大值和最小值分别为、，则函数图像的对称中心不可能是_______A．B．C．D．【答案】C【解析】设，可得为奇函数，进而得到，从而得到解析式；根据的对称中心，平移可得对称中心的坐标；再分别对应四个选项，当不是整数时，则不可能为对称中心，由此可得选项.【详解】设，则即为奇函数令则，可知的对称中心为将的图象向右平移个单位，再向上平移个单位得的图象的对称中心为当时，，不合题意，可知不可能为又当时分别对应选项，可知均为的对称中心本题正确选项：【点睛】本题考查函数性质的综合应用问题，涉及到利用奇偶性求解最值、与三角函数有关的对称中心的求解、函数图象平移变换问题，对于学生函数性质的掌握要求较高，属于偏难题.三、解答题17．已知函数.（1）求的单调增区间；（2）当时，求的最大值和最小值.【答案】（1）；（2）的最大值为2，最小值为-1【解析】（1）利用辅助角公式得：，将放入的单调递增区间中，求出的范围即可；（2）根据的范围得的范围，结合的图象可求得最值. 【详解】（1）由得：的单调增区间为（2）当时，当时，当时，的最大值为，最小值为【点睛】本题考查的单调区间的求解、函数值域的求解问题，关键是能够通过整体对应的方式，通过分析的图象求得结果.18．在中，已知，外接圆半径.（1）求角的大小；（2）试求面积的最大值.【答案】（1）（2）【解析】（1）利用二倍角公式得到关于的方程，解出，进而得到；（2）根据正弦定理求得，根据余弦定理，结合基本不等式可得，代入三角形面积公式求得面积的最大值.【详解】（1）由得：即解得：或（舍）（2）由正弦定理得：由余弦定理得当且仅当时，取得最大值，即面积的最大值为【点睛】本题考查正余弦定理解三角形、三角形面积的最值问题，关键是能够利用余弦定理构造出基本不等式的形式，从而得到积的最大值.19．已知函数的图像与轴的交点为，它在轴右侧的第一个最高点和第一个最低点的坐标分别为和.（1）求函数的解析式；（2）将函数的图像向左平移个单位后，得到的函数是奇函数，求的值.【答案】（1）（2）【解析】（1）根据最值确定振幅；再根据两对称轴之间距离为求得；代入求得；根据图象否掉的情况，从而得到结果；（2）根据图象平移得到解析式，利用求得；通过验证可知满足题意，从而确定结果.【详解】（1）由题意，，即，即或当时，函数在时先取得最小值，后取得最小值，不符合图象函数的解析式为（2）由题意得：，是奇函数又当时，满足，即为奇函数，可知满足题意【点睛】本题考查利用三角函数图象求解函数解析式、利用图象平移和函数性质求解参数的问题.本题的易错点为利用特殊值求解初相时，忽略图象最值取得的位置，从而无法舍去增根. 20．如图，制图工程师要用两个同中心的边长均为4的正方形合成一个八角形图形，由对称性，图中8个三角形都是全等的三角形，设.（1）用表示线段；（2）设，，求关于的函数解析式；（3）求八角形所覆盖面积的最大值，并指出此时的大小.【答案】（1），（2），（3）时，取得最大值【解析】（1）根据构造出与的关系，整理得到结果；（2）由（1）可得，整理化简可得结果；（3）利用将表示成，；利用换元法，可将问题转化为，根据的范围和的单调性求得最值和的取值.【详解】（1）由题意可得：，（2）由（1）得：两边平方并化简得：又，（3），令则又在上单调递增当，即时，取得最大值【点睛】本题考查利用三角函数的实际应用问题，重点考查了面积的最值求解问题，关键是能够将所求面积表示成与三角函数有关的函数关系式，从而通过换元的方式结合函数的单调性求得结果；易错点是忽略了换元后参数的取值范围.21．已知是定义在上的函数，如果存在常数，对区间的任意划分：，和式恒成立，则称为上的“绝对差有界函数”，注：.（1）求证：函数在上是“绝对差有界函数”；（2）记集合存在常数，对任意的，有成立.求证：集合中的任意函数为“绝对差有界函数”；（3）求证：函数不是上的“绝对差有界函数”.【答案】（1）见解析（2）见解析（3）见解析【解析】（1）将整理为，可知在上单调递增；可知，从而可将化简为，从而可知，得到结论；（2）取，根据，可得，从而可取得到结论；（3）取一个划分：，可将整理为；根据放缩可知只要足够大，可使得，从而得到结论.【详解】（1）当时，在区间上为单调递增函数当，时，有，所以从而对区间的任意划分：存在，使得成立综上，函数在上是“绝对差有界函数”（2）证明：任取从而对区间的任意划分：和式成立则可取所以集合中的任意函数为“绝对差有界函数”（3）取区间的一个划分：，则有：所以对任意常数，只要足够大，就有区间的一个划分：满足所以函数不是的“绝对差有界函数”【点睛】本题考查与新定义有关的证明问题，关键是能够理解新定义的具体含义，进而可通过单调性、不等关系、放缩的方式把关系式进行化简，从而可求得临界值的具体取值，再根据取值确认函数是否符合新定义，属于难题.。

复旦附中第二学期高一物理期中试卷(本试卷g取10m/s,答案做在答卷纸上)一、单选题1．如图所示，O为弹簧振子平衡位置，将振子拉至A处后放手，振子可沿水平光滑杆在A、B间作简谐振动，则振子A 从B→O回复力不断减小，速度不断减小B 在O处速度最大C 从B→O回复力不断增大，速度不断增大D 在B处速度最大2．关于机械波下列说法错误的是；A 任一质点振动每经过一个周期，波沿传播方向移动一个波长B 波的产生需要两个条件，即波源和弹性媒质C 波的传播过程是能量由近及远传递的过程D 机械波由波源向外传播的过程中，作振动的质点也同时由波源向远处运动3．曾有科学家推测，太阳系的第十颗行星就在地球的轨道上，从地球上看，它永远在太阳的背面，人类一直未能发现它，可以说是“隐居”着的地球的“孪生兄弟”.由以上信息可A. 这颗行星的公转周期与地球相等B. 这颗行星的半径等于地球的半径C. 这颗行星的密度等于地球的密度D. 这颗行星的自转周期与地球相同4．关于横波与纵波，下列说法正确的是A 纵波的传播方向与质点的振动方向相同，由于质点作变速运动，所以纵波也作变速运动B 横波可以在气体、液体、固体中传播，而纵波只能在固体中传播C 横波一定是水平传播的，纵波一定是竖直传播的。

D 横波在传播过程中一定有波峰和波谷，纵波在传播过程中一定有疏部和密部5．振动物体都有它自己的固有频率，实验表明，做受迫振动的物体，在它的固有频率等于或接近于策动力的频率时，振幅最大，这就是共振现象，如果固有频率和策动力的频率相差较大时，那么受迫振动的振幅就小，下表记录了某受迫振动的物体振幅随策动力频率变化的关系，则该振动物体的固有频率A 一定在50Hz到60Hz的范围内B 一定等于60HzC 一定在60Hz到70Hz的范围内D 一定在50Hz到70Hz的范围内二、复选题6．火星有两颗卫星，分别是火卫1和火卫2，它们的轨道近似为圆，已知火卫1的周期为7小时39分，火卫2的周期为30小时18分，则两颗卫星相比A 火卫1距火星表面较近B 火卫2的角速度较大C 火卫1的运动速度较大D 火卫2的向心加速度较大7．两列简谐波在同一种介质中传播时发生了干涉现象，则A 在振动加强区域，质点的位移总比振动减弱区域质点的位移大B 在振动加强区域，质点的振幅总比振动减弱区域质点的振幅大C 在振动加强区域，质点的位移随时间作周期性变化D 在振动加强区域，质点的振幅随时间作周期性变化8．关于声波下列说法正确的是A 在发声音叉某处听到声音特强，这是因声波反射，产生回声加强了原来的声音而形成的B 在发声音叉周围走一圈，会听到声音忽强忽弱，这是两个声波发生了干涉的结果C 隔着院墙和人讲话，虽不见其人但闻其声，这是波的衍射现象D 在门窗关闭的屋里讲话，听起来比在旷野里响，这是声音的干涉现象9．同一地点的甲、乙两个单摆的振动图象如图所示，下列说法正确的是A 甲、乙两单摆的摆长不一定相同B 甲、乙两单摆的振幅一定相同C 甲摆小球的质量一定等于乙摆小球的质量D 在1/4周期时，振子具有正向最大加速度的是乙摆10．一绳子的两端各产生一个如图所示的波形，两列波相向传播时，可能出现的波形图为三、填空题11一列在平静水面传播的正弦横波，波长80 cm，波速为4m/s，浮在水面上的一片小叶片因这列水波通过而发生上下振动，则这一振动的周期是秒。

复旦大学附属中学2019学年第一学期高一年级物理期中考试试卷考试时间60分钟，满分100 分。

一、单项选择题: （每题只有一个正确答案）（40分）1．如图所示，一个质点沿两个半径为R的半圆弧由A运动到C，规定向右方向为正方向，在此过程中，它的位移大小和路程分别为（）A．4R，2πR B．4R，﹣2πR C．﹣4R，2πR D．﹣4R，﹣2πR 2．一辆汽车停在水平地面上，有下列几种说法：（1）地面受到向下的弹力，是因为地面发生了形变；（2）地面受到了向下的弹力，是因为汽车发生了形变；（3）汽车受到了向上的弹力，是因为汽车发生了形变；（4）汽车受到了向上的弹力，是因为地面发生了形变．其中正确的是（）A．（1）（3）B．（1）（4）C．（2）（3）D．（2）（4）3．如图所示，物体以5m/s的初速度沿光滑的斜面向上做减速运动，经4s滑回原处时速度大小仍为5m/s，则物体的加速度为（）A．10m/s2，方向沿斜面向下B．0C．2.5m/s2，方向沿斜面向下D．5m/s2，方向沿斜面向下4．在力的合成中，下列关于两个分力（大小为定值）与它们的合力的关系的说法中，正确的是（）A．合力一定大于每一个分力B．合力一定小于分力C．合力的方向一定与分力的方向相同D．两个分力的夹角在0°～180°变化时，夹角越大合力越小5．小球以2m/s的初速度从离地高为3m的地方开始做竖直上抛运动，取g＝10m/s2，则小球在空中运动的时间为（）A．0.4s B．0.6s C．0.8s D．1.0s6．用轻绳系住一小球静止在光滑斜面上，如图所示．若要按力的实际作用效果来分解小球的重力，则重力的两个分力的方向分别是图中的（）A．1和5 B．2和5 C．3和5 D．3和27．P、Q、R三点在同一条直线上，一物体从P点静止开始做匀加速直线运动，经过Q点的速度为v，到R点的速度为3v，则PQ：QR等于（）A．1：8 B．1：6 C．1：5 D．1：38．如图所示是一个网球沿竖直方向运动时的频闪照片，由照片可知（）A．网球正在上升B．网球正在下降C．网球的加速度向上D．网球的加速度向下9．如图所示，直线a和曲线b分别是在平直公路上行驶的汽车a和b的位移一时间（x﹣t）图线，由图可知（）A．在时刻t l，a车与b车相遇B．在时刻t1，a．b两车运动方向相反C．在t l到t2这段时间内，a车的位移比b车小D．在t l到t2这段时间内，a车的速率一直比b车的小10．两辆完全相同的汽车，沿水平直路一前一后匀速行驶，速度均为v0，若前车突然以恒定的加速度刹车，在前车刚停住时，后车以前车刹车的加速度开始刹车，已知前车在刹车过程中所行驶的距离为s，若要保持两车在上述情况中不相撞，则两车在匀速行驶时保持的距离至少应为（）A．s B．2s C．3s D．4s11．据英国《每日邮报》2016年8月16日报道27名跳水运动员参加了科索沃年度高空跳水比赛，自某运动员离开跳台开始计时，在t2时刻运动员以v2的速度入水，选竖直向下为正方向，其速度随时间变化的规律如图所示，下列结论正确的是（）A．该运动员在0～t2的时间内加速度的大小先减小后增大，方向向上B．该运动员在t2～t3时间内加速度大小逐渐减小，方向向下C．在0～t时间内，平均速度为D．在t2～t3时间内，平均速度为12．作用于原点O的三力平衡，已知三力均位于xOy平面内，其中一个力的大小为F1，沿y轴负方向；力F2的大小未知，与x轴正方向的夹角为θ，如图所示。

2019-2020学年上海市杨浦区复旦附中高一下学期期中数学试题一、单选题1．在△ABC 中，“sin 2A >”是“34A π<”的（ ）条件A ．充分非必要B ．必要非充分C ．充要D ．既非充分又非必要 【答案】A【解析】根据三角函数的性质，得到当sin A >时，34A π<是成立的，再利用反例，得出必要性不一定成立，即求解. 【详解】在ABC ∆中，由sin 2A >，因为(0,)A π∈，可得344A ππ<<，所以当sin 2A >时，34A π<是成立的，即充分性成立；反之：例如364A ππ=<，此时1sin 22A =<，即必要性不一定成立.所以“sin A >”是“34A π<”的充分不必要条件.故选：A 【点睛】本题主要考查了充分不必要条件的判定，其中解答中熟练应用三角函数的性质，结合充分条件、必要条件的判定方法求解是解答的关键，着重考查推理与运算能力.2．以下哪个不是25lim 21nn n q q →∞-+可能的取值（ ）A ．2B ．1-C ．52-D ．7-【答案】D【解析】对q 的取值进行分类讨论，即可得答案；【详解】（1）若12q =，则0nq →，∴25lim 221n nn q q →∞-=+； （2）若2q，则n q →+∞，∴25255lim lim 12122n nn n n nq qq q→∞→∞--==-++；（3）若1q =，则1nq =，∴25lim 121nnn q q →∞-=-+； 利用排除法可得D 选项不可能， 故选：D. 【点睛】本题考查数列极限的求解，考查分类讨论思想，考查逻辑推理能力、运算求解能力. 3．若等差数列{}n a 首项为2，公差为2，其前n 项和记为n S ，则数列1{}nS 前n 项和为（ ） A ．21nn + B ．1n n + C ．1n(n 1)+D ．2(1)nn +【答案】B【解析】根据等差数列前n 项和公式求出n S ，从而得出1{}nS 的通项公式，再用裂项相消法即可求出数列1{}nS 前n 项和. 【详解】等差数列前n 项()112n n n S na d -=+，等差数列{}n a 首项为2，公差为2，代入可得()()12212n n n S n n n -=+⨯=+，所以()111111n S n n n n ==-++，所以数列1{}nS 前n 项和为111111111122334111n n T n n n n =-+-+-++-=-=+++. 故选：B 【点睛】本题主要考查等差数列前n 项和的求法，以及裂项相消法求数列前n 项和.4．已知函数()sin()f x A x ωϕ=+（其中A 、ω、ϕ均为正的常数）的最小正周期为2π，当3x π=时，函数()f x 取得最小值，则下列结论正确的是（ ）A ．(1)(1)(0)f f f <-<B ．(0)(1)(1)f f f <<-C ．(1)(0)(1)f f f -<<D ．(1)(0)(1)f f f <<-【答案】A【解析】根据周期公式可得4ω=，根据当3x π=时，函数()f x 取得最小值，可得1126k ϕππ=-，k Z ∈，所以()f x sin(4)6A x π=+，再利用诱导公式以及三角函数的性质比较大小可得答案. 【详解】 依题意得22ππω=，解得4ω=，所以()sin(4)f x A x ϕ=+，因为当3x π=时，函数()f x 取得最小值，所以4232k ππϕπ⨯+=-，k Z ∈，即1126k ϕππ=-，k Z ∈， 所以11()sin(42)6f x A x k ππ=+-11sin(4)sin(42)66A x A x πππ=-=-+sin(4)6A x π=+，因为3462πππ<+<且0A >，所以(1)sin(4)6f A π=+0<，因为(1)sin(4)sin(42)sin[(42)]666f A A A ππππππ-=-+=-++=--++11sin(4)sin(4)66A A πππ=--=-，又1104662πππ<-<<，所以110sin(4)sin 66ππ<-<， 因为0A >，所以0(1)(0)f f <-<， 综上所述：(1)(1)(0)f f f <-<. 故选：A 【点睛】本题考查了根据三角函数的性质求解析式，考查了诱导公式，考查了利用正弦函数的单调性比较大小，属于中档题.二、填空题5．一个面积为1的扇形，所对弧长也为1，则该扇形的圆心角是________弧度 【答案】12【解析】设扇形的所在圆的半径为r ，圆心角为α，应用扇形的弧长公式和面积公式，列出方程组，即可求解. 【详解】设扇形的所在圆的半径为r ，圆心角为α， 因为扇形的面积为1，弧长也为1，可得21121r r αα⎧⋅=⎪⎨⎪=⎩，即221r r αα⎧⋅=⎨=⎩，解得12,2r α==.故答案为：12【点睛】本题主要考查了扇形的弧长公式和面积公式的应用，其中解答中熟练应用扇形的弧长公式和面积公式，列出方程组是解答的关键，着重考查了运算与求解能力. 6．计算sin40sin100sin50sin10︒︒-︒︒=________ 【答案】12【解析】利用诱导公式和两角差的正弦公式，即可得到答案； 【详解】原式1sin 40cos10cos 40sin10sin 302=︒︒-︒︒=︒=， 故答案为：12. 【点睛】本题考查诱导公式和两角差的正弦公式的应用，考查转化与化归思想，考查运算求解能力.7．函数sin y x =，[,]2x ππ∈的反函数记为()g x ，则1()2g =________ 【答案】56π【解析】点51(,)62π在原函数sin y x =的图象上，根据题意两函数图象关于直线y x =对称知点15(,)26π在反函数()g x 的图象上，得解. 【详解】因为当[,]2x ππ∈时，51sin62π=，所以点51(,)62π在原函数sin y x =的图象上，因为()g x 是函数sin y x =，[,]2x ππ∈的反函数，所以点15(,)26π在反函数()g x 的图象上，则15()26g π=. 故答案为：56π【点睛】本题考查两个互为反函数的函数图象的对称性、正弦函数的图象与性质，属于基础题. 8．在△ABC中，若a =1b =，60A =︒，则B =________【答案】6π 【解析】直接利用正弦定理，结合三角形解的个数判定，即可得到答案； 【详解】11sin sin sin sin 2a bB A BB=⇒=⇒=，a b >，∴A B >，∴6B π=，故答案为：6π. 【点睛】本题考查正弦定理\三角形解的个数，考查函数与方程思想、转化与化归思想，考查逻辑推理能力、运算求解能力.9．已知等比数列{}n a 中，24a =，68a =，则10a =________ 【答案】16【解析】将等比数列的通项公式代入24a =，68a =中，可得4q ，再求10a 的值。

一、填空题1．已知，则_________. sin α=22,ππα⎛∈-⎫ ⎪⎝⎭sin 2πα⎛⎫-= ⎪⎝⎭【答案】 【分析】利用诱导公式与平方和关系求解即可.【详解】因为，所以,22ππα⎛⎫∈- ⎪⎝⎭cos α==sin cos 2παα⎛⎫-=-= ⎪⎝⎭故答案为： 2．已知i 为虚数单位，若复数是实数，则实数m 的值为__________. 1i 2iz m =++【答案】/0.215【分析】先化简复数z ，然后根据虚部为0可得. 【详解】因为为实数， ()()12i 2i 21i i i i 2i 2i 2i 555z m m m m --⎛⎫=+=+=+=+- ⎪++-⎝⎭所以，所以105m -=15m =故答案为：153．向量在向量方向上的投影为___________.()3,4a =()1,0b =- 【答案】3-【分析】由向量投影公式直接求解即可得到结果. 【详解】向量在方向上的投影为. a b 331a b b⋅-==-故答案为：.3-4．在△ABC 中，若，，，则___________.3AB =5π12B ∠=π4C ∠=BC =【分析】由三角形内角和求得，然后由正弦定理求得． A BC 【详解】由三角形内角和定理可得：， ππ3A B C =--=因为，， 3c AB ==a BC =由正弦定理可得， sin sin sin sin a c c A a A C C =⇒==. 5．已知复数z 满足（i 为虚数单位），则_________. ()22i 2i z ⋅-=+z =【分析】根据复数的四则运算化简求得复数z ，然后求模.【详解】，所以()22i2i (2i)(3+4i)211i 34i (34i)(3+4i)25252i z +++====+---z ==6．方程在区间上的所有解的和为__________. cos 2sin 0x x -=[]0,2π【答案】/ 52π52π【分析】利用倍角余弦公式得到关于的一元二次方程求解，由正弦函数值求，即可得结果. sin x x 【详解】由，即，解得或， cos 2sin 0x x -=212sin sin 0x x --=sin 1x =-1sin 2x =在，当时，当时或， []0,2πsin 1x =-32x π=1sin 2x =π6x =5π6x =所以所有解的和为. 52π故答案为：52π7．设，，且，则_______.3,sin 2a α⎛⎫= ⎪⎝⎭ 1cos ,3b α⎛⎫= ⎪⎝⎭ //a b r r tan α=【答案】1【分析】由向量平行的坐标表示，结合同角三角函数关系和商数关系可得．【详解】因为，所以. //a b r r 22231sin cos tan sin cos tan 123sin cos tan 1ααααααααα⨯===⇒=++故答案为:1.8．在△ABC 中，边a ，b ，c 满足，，则边c 的最小值为__________. 8a b +=120C ∠=︒【答案】【分析】利用基本不等式和结合余弦定理即可求解的最小值． 2()2a b ab +≤c 【详解】由余弦定理可得 当且仅当时，即取等()222222cos 264482a b c a b ab C a b ab ab +⎛⎫=+-=+-+-= ⎪⎝⎭≥a b =4a b ==号，所以c ≥故答案为：9．在直角三角形中，，，，点是外接圆上的任意一点，则ABC 5AB =12AC =13BC =M ABC A 的最大值是___________. AB AM ⋅【答案】45【分析】建立平面直角坐标系，用圆的方程设点的坐标，计算的最大值． M AB AM ⋅【详解】建立平面直角坐标系，如图所示：，，，(0,0)A (5,0)B (0,12)C 外接圆，ABC A 225169()(6)24x y -+-=设，，M 513(cos 22θ+136sin )2θ+则，，513(cos 22AM θ=+ 136sin )2θ+，，当且仅当时取等号． (5,0)AB =2565cos 4522AM AB θ⋅=+ …cos 1θ=所以的最大值是45． AB AM ⋅故答案为：45．10．在锐角三角形ABC 中，O 为△ABC 的外心，则的cos A =32OA OB OC ++取值范围为__________.【答案】3⎡⎣【分析】三角形外接圆的性质、正弦定理得、，、，π2BOC ∠=3π22AOB B ∠=-2AOC B ∠=1R =利用向量数量积的运算律转化求.32OA OB OC ++【详解】，222232941264OA OB OC OA OB OC OA OB OA OC OB OC ++=+++⋅+⋅+⋅因为锐角三角形中，，cos A =π4A =π2BOC ∠=所以，，又，即， 3π22AOB B ∠=-2AOC B ∠=22sin aR A ==1R =则且， ()()232146cos 22sin 2142OA OB OC B B B ϕ++=+-=++ tan 2ϕ=则，即. 23214OA OB OC ⎡++∈-+⎣ 323OA OB OC ⎡++∈+⎣故答案为：3⎡⎣11．如图所示，在直角梯形ABCD 中，已知，，，，M 为//AD BC 2ABC π∠=1AB AD ==2BC =BD 的中点，设P 、Q 分别为线段AB 、CD 上的动点，若P 、M 、Q 三点共线，则的最大值AQ CP ⋅为__.【答案】2-【分析】建立直角坐标系，设，，由P 、M 、Q 三点共线，设(0,)P m [0,1]m ∈，求得，代入计算知11(1)(2,2),2BM BQ BP k k m m λλλλλλ⎛⎫=+-=-+-= ⎪⎝⎭u u u r u u u r u u r 2322m k m -=+AQ CP⋅ ，构造函数，，结合函数的单调性求得51(1)221m m ⎡⎤=-+-⎢⎥+⎣⎦51()(1)221f m m m ⎡⎤=-+-⎢⎥+⎣⎦[0,1]m ∈最值.【详解】如图所示，建立直角坐标系，则，，，，，(0,0)B (2,0)C (0,1)A (1,1)D 11,22M ⎛⎫⎪⎝⎭又Q 是线段CD 上的动点，设，CQ kCD =u u u r u u u r[0,1]k ∈则，可得 (2,0)(1,1)(2,)BQ BC kCD k k k =+=+-=-u u u r u u u r u u u r(2,)Q k k -设，，(0,)P m [0,1]m ∈由P 、M 、Q 三点共线，设11(1)(2,2),2BM BQ BP k k m m λλλλλλ⎛⎫=+-=-+-= ⎪⎝⎭u u u r u u u r u u r112,.22k k m m λλλλ∴-=+-=利用向量相等消去可得：， λ2322mk m -=+ 23(2,1)(2,)424(2)22m AQ CP k k m k mk m m m m -⋅=--⋅-=-++-=-++⨯-+u u u r u u r 51(1)221m m ⎡⎤=-+-⎢⎥+⎣⎦令，，则在上单调递减， 51()(1)221f m m m ⎡⎤=-+-⎢⎥+⎣⎦[0,1]m ∈()f m [0,1]m ∈故当时，取得最大值 0m =()f m (0)2f =-故答案为：2-【点睛】方法点睛：本题考查向量的坐标运算，求解向量坐标运算问题的一般思路：向量的坐标化：向量的坐标运算，使得向量的线性运算可用坐标进行，实现了向量坐标运算完全代数化，将数与形紧密的结合起来，建立直角坐标系，使几何问题转化为数数量运算，考查学生的逻辑思维与运算能力，属于较难题.12．设函数，若恰有个零点，.()()[]0,0,0,26f x Asin x A x πωωπ⎛⎫=->>∈ ⎪⎝⎭()f x 4则下述结论中:①若恒成立，则的值有且仅有个；()()0f x f x ≥0x 2②在上单调递增；()f x 80,19π⎡⎤⎢⎥⎣⎦③存在和，使得对任意恒成立；ω1x ()()11()2f f x f x x π≤+≤[]0,2x π∈④“”是“方程在恰有五个解”的必要条件. 1A ≥()12f x =-[0,2]π内所有正确结论的编号是______________； 【答案】①③④【解析】根据条件画出的图像，结合图像和()()[]0,0,0,26f x Asin x A x πωωπ⎛⎫=->>∈ ⎪⎝⎭逐一判断即可. 1925,1212ω⎡⎫∈⎪⎢⎣⎭【详解】恰有个零点，，，函数的图像如图： ()f x 4∴3246πππωπ≤-<∴1925,1212ω⎡⎫∈⎪⎢⎣⎭①如图，即有两个交点，正确； ()f x A =②结合右图，且当时，在递增，错误； 2512ω=()f x 80,25π⎡⎤⎢⎥⎣⎦③，，1925,1212ω⎡⎫∈⎪⎢⎣⎭∴1212,22519T πππω⎛⎤=∈ ⎥⎝⎦，存在为最小值，为最大值，正确；1212,22519πππ⎛⎤∈ ⎥⎝⎦∴()1f x 12f x π⎛⎫+ ⎪⎝⎭④结合右图，若方程在内恰有五个解，需满足，即，同时结合左()12f x =-[]0,2π()102f ≤-1A ≥图，当，不一定有五个解，正确. 1A ≥()12f x =-故答案为：①③④.【点睛】本题考查了三角函数的图像和性质，考查了数形结合思想和分类讨论思想，属于难题.二、单选题13．已知，则“为纯虚数”是“”的（ ） C z ∈z 0z z +=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】A【分析】根据纯虚数的定义判断充分性，再举反例判断必要性即可 【详解】由题意，为纯虚数则设，则；z ()i ,0z b b b =∈≠R i i 0z z b b +=-=当时，可取，则为纯虚数不成立.故“为纯虚数”是“”的充分非必要条件 0z z +=0z z ==z z 0z z +=故选：A14．已知顶点在原点的锐角，始边在x 轴的非负半轴，始终绕原点逆时针转过后交单位圆于α3π，则的值为（ ）1(,)3P y -sin αA B C D【答案】B【分析】根据任意角的三角函数的定义求出，然后凑角结合两角差的正弦公式求出1cos()33πα+=-.sin α【详解】由题意得(为锐角)1cos(33πα+=-α∵为锐角，∴，∴α5336πππα<+<sin()03πα+>sin(sin sin (333πππααα⎡⎤⇒+=⇒=+-⎢⎥⎣⎦1123⎛⎫=--= ⎪⎝⎭故选：B15．某港口某天0时至24时的水深（米）随时间（时）变化曲线近似满足如下函数模型y x （）.若该港口在该天0时至24时内，有且只有3个时刻水深为30.5sin 3.246y x πωπ⎛⎫=++ ⎪⎝⎭0ω>米，则该港口该天水最深的时刻不可能为（ ） A ．16时 B ．17时C ．18时D ．19时【答案】D【分析】本题是单选题，利用回代验证法，结合五点法作图以及函数的最值的位置，判断即可．【详解】解：由题意可知，时，，0x =0.5sin 0 3.24 3.496y πωπ⎛⎫=⨯++= ⎪⎝⎭由五点法作图可知：如果当时，函数取得最小值可得：，可得，16x =51662ππωπ+=748ω=此时函数，函数的周期为：， 70.5sin 3.24486y x ππ⎛⎫=++ ⎪⎝⎭296147748T ππ==≈该港口在该天0时至24时内，有且只有3个时刻水深为3米，满足， 如果当时，函数取得最小值可得：，可得，19x =51962ππωπ+=757ω=此时函数，函数的周期为：，70.5sin 3.24576y x ππ⎛⎫=++ ⎪⎝⎭21147757T ππ==时，，如图：24x =70.5sin 24 3.243576y ππ⎛⎫=⨯++> ⎪⎝⎭该港口在该天0时至24时内，有且只有3个时刻水深为3米，不满足， 故选：D ．【点睛】本题考查三角函数的模型以及应用，三角函数的周期的判断与函数的最值的求法，考查转化思想以及数形结合思想的应用，是难题．16．设是的垂心，且，则的值为（ ） H ABC A 3450HA HB HC ++=cos BHC ∠A ．B ．C ．D ．【答案】D【分析】由三角形垂心性质及已知条件可求得，HB = HC = 求解．【详解】由三角形垂心性质可得，，不妨设HA HB HB HC HC HA ⋅=⋅=⋅x ，HA HB HB HC HC HA ⋅=⋅=⋅=∵345， HA + HB +0HC = ∴，23450HA HB HB HC HB ⋅++⋅=∴，同理可求得HB = HC =∴HB HC cos BHC HB HC ⋅∠== 故选：D ．【点睛】本题考查平面向量的运用及向量的夹角公式，解题的关键是由三角形的垂心性质，进而用同一变量表示出，要求学生有较充实的知识储备，属于中档题． HB HC，三、解答题17．已知关于x 的实系数一元二次方程.290x mx ++=(1)若复数z 是该方程的一个虚根，且，求m 的值； 4z z +=-(2)记方程的两根为和，若，求m 的值. 1x 2x 12x x -=【答案】(1)－2(2) ±±【分析】（1）利用，结合韦达定理可求解.2z z z =⋅（2）分讨论方程的两根为实根还是虚数根两种情况讨论，结合韦达定理可求解.【详解】（1）解：因为，所以，因为，所以，29z z z =⋅=3z=4z z +=-1z =-所以，由韦达定理可得，所以；1z =+2m z z -=+=2m =-（2）解：若方程的两根为实数根，则12x x -===解得，m =±若方程的两根为虚数根，则设，，可得1i x a b =+2i,,R x a ba b =-∈122x x b -==则，，，所以，所以1x a =2x a =21239x x a =+=26a =a =由韦达定理可得，所以12m xx -=+=±m =±此时，满足题意， 2360m ∆=-<综上，m =±±18．已知向量，，函数. cos sin 2x x m ⎫-=⎪⎭ ()2cos ,sin cos n x x x =+ ()f x m n =⋅(1)求函数的严格减区间与对称轴方程；()y f x =(2)若，关于x 的方程恰有三个不同的实数根，π2π,63x ⎡⎤∈-⎢⎥⎣⎦()()1sin 6πf x x R λλλ⎛⎫+++=∈ ⎪⎝⎭1x 2x ，求实数的取值范围及的值.3x λ123x x x ++【答案】(1)，；，π2ππ,π63⎡⎤++⎢⎥⎣⎦k k Z k ∈ππ62k x =+Z k ∈(2)， )1,33π2【分析】（1）由数量积的坐标表示求得，结合正弦函数的基准减区间和对称轴求得的严()f x ()f x格减区间和对称轴；（2）方程化简得和，由正弦函数性质和的范围，同时得出和，求得sin 1x =1sin 2x λ-=λ1x 23x x +结论．【详解】（1） ()22cos sin cos 2x xf x m n x x -=⋅=+1π2cos 2sin 226x x x ⎛⎫=+=+ ⎪⎝⎭，解得， ππ3π2π22π262k x k +≤+≤+π2πππ63k x k +≤≤+令，解得，ππ2π62x k +=+ππ62k x =+所以函数的严格减区间为，， π2ππ,π63⎡⎤++⎢⎥⎣⎦k k Z k ∈对称轴方程为； ππ62k x =+Z k ∈（2）， 2sin 2cos 2π12sin 62πf x x x x ⎛⎫⎛⎫+=+==- ⎪ ⎪⎝⎭⎝⎭即，形为，()212sin 1sin x x λλ-++=()22sin 1sin 10x x λλ-++-=所以，()()2sin 1sin 10x x λ⎡⎤---=⎣⎦当，有一个解，不妨设为，π2π,63x ⎡⎤∈-⎢⎥⎣⎦sin 10x -=1π2x =则，即有不同于的两个解，()2sin 10x λ--=1sin 2x λ-=12x π=因为，所以，π2π,63x ⎡⎤∈-⎢⎥⎣⎦1sin ,12y x ⎡⎤=∈-⎢⎥⎣⎦且在上严格递增，在上严格递减，ππ,62x ⎡⎤∈-⎢⎥⎣⎦sin y x =π2π,23x ⎡⎤∈⎢⎥⎣⎦sin y x =要想有不同于的两个解，则，解得， 1sin 2x λ-=1π2x =12λ⎫-∈⎪⎪⎭)1,3λ∈此时的两根关于对称，则， 1sin 2x λ-=π2x =23πx x +=所以. 1233π2x x x ++=19．近年来,为“加大城市公园绿地建设力度,形成布局合理的公园体系”,许多城市陆续建起众多“口袋公园”、现计划在一块边长为200米的正方形的空地上按以下要求建造“口袋公园”、如图所示,以中点A 为圆心,为半径的扇形草坪区,点在弧BC 上（不与端点重合）,AB 、弧BC 、EF FG ABC P CA 、PQ 、PR 、RQ 为步行道,其中PQ 与AB 垂直,PR 与AC 垂直.设.PAB θ∠=(1)如果点P 位于弧BC 的中点,求三条步行道PQ 、PR 、RQ 的总长度;(2)“地摊经济”对于“拉动灵活就业、增加多源收入、便利居民生活”等都有积极作用.为此街道允许在步行道PQ 、PR 、RQ 开辟临时摊点,积极推进“地摊经济”发展,预计每年能产生的经济效益分别为每米5万元、5万元及5.9万元.则这三条步行道每年能产生的经济总效益最高为多少?（精确到1万元）【答案】(1)(米)200+(2)2022万元【分析】(1)根据图依次求出三条线段长度即可求出总长度;(2)将PQ 、PR 、RQ 三边通过图中的关系用关于的等式表示,再记经济总效益,将进行表示,通θW W 过辅助角公式化简求出最值即可.【详解】（1）解:由题200,100,AC EA EC ==∴=,同理,故, π3EAC ∴∠=π3FAB ∴∠=π3BAC ∠=由于点P 位于弧BC 的中点,所以点P 位于的角平分线上,BAC ∠则, πsin 200sin1006PQ PR PA PAB ==⋅∠=⨯=, cos 200AQ AP PAB =∠==因为,, π3BAC ∠=AQ AR ==所以为等边三角形,ARQ A 则,100RQ AQ ==因此三条街道的总长度为.100100200l PQ PR RQ =++=++=+（2）由图可知,sin 200sin PQ AP θθ==, sin 200sin 100sin 33PR AP ππθθθθ⎛⎫⎛⎫=-=-=- ⎪ ⎪⎝⎭⎝⎭,cos 200cos AQ AP θθ==, cos 200cos 100cos 33AR AP ππθθθθ⎛⎫⎛⎫=-=-=+ ⎪ ⎪⎝⎭⎝⎭在中由余弦定理可知:ARQ A 222π2cos3RQ AQ AR AQ AR =+-()()22200cos 100cos θθθ=++()2200cos 100cos cos3πθθθ-⨯+, 30000=则100RQ =设三条步行道每年能产生的经济总效益,则W ()5 5.9W PQ PR RQ =+⨯+⨯()200sin 100sin 5θθθ=+-⨯+, π1000sin 3θ⎛⎫=++ ⎪⎝⎭当即时取最大值, sin 13πθ⎛⎫+= ⎪⎝⎭π6θ=W最大值为.10002022+≈答:三条步行道每年能产生的经济总效益最高约为2022万元.20．在平面直角坐标系中，，设点，是线段AB 的n 等分点，其中()1,0A ()0,1B 1P 12,,n PP -⋅⋅⋅n ∈N ，. 2n ≥(1)当时，使用，表示，； 3n =OA OB 1OP 2OP (2)当时，求的值；2023n =121n OP OP OP -+++ (3)当时，求（，，i ，）的最小值. 10n =()i i j OP OP OP ⋅+ 1i ≤1j n -≤j ∈N【答案】(1)， 12133OP OA OB =+ 21233OP OA OB =+(2)(3)2325【分析】（1）根据题意结合向量的线性运算求解；（2）根据向量的坐标运算求解； （3）据向量的坐标运算可得，结合函数分析求解. ()()251510050i i j i j i i OP OP OP -+-+⋅+=u u u r u u u r u u u r 【详解】（1）由题意可得：， ()1i i i i i i OP OA AP OA AB OA OB OA OA OB n n n n ⎛⎫=+=+=+-=-+ ⎪⎝⎭u u u r u u r u u u r u u r u u u r u u r u u u r u u r u u r u u u r 当时，所以，. 3n =12133OP OA OB =+ 21233OP OA OB =+ （2）因为，则， ()()1,0,0,1A B ()()1,0,0,1OA OB ==u u r u u u r 由（1）可得：， 11,i i i i i OP OA OB n n n n ⎛⎫⎛⎫=-+=- ⎪ ⎪⎝⎭⎝⎭u u u r u u r u u u r 当时，则，， 2023n =2023,20232023i i i OP -⎛⎫= ⎪⎝⎭1,2,,1i n =⋅⋅⋅-所以 121202220211122022,20232023n OP OP OP -++++++⎛⎫+++= ⎪⎝⎭因为， ()20221202212202220222++++== 所以，()1212022,2022n OP OP OP -+++=121n OP OP OP -++⋅⋅⋅+==u u u r u u u r u u u u r （3）当时，，， 10n =10,1010i i i OP -⎛⎫= ⎪⎝⎭10,1010j j j OP -⎛⎫= ⎪⎝⎭ 可得， 101055501010101050i j i j i j i j i j OP OP --⋅--+⋅=⨯+⨯=u u u r u u u r ， 2222101050101050i i i i i OP --+⎛⎫⎛⎫=+= ⎪ ⎪⎝⎭⎝⎭， ()2i i j i i j OP OP OP OP OP OP ⋅+=+⋅ ()22515100105055505050i j i i i i i j i j -+-+-++⋅--+==构建， ()()251510050i j i i M j -+-+=①当，7，8，9时，， 6i =()()()225115100149515050i i i i i M j M -⋅+-+-+==≥可得当时，上式有最小值； 7i =2325②当时，， 5i =()2575100150M j -+==③当，2，3，4时，， 1i =()()()22591510065595050i i i i i M j M -⋅+-+-+==≥可得当时，上式有最小值； 3i =2325综上所述：的最小值为. ()i i j OP OP OP ⋅+ 232521．对于函数，，如果存在一组常数，，…，（其中k 为正整数，且()y f x =x ∈R 1t 2t k t ）使得当x 取任意值时，有则称函数120k t t t =<<< ()()()120k f x t f x t f x t ++++++= 为“k 级周天函数”.()y f x =(1)判断下列函数是否是“2级周天函数”，并说明理由：①；②；()1sin f x x =()22f x x =+(2)求证：当时，是“3级周天函数”；()32n n ω=+∈Z ()()cos g x x ω=(3)设函数，其中b ，c ，d 是不全为0的实数且存在，使()cos 2cos5cos8h x a b x c x d x =+++R m ∈得，证明：存在，使得.()4h m a =n ∈R ()0h n <【答案】(1)是，不是；理由见解析()1f x ()2f x (2)证明见解析(3)证明见解析【分析】（1）令，，然后化简，根据定义可知；10t =2πt =（2）令，，，然后化简，从而得证； 10t =22π3t =34π3t =（3）若，则，取，则；若，则利用反证法证明即可；若0a <()40h m a =<n m =()0h n <0a =时，由，可得，从而可得0a >()2π4π333h m h m h m a ⎛⎫⎛⎫++++= ⎪ ⎪⎝⎭⎝⎭2π4π033h m h m a ⎛⎫⎛⎫+++=-< ⎪ ⎪⎝⎭⎝⎭结论【详解】（1）令，，则， 10t =2πt =()()()1112sin sin sin sin 0f x t f x t x x x x π+++=++=-=所以是“2级周天函数”；()1sin f x x =，不对任意x 都成立，()()()()212212222240f x t f x t x t x t x t +++=+++++=++=所以不是“2级周天函数”；()22f x x =+（2）令，，，则 10t =22π3t =34π3t =()()()123g x t g x t g x t +++++ ()()()4π8πcos 32cos 322πcos 324π33n x n x n n x n ⎡⎤⎡⎤⎡⎤=+++++++++⎣⎦⎢⎥⎢⎥⎣⎦⎣⎦()()()4π8πcos 32cos 32cos 3233n x n x n x ⎡⎤⎡⎤⎡⎤=+++++++⎣⎦⎢⎥⎢⎥⎣⎦⎣⎦()()()π2πcos 32cos 32cos 3233n x n x n x ⎡⎤⎡⎤⎡⎤=+-+++++⎣⎦⎢⎥⎢⎥⎣⎦⎣⎦()()()ππcos 32cos 32cos sin 32sin 33n x n x n x ⎡⎤⎡⎤⎡⎤=+-+-+⎣⎦⎣⎦⎣⎦ ()()2π2πcos 32cos sin 32sin 33n x n x ⎡⎤⎡⎤++++⎣⎦⎣⎦()()()1cos 32cos 32322n x n x n x ⎡⎤⎡⎤⎡⎤=+-++⎣⎦⎣⎦⎣⎦()()1cos 32322n x n x ⎡⎤⎡⎤-++⎣⎦⎣⎦()()cos 32cos 320n x n x ⎡⎤⎡⎤=+-+=⎣⎦⎣⎦所以是“3级周天函数”；()()cos g x x ω=（3）对其进行分类讨论：1°若，则，此时取，则；0a <()40h m a =<n m =()0h n <2°若，采用反证法，若不存在，使得，则恒成立， 0a =n ∈R ()0h n <()0h x ≥由（2）可知是“3级周天函数”，()cos 2cos5cos8t x b x c x d x =++所以， ()2π4π033t x t x t x ⎛⎫⎛⎫++++= ⎪ ⎪⎝⎭⎝⎭所以， ()2π4π3033h x h x h x a ⎛⎫⎛⎫++++== ⎪ ⎪⎝⎭⎝⎭因为，，， ()0h x ≥2π03h x ⎛⎫+≥ ⎪⎝⎭4π03h x ⎛⎫+≥ ⎪⎝⎭所以， ()2π4π033h x h x h x ⎛⎫⎛⎫=+=+= ⎪ ⎪⎝⎭⎝⎭再由恒成立，()()π0cos 2cos80h x h x b x d x ++=⇒+=所以，0b d ==进而可得，这与b ，c ，d 是不全为0矛盾，0c =故存在，使得；n ∈R ()0h n <3°若，由，， 0a >()2π4π333h m h m h m a ⎛⎫⎛⎫++++= ⎪ ⎪⎝⎭⎝⎭()4h m a =得， 2π4π033h m h m a ⎛⎫⎛⎫+++=-< ⎪ ⎪⎝⎭⎝⎭所以存在，使得， n ∈R ()0h n <所以命题成立.。

浦东复旦附中分校2022 届高一（下)拓展考April 23rd, 2020I.Listening Comprehension (20 points)Section A Short ConversationsDirections：In Section 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. Spend more time working on chemistry problems。

B.Talk to his advisor about dropping the course.C.Work on the assignment with a classmate.D.Ask graduate assistants for help。

2. A. Go home to get the washing.B.Send dirty clothes to the laundry。

C.Pick up the washing.D.Remind the woman to hurry up.3.A。

The woman could use his calculator.B.He’ll finish the adding up for the woman.C.The woman’s calculator is better than his。

浦东复旦附中分校2022届高一（下）拓展考April23rd,2020I.Listening Comprehension(20points)Section A Short ConversationsDirections:In Section 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.Spend more time working on chemistry problems.B.Talk to his advisor about dropping the course.C.Work on the assignment with a classmate.D.Ask graduate assistants for help.2. A.Go home to get the washing.B.Send dirty clothes to the laundry.C.Pick up the washing.D.Remind the woman to hurry up.3. A.The woman could use his calculator.B.He’ll finish the adding up for the woman.C.The woman’s calculator is better than his.D.He’s faster at adding numbers up than the woman.4. A.The final will begin next week.B.The man should talk with his doctor again.C.She hopes the man will be able to play in the final.D.She wants the man to watch the soccer game with her.5. A.The advisor will approve of the man’s class schedule.B.The advisor is not easy to make an appointment with.C.The man should work harder next semester.D.The man should go to see his advisor.6. A.An educational policy.B.An economic issue.C.A heated debate.D.A famous economist.7. A.The former is more valued than the latter.B.Views on them vary from person to person.C.The former is less valued than the latter.D.Neither of them should be valued so much.8. A.Disappointed.B.Puzzled.C.Astonished.D.Relieved.9. A.Why the woman took up this job.B.Why there are ice-cream tasting schools.C.How the man went into the ice-cream business.D.How it feels to work as an ice-cream taster.10.A.They can go to the theatre on foot.B.The theatre is on the other side of the town.C.This town is big enough to walk around.D.Everything in the town is worth visiting.Section BDirections:In Section B,you will hear two short passages,and you will be asked to 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.Questions11through13are based on the following passage.11. A.A doctor.B.A model.C.A TV host.D.A magazine editor.12.A.Kirsty is going to develop an eating disorder.B.Teen programs talk a lot about appearance.C.Being thin is a common trend.D.Magazines are informative.13.A.How to develop a healthy diet.B.Whether the media are a bad influence.C.How to follow famous people’s lifestyles.D.Whether your appearance should matter or not.Questions14through16are based on the following passage.14.A.Peary wasn’t an experienced explorer.B.He had reached the pole before Peary did.C.Peary had announced his success too late.D.Peary’s exploration wasn’t thoroughly investigated.15.A.They interviewed Peary himself.B.They examined the tools that Peary used.C.They talked to one of Peary’s companions.D.They conducted a computer analysis of his photographs.16.A.Exploration of space.B.Doctor Cook’s exploration.C.Exploration of the South Pole.D.The Navigation Foundation’s conclusions.Section CDirections:In Section C,you will hear a longer conversation.The conversation will be read twice.After you hear the conversation,you are required to answer the following questions.Questions17through20are based on the following conversation.17.A.A biologist.B.A psychologist.C.An artist.D.A reporter.18.A.Talents play a role in children’s overall development.B.Typical parental involvement is not as important as thought.C.Arts training will improve children’s performance in other subjects.D.Children need to be assigned to different groups according to their ability.19.A.Its subjects were young children.B.It showed what skill is essential to maths.C.It illustrated why abstract reasoning is important.D.Its results helped explain why parents turn to arts education.20.A.By helping activate children’s brains.B.By helping raise educators’awareness.C.By helping children learn how to learn.D.By helping scientists find a rich environment.II.Multiple Choice.(20points)21.Rafael Nadal is a tennis champion equipped with a big heart,to win and resolution to defeat all his opponents.A.persuasionB.determinationC.fascinationD.accumulationaround 22.The winners of China’s Got Talent are planning to a talent-showthe world next month.A.carry on…voyageB.carry out…tourC complete…trip D.fulfill…journey23.The captain and were accused of abandoning passengers in South Korea ferry disaster.A.membersB.crewC.partnersD.team24.Chichester covered anyone had previously sailed alone.A.more than twice the distanceB.more than twice longerC.over twice longer the distance asD.as over twice long as25.Steve Jobs was fired by the person he had hired and trusted the fact he was the one who co-established the company.A.despite…whichB.regardless…thatC.in spite of…thatD.against…in which26.The parents attempted to marrying the boy,though motivated but without ahandsome pay,a decent car or a respectable apartment.A.encourage their daughter fromB.discourage their daughter intoC.dissuade their daughter intoD.persuade their daughter out of27.So that the boat almost.A.tough the sea became…turned downB.rough did the sea become…turned overC.rough the sea became…turned overD.tough did the sea become…turned down28.After around Cape Horn,Chichester sent the message of victory to London.A.succeeding in sailingB.successfully sailedC.succeeded in sailingD.being successfully sailed29.Although the boss appears to wear a smile most of the time,his smile is by no means sincere orgenuine,instead,it could be somewhat and hostile at times.A.hospitableB.friendlyC.toughD.sinister30.With much work that remains to be completed,the manager found their project.A.more than satisfyingB.not at all satisfiedC.far from satisfactoryD.with more satisfaction than not31.My friends thought“I had lost my mind”,which means they thought.A.I was very much depressedB.I had lost my heartC.I was in severe lack of confidenceD.I had been mad and crazy32.When challenges and setbacks,he remains,brave and firm.A.faced with…hesitantB.facing…resoluteC.in face of…indecisiveD.facing with…determined33.The stormy and treacherous weather difficulty and uncertainty in the search forthe missing passengers.A.added toB.added up toC.added upD.added34.Marathon is intended to put the athletes to a challenging test of and willpower.A.insistenceB.enduranceC.preservationD.bravery35.The devoted and determined single mother brought up her five children single-handed.Which of the following is Wrong?A.without others’assistanceB.independent of other’s helppletely of her ownD.all by herself36.Not until Unit One of College English on learning strategies the importance of distinguishing active vocabulary from passive ones in vocabulary accumulation. A.did he learn…he realized B.he learned…did he realizeC.had he learned…did he realizeD.he learned…didn’t he realize37.Only when he conquered his inner fear a newfound sense of accomplishment andself-confidenceA.did he gainB.he had gainedC.he gainedD.has he gained38.–I learned that Francis Chichester was knighted by Queen Elizabeth II.--Yes,so and so.A.he was…was Sir Francis DrakeB.was he…Sir Francis Drake wasC.he did…the same with Sir Francis DrakeD.he was…Sir Francis Drake was the same39.No sooner a shelter it began to pour.A.we found…thanB.had we found…whenC.had we found…thanD.we found…when40.Out to the finishing line…!And here!A.he dashes…comes our champion and heroB.dashes he…our champion and hero comeC.he dashes…come our champion and heroD.dashes he…comes our champion and hero III.Grammar and Vocabulary(20points)Section ADirections: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.Over-dried EarthThe south-west of the United States,together with some parts ofMexico across the Rio Grande,is one of the driest parts of the NorthAmerican continent.But,over the past two decades,even that expecteddryness(41)(take)to the limit.According to Park Williams,who works at Columbia University’s Lamont-Doherty EarthObservatory,the current lack of rainfall in the area constitutes amegadrought of a severity(42)(see)on only four otheroccasions in the past1,200years.Dr Williams studied the annual growth rings of1,586ancientA.essentialB.adventureC.adoptedD.entertainingE.routeF.disastrousK.religious trees,in order to reconstruct soil-moisture patterns going back to 800A.D.During warm,wet years trees grow fast,producing wide rings.During cold,dry (43)they grow more slowly,producing narrow rings.During a drought,a tree (44)not grow much at all.(45)they describe in this week’s Science ,the team identified dozens of droughts over the centuries in question.But four stood out.They then took the average soil-moisture value for the current drought and compared it with sequential (连续的)19-year averages with the previous four,one of them (46)(last)nearly a century.This showed that the region is already drier than it was during the first three of the previous megadroughts,and is equivalent to the event of 1575-1603.In a world (47)human actions are driving temperatures up,Dr Parker and hiscolleagues wondered how much people are (48)estimate that,they turned to climate modelling.Climate models are able to re-run the past with and (49)(blame)for this state of affairs.To the warming effects of human activity,offering a way to compare what actually happened with what might have done.In their simulated world in which anthropogenic (人类起源的)emissions had not increased the greenhouse-gas effect,the team found that a drought did indeed still influence the western reaches of North America during the first two decades of the 21st century.But this imaginary dry spell was considerably (50)(severe)than the real one—ranking 11th rather than 2nd in the period under study (see chart).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."Life of Pi"is narrated by an older Pi and takes place in thepresent as he tells his life story to a writer.Pi begins with how he got his name and how his family owned a zoo,but eventually leads into his family needing to leave India for financial 51.The heart of the story lies in Pi as a young man and the time he spent out at sea on a lifeboat with a tiger named Richard Parker,but getting to that point is quite an ordeal(苦难考验).Pi loses everything and everyone close to him in the blink of an eye and he doesn't 52on making what feels like an unbreakable bond with a Bengal tiger.The 53film offers a rather marvelous message about religion as Pi discusses the experimentation with several different religions as a young boy.Even though it seems kind of silly that a boyhas 54so many religions at one point in his life,it portrays a/an 55message in a light that doesn't seem too preachy (说教的).But the religious aspect is really only a backdrop for being on your own 56to finding who you really are.Feeling so alone for so long and having Pi talk about not being alone and not getting to say goodbye to those he cared about most is one of those things that just hits you really hard.While witnessing Pi attempt to tame Richard Parker is really 57,there are so many unbelievable sights to witness in "Life of Pi".There's the whale,the jellyfish,the flying fish,that58storm,and that seaweed island with all of the meerkats (a kind of monkey)."Life of Pi"never really lets up on being 59breathtaking and amazing from beginning to end."Life of Pi"is pleasantly satisfying.With a heartfelt message,beautiful visuals,and stunning camera work,"Life of Pi"is truly a sight to behold and illustrates how 603D can be to a film that isn't just a summer hit orblockbuster.IV.Reading ComprehensionSection A(15points)Directions:For each blank in the following passage there are four words or phrases marked A,B, C and D.Fill in each blank with the word or phrase that best fits the context.The Old Man and the Sea is the story of an epic struggle between an old,seasoned fisherman and the greatest catch of his life.For eighty-four days,Santiago,an aged Cuban fisherman,has set out to sea and returned61-handed.So conspicuously unlucky is he that the parents of his young, devoted apprentice(徒弟)and friend,Manolin,have62the boy to leave the old man in order to fish in a more prosperous boat.63,the boy continues to care for the old man upon his return each night.He helps the old man haul his gear to his ramshackle hut（破旧不堪的陋室）, secures food for him,and discusses the latest developments in American baseball,especially the trials of the old man’s hero,Joe DiMaggio.Santiago is64that his unproductive streak of failure will soon come to an end,and he65to sail out farther than usual the following day.On the eighty-fifth day of his unlucky streak,Santiago doesas promised,sailing his skiff(小艇)far beyond the island’sshallow coastal waters and66into the Gulf Stream.Heprepares his lines and drops them.At noon,a big fish,which heknows is a marlin(青枪鱼),takes the bait(鱼饵)that Santiago hasplaced one hundred fathoms deep in the waters.The old manexpertly hooks the fish,but he cannot pull it in.Instead,the fishbegins to pull the boat.Unable to tie the line67to the boat for fear the fishwould snap a taut line,the old man bears the strain of the linewith his shoulders,back,and hands,ready to give slack shouldthe marlin make a run.The fish68the boat all through theday,through the night,through another day,and through another night.The entire time,Santiago endured69pain from the fishing line.Whenever the fish lunges,leaps,or makes a dash for70,the cord(绳)cuts Santiago badly.Although wounded and weary,the old man feels a deep empathy and admiration,for the marlin,his brother in suffering,strength,and determination.As Santiago sails on with the fish,the marlin’s blood leaves a trail in the water and attracts sharks.The first to attack is a great mako shark,which Santiago manages to kill with the harpoon (鱼叉).In the71,the old man loses the harpoon and lengths of valuable rope,which leaves him vulnerable to other shark attacks.Although he kills several sharks,more and more appear. They devour(吞噬)the marlin’s precious meat,leaving only skeleton,head,and tail.Santiago punished himself for going“out too far,”and for sacrificing his great and worthy72.The next morning,a crowd of73fishermen gathers around the skeletal carcass of the fish, which is still lashed（紧系）to the boat.Knowing nothing of the old man’s struggle,tourists at a nearby caféobserve the74of the giant marlin and mistake it for a shark.Manolin,who has been worried over the old man’s absence,is moved to tears when he finds Santiago safe in his bed.The boy fetches the old man some coffee and the daily papers with the baseball scores,and watches him sleep.When the old man wakes,the two agree to fish as75once more.The old man returns to sleep and dreams his usual dream of lions at play on the beaches of Africa.61. A.second B.single C.empty D.first62. A.made B.encouraged C.dissuaded D.forced63. A.Nevertheless B.Therefore C.Furthermore D.Besides64. A.confident B.depressed C.pessimistic D.proud65. A.determines B.fails C.considers D.favors66. A.venturing B.diving C.drowning D.securing67. A.slow B.fast C.quick D.loose68. A.drives B.steers C.pushes D.pulls69. A.sustainable B.temporary C.constant D.instant70. A.attempt B.control C.freedom D.damage71. A.journey B.shock C.quarrel D.struggle72. A.opponent B.master C.acquaintance D.hero73. A.disappointed B.amazed C.terrified D.accomplished74. A.remains B.meats C.ruins D.rests75. A.seniors panies C.coaches D.partners Section B(10points)Directions:Read the following passages.Each passage is followed by several questions or unfinished statement.For each of them there are four choices marked A,B,C and D.Choose the one that first best according to the information given in the passage you have just read.Like many of my generation，I have a weakness for hero worship.At some point，however，we all begin to question our heroes and our need for them.This leads us to ask：What is a hero?Despite immense differences in cultures，heroes around the world generally share a number of characteristics that instruct and inspire people.A hero does something worth talking about.A hero has a story of adventure to tell and a community who will listen.But a hero goes beyond mere fame.Heroes serve powers or principles larger than themselves.Like high-voltage(电压) transformers，heroes take the energy of higher powers and step it down so that it can be used by ordinary people.The hero lives a life worthy of imitation.Those who imitate a genuine hero experience life with new depth，enthusiasm，and meaning.A sure test for would-be heroes is what or whom do they serve?What are they willing to live and die for?If the answer or evidence suggests they serve only their own fame，they may be famous persons but not heroes.Madonna and Michael Jackson are famous，but who would claim that their fans find life more abundant?Heroes are catalysts(催化剂)for change.They have a vision from the mountain top.They have the skill and the charm to move the masses.They create new possibilities.Without Gandhi，India might still be part of the British Empire.Without Rosa Parks and Martin Luther King，Jr.，we might still have segregated(隔离的)buses，restaurants，and parks.It may be possible for large-scale change to occur without leaders with magnetic personalities，but the pace of change would be slow，the vision uncertain and the committee meetings endless.76.Although heroes may come from different cultures，they.A．generally possess certain inspiring characteristicsB.probably share some weaknesses of ordinary peopleC.are often influenced by previous generationsD.all unknowingly attract a large number of fans77.According to the passage，heroes are compared to high-voltage transformers in that. A．they have a vision from the mountaintopB.they have warm feelings and emotionsC.they can serve as empowering examples of noble principlesD.they can make all people feel stronger and more confident78.Madonna and Michael Jackson are not considered heroes because.A．they are popular only among certain groups of peopleB.their performances do not improve their fans morallyC.their primary concern is their own financial interestsD.they are not clear about the principles they should follow79.Gandhi and Martin Luther King are typical examples of outstanding leaders who. A．are good at demonstrating their charming charactersB.can move the masses with the skill and the charmC.are capable of meeting all challenges and hardshipsD.can provide an answer to the problems of their people80.The author concludes that historical changes would.A．be delayed without leaders with inspiring personal qualitiesB.not happen without heroes making the necessary sacrificesC.take place if there were heroes to lead the peopleD.produce leaders with attractive personalitiesV.Translation(3,3,4,5)Directions:Translate the following sentences into English,using the words given in the brackets.81.他说起意大利就好像亲自去过似的。

浦东复旦附中分校2022届高一（下）拓展考I. Listening Comprehension (20 points)Section A Short ConversationsDirections: In Section A, you will hear ten short conversations between two speakers. At theend of each conversation, a question will be asked about what was said. The conversationsand the questions will be spoken only once. After you hear a conversation and the questionabout 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. Spend more time working on chemistry problems.B. Talk to his advisor about dropping the course.C. Work on the assignment with a classmate.D. Ask graduate assistants for help.2. A. Go home to get the washing.B. Send dirty clothes to the laundry.C. Pick up the washing.D. Remind the woman to hurry up.3. A. The woman could use his calculator.B He’ll finish the adding up for the woman.C. The woman’s calculator is better than his.D. He’s faster at adding numbers up than the woman.4. A. The final will begin next week.B. The man should talk with his doctor again.C. She hopes the man will be able to play in the final.D. She wants the man to watch the soccer game with her.5. A. The advisor will approve of the man’s class schedule.B. The advisor is not easy to make an appointment with.C. The man should work harder next semester.D. The man should go to see his advisor.6. A. An educational policy..B. An economic issue.C A heated debate.D. A famous economist.7. A. The former is more valued than the latter.B. Views on them vary from person to person.C. The former is less valued than the latter.D. Neither of them should be valued so much.8. A. Disappointed.B. Puzzled.C. Astonished.D. Relieved.9. A. Why the woman took up this job.B. Why there are ice-cream tasting schools.C. How the man went into the ice-cream business.D. How it feels to work as an ice-cream taster.10. A. They can go to the theatre on foot.B. The theatre is on the other side of the town.C. This town is big enough to walk around.D. Everything in the town is worth visiting.Section BDirections: In Section B, you will hear two short passages, and you will be asked to 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. A doctor.B. A model.C. A TV host.D. A magazine editor.12. A. Kirsty is going to develop an eating disorder.B. Teen programs talk a lot about appearance.C. Being thin is a common trend.D. Magazines are informative.13. A. How to develop a healthy diet.B. Whether the media are a bad influence.C. How to follow famous people’s lifestyles.D. Whether your appearance should matter or not.Questions 14 through 16 are based on the following passage.14. A. Peary wasn’t an experienced explorer.B. He had reached the pole before Peary did.C. Peary had announced his success too late.D. Peary’s exploration wasn’t thoroughly investigated.15. A. They interviewed Peary himself.B. They examined the tools that Peary used.C. They talked to one of Peary’s companions.D. They conducted a computer analysis of his photographs.16. A. Exploration of space.B. Doctor Cook’s exploration.C. Exploration of the South Pole.D. The Navigation Foundation’s conclusions.Section CDirections: In Section C, you will hear a longer conversation. The conversation will be read twice. After you hear the conversation, you are required to answer the following questions. Questions 17 through 20 are based on the following conversation.17. A. A biologist.B. A psychologist.C. An artist.D. A reporter.18. A. Talents play a role in children’s overall development.B. Typical parental involvement is not as important as thought.C. Arts training will improve children’s performance in other subjects.D. Children need to be assigned to different groups according to their ability.19. A. Its subjects were young children.B. It showed what skill is essential to maths.C. It illustrated why abstract reasoning is important.D. Its results helped explain why parents turn to arts education.20. A. By helping activate children’s brains.B. By helping raise educators’ awareness.C. By helping children learn how to learn.D. By helping scientists find a rich environment.II. Multiple Choice. (20 points)1.Rafael Nadal is a tennis champion equipped with a big heart, to win and resolution to defeat all his opponents.A. persuasionB. determinationC. fascinationD. accumulation【答案】B【解析】【详解】考查名词词义辨析。

2019-2020学年复旦大学附属中学高三英语下学期期中考试试题及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AFor some people, there’s no better companion than mans best friend-a dog. This four-legged pet can bring comfort and joy and provide much- needed exercise for you when it needs walkies! This probably explains why dog ownership increased last year because people spent more time at home during he CovID-I9 lockdown.However, as demand for a new dog increased, so did the price tag. Popular breeds, such as Cockapoos and Cocker Spaniels, saw even sharper price increases, and puppies have been selling for $3,000 or more.Animal welfare charities fearthat high prices could encourage puppy farming, smuggling (走私) or dog theft. An investigation found some breeders have been selling puppies and kittens on social media sites--something charities have called “extremely irresponsible”.But despite some new owners purchasing a dog legally, maybe from a rescue center or registered breeder, they’ve proved to be ill-prepared for life with a new pet, and the pet itself has found it hard tocome to terms withlife in a new home.Looking to the future, there are concerns about the welfare of these much-loved pets. Lan Alkin manager of the Oxfordshire Animal Sanct uary in the UK, notes: “At the moment, the dogs are having a great time, but separation anxiety could still surface when people go back to work.” And Cliare Calder from the UKs Dogs Trust rescue charity says, “The economic situation also means that some people may find they can’t afford to look aftera dog.” The message is not to buy a dog in haste and to pick one that fits into our lifestyle.1. The greater demand for dogs can cause the following problems except ________.A. illegal trade of dogsB. less dog farmingC. high prices of dogsD. online sale of dogs2. What does the underlined phrase"come to terms with"in paragraph 4 mean?A. Fit in withB. Go in forC. Make up for.D. End up with3. What can we learn from the last paragraph?A. Despite the problems, dogs are living happily.B. The writer has a positive attitude towards dogs future.C. Experts are worried that dogs will be unaffordable to people.D. The writer advises people to think twice before keeping dogs as pets.BCalifornia's August Complex Fire tore through more than 1,600 square miles of forest last summer,burning nearly every tree in its path. It was the largest wildfire in the state's recorded history, breaking the record previously set in 2018. After the fire, land managers must determine where to most efficiently plant new trees.A predictive mapping model called the Postfire Spatial Conifer Restoration Planning Tool recently described in Ecological Applications could inform these decisions, saving time and expense. The tool can “show where young trees are needed most, where the forest isn't going to come back on its own, where we need to intervene（干预）if we want to maintain forests," says lead author Joseph Stewart, an ecologist at the University of California, Davis.To develop the model, Stewart and his colleagues classified data collected from more than 1,200 study plots in 19 areas that burned between 2004 and 2012. They combined these data with information on rainfall, geography, climate, forest composition and bum severity.Theyalso included how many seeds sample conifer trees （针叶树）produced in 216locations over 18 years, assessing whether the trees release different numbers of seeds after a fire.The tool's potential benefits are significant, says Kimberley Davis, a conservation scientist at theUniversityofMontana, who was not involved in the study. Those managers will still have to make hard decisions, such as which species to plant in areas that may experience warmer and drier conditions resulting from climate change, but the model provides some research-based guidance to help the forests recover.4. What challenge do land managers face after the wildfire?A. Lack of wood supplies.B. Where to plant new trees best.C. How to save the burned trees.D. Loss of trees and wild animals.5. What's the main idea of paragraph 2?A. The function of the tool.B. The disadvantages of the tool.C. The improvement of the tool.D. The development of the tool.6. What does the underlined word "They" refer to?A. The study plots.B. The data.C. Stewart and his colleagues.D. The seeds.7. What isDavis' attitude towards the tool?A. Skeptical.B. Ambiguous.C. Tolerant.D. Optimistic.CMore than 10,000 people were made homeless in Ternang when the Sungai Mas overflowed its banks yesterday after six days of continuous heavy rain.The wooden bridge across the river has been washed away. The town is cut off by flood waters. At the fifth mile, Jalan Tengkn, the water is two meters deep. It is closed to all traffic. Flooding first happened at mid-afternoon yesterday along the river banks. People trying to get to higher ground were just in time to escape the destroying of the flood. Most of the flood victims(受害者) had to leave all their things behind.The National Flood Relief(救济) Center was reported to give its help and by early evening the whole town was moved out, helped by the army, police, Red Cross Society and volunteers(志愿者).The flood victims are now housed in different simple relief centers in the nearby town of Ternang. “Everything possible is being done to help the unlucky people,” a government spokesman said, “In fact, money, food and clothing have begun to come in from public organizations and helpful people. A Disaster Relief Fund(救灾基金会) will be started as soon as possible.”According to the latest reports it is still raining heavily at Ternang. The whole town is expected to be wholly covered by the water. So far no deaths have been reported.8. This passage is probably taken from______.A. a storybookB. a textbookC. a magazineD. a newspaper9. It seems that the flood happened just because ______.A. the banks were too lowB. the wooden bridge is solid (坚固)C. the river was too narrowD. they had had wet days for some time10.From the passage we know that ______.A. though the town was flooded, you could still go there by busB. as soon as the water flowed over the banks, people began to go to higher landsC. the government and the whole society are taking great care of the flood victimsD. the flood hasn't brought a large loss to the town as nobody died in the floodA. A Big FloodB. Ternang Is destroyedC. Over 10,000 People Are HomelessD. Rescue Team ArrivedDPeople saved a 20-foot orca (虎鲸) that was stuck between rocks on an Alaskan shore by continuously pouring water over it and protecting it from birds who circled above the defenseless whale.The whale was ultimately saved after a six-hour, labor-intensive life-saving operation. Someone spotted the large whale on the Prince of Wales Island near the coast of British Columbia on the morning of July 29th. The Coast Guard was called around 9 a.m. local time. Chance Strickland, the captain of a private yacht in Alaska, and his crew anchored and began life-saving action that were videoed by Aroon Melane and posted on the Internet.Strickland could hear the orca calling out to killer whales swimming in the area. People on other boats stopped with water and buckets to pour water over the animal. “There were tears coming out of its eyes,” Mr. Strickland told the local newspaper. “It was pretty sad.”The group of people formed a chain that passed buckets of seawater back and forth and poured the water on the orca, which seemed to liven it up. It made a noise and raised its tail when it got water.The National Oceanic and Atmospheric Administration (NOAA) was called in, which can be seen on the video using a machine to spray amist of seawater on the orca, which doubled as a way to keep the whale cool and scare the large group of birds that were hoping to feast on the beast.Melane said in her video that the orca was stranded (搁浅) for about six hours until the tide came in andswept it back into the ocean. The group efforts of Strickland’s crew and the NOAA saved the 13-year-old killer whale.12. Why did birds circle above the orca?A. They were eager to eat it.B. They wished to protect it.C. They were attracted by the people.D. They wanted to find a place to rest.13. What did Strickland do immediately after finding the whale?A. Posted pictures online.B. Called friends for help.C. Took action to save it.D. Videoed the trapped animal.14. Why did the whale make a noise and raise its tail?A. To express its eager for water.B. To extend its thanks to people.C. To call out to its fellow whales.D. To show its power and sadness.A. Killer Whale Got SavedB. The Orca Inspired KindnessC. Combined Efforts WantedD. Animals and Humans United第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。