《附15套高考模拟卷》上海市第二中学2019-2020学年高三9月初态测试物理试卷含解析

第二学期期中高三年级数学学科教学质量监测试卷（满分150分，时间120分钟）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）考生应在答题纸的相应位置直接填写结果． 1. 若集合{}0A x x =>，{}1B x x =<，则AB = ．2. 已知复数z 满1z i ⋅=+（i 为虚数单位），则z = ．3. 函数()sinx cosxf x cosx sinx=的最小正周期是 ．4. 已知双曲线222181x y a -=（0a >）的一条渐近线方程为3y x =，则a = ．5. 若圆柱的侧面展开图是边长为4的正方形，则圆柱的体积为 ．6. 已知x y ，满足0220x y x y x -≤⎧⎪+≤⎨⎪+≥⎩，则2z x y =+的最大值是 ． 7. 直线12x t y t =-⎧⎨=-⎩（t 为参数）与曲线32x cos y sin θθ=⎧⎨=⎩（θ为参数）的交点个数是 ．8. 已知函数()()220()01xx f x log x x ⎧≤⎪=⎨<≤⎪⎩ 的反函数是1()f x -，则11()2f -= ．9. 设多项式231(1)(1)(1)nx x x x ++++++++（*0x n N ≠∈，）的展开式中x 项的系数为n T ，则2nn T limn →∞= ．10. 生产零件需要经过两道工序，在第一、第二道工序中产生废品的概率分别为0.01和p ，每道工序产生废品相互独立．若经过两道工序后得到的零件不是废品的概率是0.9603，则p = ．11. 设向量m ()x y =，，n ()x y =-，，P 为曲线1m n ⋅=（0x >）上的一个动点，若点P 到直线10x y -+=的距离大于λ恒成立，则实数λ的最大值为 ．12. 设1210x x x ，，，为1210，，，的一个排列，则满足对任意正整数m n ，，且110m n ≤<≤，都有m n x m x n +≤+成立的不同排列的个数为 ．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．考生应在答题纸的相应位置，将代表正确选项的小方格涂黑． 13. 设a b R ∈，，则“4a b +>”是“1a >且3b >”的………………………（ ）（A ）充分而不必要条件 （B ）必要而不充分条件（C ）充要条件（D ）既不充分又不必要条件14. 如图，P 为正方体1111ABCD A BC D -中1AC 与1BD 的交点，则PAC ∆在该正方体各个面上的射影可能是 …………………………………………………………………（ ）（A ）①②③④ （B ）①③ （C ）①④ （D ）②④ 15. 如图，在同一平面内，点P 位于两平行直线12l l ，同侧，且P 到12l l ，的距离分别为13，．点M N ，分别在12l l ，上，8PM PN +=，则PM PN ⋅的最大值为…………………（ ）（A ）15 （B ）12 （C ）10 （D ）9 16. 若存在t R ∈与正数m ，使()()F t m F t m -=+成立，则称“函数()F x 在x t =处存在距离为2m 的对称点”．设2()x f x xλ+=（0x >），若对于任意t ∈，总存在正数m ，使得“函数()f x 在x t =处存在距离为2m 的对称点”，则实数λ的取值范围是…………………………………………………………………………………………（ ）（A ）(]02， （B ）(]12，（C ）[]12， （D ）[]14， 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8分，第2小题满分6分）如图，在正方体1111ABCD A BC D -中，E F 、分别是线段1BC CD 、的中点．（1）求异面直线EF 与1AA 所成角的大小； （2）求直线EF 与平面11AA B B 所成角的大小．18. （本题满分14分，第1小题6分，第2小题8分）已知抛物线22y px =（0p >），其准线方程为10x +=，直线l 过点(0)T t ，（0t >）且与抛物线交于A B 、两点，O 为坐标原点．（1）求抛物线方程，并证明：OB OA ⋅的值与直线l 倾斜角的大小无关； （2）若P 为抛物线上的动点，记||PT 的最小值为函数()d t ，求()d t 的解析式．19. （本题满分14分，第1小题6分，第2小题8分）对于定义域为D 的函数()y f x =，如果存在区间[]m n D ⊆，（m n <），同时满足： ①()f x 在[]m n ，内是单调函数；②当定义域是[]m n ，时，()f x 的值域也是[]m n ，．则称函数()f x 是区间[]m n ，上的“保值函数”． （1）求证：函数2()2g x x x =-不是定义域[01]，上的“保值函数”； （2）已知211()2f x a a x=+-（0a R a ∈≠，）是区间[]m n ，上的“保值函数”，求a 的取值范围．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）数列{}n a 中，已知12121()n n n a a a a k a a ++===+，，对任意*n N ∈都成立，数列{}n a 的前n 项和为n S ．（这里a k ，均为实数） （1）若{}n a 是等差数列，求k 的值；（2）若112a k ==-，，求n S ； （3）是否存在实数k ，使数列{}n a 是公比不为1的等比数列，且任意相邻三项12m m m a a a ++，，按某顺序排列后成等差数列？若存在，求出所有k 的值；若不存在，请说明理由．21. （本题满分18分，第1小题满分4分，第2小题满分6分，第3小题满分8分）设T R ⊂≠，若存在常数0M >，使得对任意t T ∈，均有t M ≤，则称T 为有界集合，同时称M 为集合T 的上界．（1）设12121x x A y y x R ⎧⎫-⎪⎪==∈⎨⎬+⎪⎪⎩⎭，、212A x sinx ⎧⎫=>⎨⎬⎩⎭，试判断1A 、2A 是否为有界集合，并说明理由； （2）已知2()f x x u =+，记11()()()(())n n f x f x f x f f x -==，（23n =，，）．若m R ∈，1[)4u ∈+∞，，且{}()n B f m n N *=∈为有界集合，求u 的值及m 的取值范围；（3）设a b c 、、均为正数，将222()()()a b b c c a ---、、中的最小数记为d ．是否存在正数(01)λ∈，，使得λ为有界集合222{|dC y y a b c==++，a b c 、、均为正数}的上界，若存在，试求λ的最小值；若不存在，请说明理由．参考答案及评分标准一、填空题（本大题共有12题，满分54分） 1、()0,1 2、1 3、π 4、3 5、16π6、37、28、1-9、1210、0.03 1112、512 二、选择题（本大题共有4题，满分20分） 13、B 14、C 15、A 16、A三、解答题（本大题共有5题，满分76分）17. 解：（1）方法一：设正方体棱长为2，以D 为原点，直线DA ，DC ，1DD 为x ，y ，z 轴，建立空间直角坐标系，则(000)D ，，，(220)B ，，，(020)C ，，，1(002)D ，，，故(12E ，，，(011)F ，，，()111EF =--，，，()1002AA =，，， …………………4/设异面直线EF 与1AA 所成角的大小为α，向量EF 与1AA 所成角为β，则11EF AA cos cos EF AA αβ⋅==⋅…… 6/3==，……7/注意到02πα⎛⎤∈ ⎥⎝⎦，，故3arccosα=，即异面直线EF 与1AA 所成角的大小为3arccos．…………………8/ （2）由（1）可知，平面11AA B B 的一个法向量是(100)n =，，，…………………10/设直线EF 与平面11AA B B 所成角的大小是θ，向量EF 与n 所成角为γ，则EF n sin cos EF nθγ⋅==⋅………12/3=13/1又02πθ⎡⎤∈⎢⎥⎣⎦，，θ∴=线EF 与平面11AA B B 所成角的大小为．………………14/方法二：设正方体棱长为2．（1）在面11CC D D 内，作FH CD ⊥于H ，联结HE ．因为正方体1111ABCD A BC D -，所以1AA ∥1DD ；在面11CC D D 内，有FH ∥1DD ，故异面直线EF 与1AA 所成的角就是EFH ∠（或其补角）．………………………4/由已知及作图可知，H 为CD 的中点，于是，在Rt EFH ∆中，易得1FH =，HE=，故HE tanEFH FH∠=， ………………………………………… 6/== 7/ 又(0)2EFH π∠∈，，所以EFH∠=从而异面直线EF 与1AA 所成角的大小为8/（2）因为正方体1111ABCD A BC D -，所以平面11AA B B ∥平面11CC D D ，故直线EF 与平面11AA B B 所成角的大小就是直线EF 与平面11CC D D 所成角．注意到BC ⊥平面11CC D D ，即EC ⊥平面11CC D D ，所以直线EF 与平面11AA B B所成角的大小即为EFC∠． ………………………………10/在Rt EFC∆中，易得1EC FC ==，，故ECtan EFCFC∠=……………………12/2==，………………13/又(0)2EFCπ∠∈，，故2E F C a r c ta n∠=，即直线EF与平面11AA B B所成角的大小为……14/18.解：（1）方法一：由题意，2=p，所以抛物线的方程为xy42=．……………2/当直线l的斜率不存在时，直线l的方程为tx=，则(A t，(B t-，，ttOBOA42-=⋅．…………3/当直线l的斜率k存在时，则0≠k，设l的方程为)(txky-=，11()A x y，，22()B x y，，由24()y xy k x t⎧=⎨=-⎩消去x，得0442=--ktyky，故121244y yky y t⎧+=⎪⎨⎪=-⎩，所以，ttyyyyyyxx41622122212121-=+=+=⋅．…………………………………………5/综上，OBOA⋅的值与直线l倾斜角的大小无关．…………………………………………6/方法二：由题意，2=p，所以抛物线的方程为xy42=．………………………………2/依题意，可设直线l 的方程为x my t =+（m R ∈），11()A x y ，，22()B x y ，，由24y x x my t ⎧=⎨=+⎩得2440y my t --=， 故121244y y my y t+=⎧⎨=-⎩， 所以，12121212()()OA OB x x y y my t my t y y ⋅=+=+++221212(1)()m y y mt y y t =++++ …………………………5/22(1)(4)4m t mt m t =+-+⋅+24t t =-综上，OB OA ⋅的值与直线l倾斜角的大小无关． …………………………6/（2）设00()P x y ，，则0204x y =，||PT =， ……………………… (8)/注意到00≥x ，所以，若20t -≥，即2t ≥，则当02x t =-时，||PT 取得最小值，即()2)d t t =≥；………10/若20t -<，即有02t <<，则当00x =时，||PT 取得最小值，即()(02)d t t t =<<；………12/综上所述，()()2()02t d t tt ⎧≥⎪=⎨<<⎪⎩…………………………………………………14/19.解：（1）函数2()2g x x x =-在[01]x ∈，时的值域为[10]-，，…………………………4/不满足“保值函数”的定义，因此函数2()2g x x x =-不是定义域[01]，上的“保值函数”．………………………6/（2）因xa a x f 2112)(-+=在[]m n ，内是单调增函数，故()()f m mf n n ==，，……8/这说明m n ，是方程x xa a =-+2112的两个不相等的实根， ………………………………10/其等价于方程1)2(222=++-x a a x a 有两个不相等的实根，……………………………11/由222(2)40a a a ∆=+->解得23-<a 或21>a ． ………………………………………13/ 故a的取值范围为3122⎛⎫⎛⎫-∞-+∞ ⎪ ⎪⎝⎭⎝⎭，，． ………………………………………………14/20.解：（1）若{}n a 是等差数列，则对任意*n N ∈，有122n n n a a a ++=+，………………2/即121()2n n n a a a ++=+，………………………………………………………………………3/故12k =．………………………………………………………………………………………4/（2）当12k =-时，121()2n n n a a a ++=-+，即122n n n a a a ++=--， 211()n n n n a a a a ++++=-+，故32211()n n n n n n a a a a a a ++++++=-+=+． …………………………………………5/所以，当n 是偶数时，1234112()(11)22n n n n nS a a a a a a a a n -=++++++=+=+=；……………………7/当n 是奇数时，2312()2a a a a +=-+=-，12341n n n S a a a a a a -=++++++123451()()()n n a a a a a a a -=+++++++11(2)22n n -=+⨯-=-． ……………9/综上，()()222n n n S nn-=⎧⎪=⎨=⎪⎩（*k N ∈）． …………………………………………10/（3）若}{n a 是等比数列 ，则公比a a a q ==12，由题意1≠a ，故1-=m m a a ，m m a a =+1，12++=m m a a ．……11/① 若1m a +为等差中项，则122m m m a a a ++=+，即112m m m a a a -+=+⇔221a a =+，解得1=a （舍去）；……12/② 若ma 为等差中项，则122m m m a a a ++=+，即112m m m a a a -+=+⇔22a a =+，因1≠a ，故解得，2a =-，11122215m m m m m m a a a k a a a a a +-++====-+++； ……………………………14/③ 若2m a +为等差中项，则212m m m a a a ++=+，即112221m mma a aa a+-=+⇔=+， 因为1≠a ，解得212215a a k a =-==-+，． …………………………………………15/综上，存在实数k满足题意，25k =-．…………………………………………………16/21.解：（1）对于1A ，由2121x xy -=+得1201x y y +=>-，解得11y -<<，………………2/1A ∴为有界集合； …………………………………………3/显然252266A x k x k k Z ππππ⎧⎫=+<<+∈⎨⎬⎭⎩，不是有界集合． ………………………4/（2）记()n n a f m =，则21n n a a u +=+．若14u =，则21()4f m m =+，22111()42n n n n n a a a a a +=+=-+≥，即1n n a a +≥，且211111()()2422n n n n a a a a +-=-=-+，从而1111222n n n a a a +-=-⋅+． （ⅰ）当12m =时，1()2n n f m a ==，所以1{}2B =，从而B 为有界集合．…………5/（ⅱ）当12m <时，由2114n n a a +=+，2111()()4a f m f m m ===+，显然，此时0n a >，利用数学归纳法可得12n a <，故B 为有界集合．…………………………………………6/（ⅲ）当12m >时，211111()()42n n a a a f m f m m m +≥≥≥===+≥>，2114n n n n a a a a +-=-+21()2n a =- 211()2a ≥-，即2111()2n n a a a +-≥-，由累加法得2111(1)()2n a a n a ≥+--→+∞，故B 不是有界集合．因此，当14u =，且12m ≤时，B 为有界集合；当14u =，且12m >时，B 不是有界集合； 若14u >，则211()()a f m f m m u u ===+≥，即114a u ≥>， 又2114n n a a u u +=+>>（n N *∈）， 即14n a >（n N *∈）． 于是，对任意n N *∈，均有221111()244n n n n n a a a a u a u u +-=-+=-+-≥-，即114n n a a u +-≥-（n N *∈），再由累加法得11(1)()4n a a n u ≥+--→+∞，故B 不是有界集合．………8/综上，当14u =，且12m ≤时，B 为有界集合；当14u =，且12m >时，B 不是有界集合；当14u >(m R ∈)时，B 不是有界集合． 故，满足题设的实数u 的值为14，且实数m 的取值范围是11[]22-，．………………10/ （3）存在．………………………………………………………………………11/不妨设a b c ≥≥．若2a cb +≤，则2a b c ≥-，且2()d b c =-． 故22222225()5()()d a b c b c a b c -++=--++22225()[(2)]b c b c b c ≤---++3(2)0c c b =-<，即22222215()05d d a b c a b c -++<⇔<++；…………13/若2a cb +>，则2a ac b <+<，即220a b a b <⇔-<， 又2a cb bc a b +>⇔->-，故2()d a b =-，又 22222225()5()()d a b c a b a b c -++=--++22(2)(2)0a b a b c =---<，即 2225()0d a b c -++<22215d a b c ⇔<++，因此，15是有界集合C 的一个上界．…………………………15/下证：上界15λ<不可能出现． 假设正数15λ<出现，取2a c b +=，1()05c a λ=->，则22a c d -⎛⎫= ⎪⎝⎭，此时，d22222213()()()55a b c a b c acλλ=+++-++-22221()()5a b c a acλλ>+++--222()a b c λ=++（*）…17/由式（*）可得222222()dd a b c a b c λλ>++⇔>++，与λ是C 的一个上界矛盾！．综上所述，满足题设的最小正数λ的值为15． …………………………………………18/。

+■B+ -Nl+ A T 3+ g M5 22019-2020年高三第二次模拟考试物理试题含解析一、单项选择题1（共 16分，每小题2分•每小题只有一个正确选项.）1. （ 2分）（2015?奉贤区二模）下列属于标量的物理量是（A. 平均速度B .电场强度C .磁感应强度 D .电流强度【考点】：矢量和标量.【分析】：即有大小又有方向，相加时遵循平行四边形定则的物理量是矢量，如力、速度、 加速度、位移、动量等都是矢量；只有大小，没有方向的物理量是标量，如路程、时间、质量、速率等都是标量.【解析】：解：A 、平均速度的位移与时间的比值，是有大小又有方向的物理量是矢量.故 A错误；B 、 电场强度有大小又有方向的物理量是矢量.故 B 错误；C 、 磁感应强度有大小又有方向的物理量是矢量.故 C 错误；D 、 电流强度只有大小，没有方向的物理量是标量，故D 正确.故选：D .【点评】：该题考查矢量与标量的区别，如力、速度、加速度、位移、动量、电场强度、磁 感应强度等都是矢量；如路程、时间、质量、速率等都是标量.基础题目.2. （2分）（2015?奉贤区二模）二十世纪初，为研究物质的内部结构，物理学家做了大量的实【考点】：粒子散射实验.【专题】：原子的核式结构及其组成.【分析】：解答本题应抓住：该实验是卢瑟福和他的助手们做的 a 粒子散射实验，根据这个 实验的结果，卢瑟福提出了原子的核式结构模型.【解析】：解：本实验是a 粒子散射实验，卢瑟福根据极少数 a 粒子产生大角度偏转，提出了原子的核式结构模型.故 A 正确. 故选：A【点评】：本实验是a 粒子散射实验装置，是卢瑟福和他的助手们做的，是物理上著名的实 验，要加强记忆.3. （ 2分）（2015?奉贤区二模）下列核反应属于a 衰变的是（ ）验，如图装置的实验是（A . a 粒子散射实验 C .发现电子的实验）B .发现质子的实验 D .发现中子的实验C .Th i • ；Ra+ -He SO S3 2【考点】：原子核衰变及半衰期、衰变速度. 【专题】：衰变和半衰期专题.【分析】：a 衰变是指原子核分裂并只放射出氦原子核的反应过程，根据这一特定即可判断. 【解析】：解：A 、方程 「B+ -Hei 1 -N+ -n ;是人工核反应方程，是方向中子的核反 5 2 7 0 应方程.故A 错误；D 、方程 :H+ : H i :.-He 是轻核的聚变反应•故 D 错误.故选：C【点评】：该题考查a 衰变的本质与常见的核反应方程，比较简单，在平时学习中要掌握衰 变，裂变，聚变和几种粒子发现的方程式.4. （2分）（2015?奉贤区二模）一颗恒星的寿命取决于（）A. 体积B .质量C .温度D .离地球的距离【考点】：恒星的演化.【分析】：知道恒星的寿命和它的质量有关，质量越大的恒星寿命越短，反之亦然.【解析】：解：恒星的寿命和它的质量有关，质量越大的恒星寿命越短，这是因为质量越大 压力就越大，这种情况下恒星内部的核反应就更加剧烈，故 ACD 错误，B 正确.故选：B .【点评】：本题考查学生对恒星的理解，这要求在平时的学习中要对积累相关的知识，但此 题极易出错.5. （2分）（2015?奉贤区二模）如图所示的 4种明暗相间的条纹，分别是红光、蓝光各自通过 同一个双缝干涉仪器和同一单缝衍射仪器形成的图样 （灰黑色部分表示亮纹）.则属于蓝光的干涉图样的是（）IIIIIL MIIIL IIIIIIL IIIII【考点】：双缝干涉的条纹间距与波长的关系. 【专题】：光的干涉专题.【分析】：根据双缝干涉条纹间距 △<-■入可判定哪个图样是双缝干涉，它们的波长又有什么d关系；根据单缝衍射条纹是中间亮条纹明亮且宽大，越向两侧宽度越小，而波长越大，中央 亮条纹越粗进行判断B 、 方程C 、 方程C 正确；+ g M5 2一-He ,该反应是人工核反应方程.故 B 错误;OTh3 O 2 9=」Ra+ :He ,是二川Th 原子核分裂并只放射出氦原子核的反应过程.故u 9 2 9 UAl+【解析】：解：双缝干涉的图样是明暗相间的干涉条纹，所有条纹宽度相同且等间距，故从左边两个是双缝干涉现象，根据双缝干涉条纹间距△=、入可知波长入越大，△越大，故左边d第一个是红光，第三个是蓝光.单缝衍射条纹是中间明亮且宽大，越向两侧宽度越小越暗，而波长越大，中央亮条纹越粗，故从左向右依次是红光（双缝干涉）、蓝光（单缝衍射）、蓝光（双缝干涉）和红光（单缝衍射）.以上分析可知，故ABD错误，C正确.故选：C.【点评】：掌握单缝衍射和双缝干涉的图样的特点和图样与波长的关系是解决此题的唯一途径，故要加强对基础知识的记忆.6. （2分）（2015?奉贤区二模）两个分子相距r i时，分子力表现为引力，相距匕时，分子力表现为斥力，则（）A .相距r i时，分子间没有斥力存在B. 相距「2时的斥力大于相距r i时的斥力C. 相距「2时，分子间没有引力存在D. 相距r i时的引力大于相距 s时的引力【考点】：分子间的相互作用力.【专题】：分子间相互作用力与分子间距离的关系.【分析】：分子间的相互作用力由引力f引和斥力f斥两部分组成，这两种力同时存在，实际的分子力是引力和斥力的合力. 引力f引和斥力f斥随着分子间距离增大而减小，随着分子间距离减小而增大.【解析】：解：A、分子间的相互作用力由引力f引和斥力f斥两部分组成，这两种力同时存在，实际的分子力是引力和斥力的合力，故A错误；B、两个分子相距为r i,分子间的相互作用力表现为引力，相距为r2时，表现为斥力，故r i >S；分子间的引力和斥力随着分子间距的增加而减小，故相距为r2时，分子间的斥力大于相距为r i时的斥力，故B正确；C、分子间的相互作用力由引力f引和斥力f斥两部分组成，这两种力同时存在，实际的分子力是引力和斥力的合力，故C错误；D、两个分子相距为r i,分子间的相互作用力表现为引力，相距为r2时，表现为斥力，故r i >r2；分子间的引力和斥力随着分子间距的增加而减小，故相距为r i时，分子间的引力小于相距为r2时的引力，故D错误；故选：B.【点评】：对于分子力的特点，要抓住三个同”：分子间的引力和斥力是同时存在的；随着分子间距离变化，分子引力和斥力是同增同减的.7. （2分）（2015?奉贤区二模）做简谐振动的单摆摆长不变，若摆球质量减小为原来的球经过平衡位置时速度增大为原来的2倍，则单摆振动的（）A .频率、振幅都不变B .频率、振幅都改变C .频率不变、振幅改变D .频率改变、振幅不变【考点】：单摆周期公式. 单摆的能量的变化，从而可以判断振幅的变化. 振幅A 是反映单摆运动过程中的能量大小的物理量，据动能公式可知，摆球经过平衡位置时 的动能不变，但质量增加，所以高度减小，因此振幅改变，故ABD 错误，C 正确.故选：C .【点评】：知道单摆的摆长和重力加速度的大小决定单摆的周期的大小，单摆的能量决定单 摆的振幅的大小；从机械能守恒的角度分析振幅的变化.&（ 2分）（2015?奉贤区二模）如图，在斜面上木块A 与B 的接触面是水平的，绳子呈水平状态，两木块均保持静止•则关于木块A 和木块B 受力个数不可能的（）【考点】：物体的弹性和弹力. 【专题】：受力分析方法专题.【分析】：先对A 分析，可能受到2个力，重力与支持力；也可能受到4个力，增加拉力与B 对A 有向右的静摩擦力；再分析 B 受力，除受到重力、支持力、 A 对B 的静摩擦力，A 对 B 的压力，可能受到斜面的静摩擦力，也可能不受斜面的静摩擦力，B 受力情况有两种可能.【解析】：解：B 至少受到重力、A 对B 的压力和静摩擦力、斜面的支持力四个力•斜面对 物体B 可能有静摩擦力，也有可能没有静摩擦力，因此B 受到4个力或5个力；而A 受到力支持力与重力外，可能受到拉力与B 对A 的摩擦力.因此 A 可能受到2个力或4个力.故ACD 正确，B 错误； 题目要求选不可能的，故选： B .【点评】：本题关键先对 A 分析，根据平衡条件得到 B 对A 有向左的静摩擦力，然后根据牛 顿第三定律得到 A 对B 有向右的静摩擦力；再按照重力、弹力、摩擦力的顺序找力.二、单项选择题H （共 24分，每小题3分.每小题只有一个正确选项.）9. （3分）（2015?奉贤区二模）关于光的说法正确的是（）A .有的光是波，有的光是粒子B .光子可以被电场加速C .光的波长越长，其波动性越显著D .大量光子产生的效果往往显示粒子性【专题】：【分单摆问题. 由单摆的周期公式 可以判断单摆的周期的变化，由机械能守恒可以判断【解析】：解：由单摆的周期公式A . 2个和4个B . 3个和4个C . 4个和4个D . 4个和5个T=2T=2则周期不变，频率不变;【考点】：光子.【分析】：光子是电磁波，不带电；光既有波动性，又有粒子性，个别光子的作用效果往往表现为粒子性；大量光子的作用效果往往表现为波动性；根据E=h 丫判断出光子的能量与什么因素有关，可知波长与频率成反比.【解析】：解：A、光既有波动性，又有粒子性，不能说有的光是波，有的光是粒子•故A错误；B、光子不带电，不可以被电场加速•故B错误，C、光的波长越长，光的波动性越显著，频率越高，贝恍的粒子性越显著.故C正确；D、个别光子的作用效果往往表现为粒子性；大量光子的作用效果往往表现为波动性. 故D错误.故选：C.【点评】：该题考查光子的特性与本质，解决本题的关键知道光子的能量E=h Y与频率成正比.同时知道光有波粒两象性.10. （3分）（2015?奉贤区二模）趣味投篮比赛中，运动员站在一个旋转较快的大平台边缘上，相对平台静止，向平台圆心处的球筐内投篮球.则如图（各俯视图）篮球可能被投入球筐（图中箭头指向表示投篮方向）的是（）【考点】：运动的合成和分解.【专题】：运动的合成和分解专题.【分析】：球参与了沿圆周切线方向运动和出手方向的运动，根据平行四边形定则确定合速度的方向，从而确定篮球能否被投入球框.【解析】：解：当沿圆周切线方向的速度和出手速度的合速度沿篮筐方向，球就会被投入篮筐.故C正确，A、B、D错误.故选：C.【点评】：解决本题的关键知道球参与了两个方向的运动，通过平行四边形定则进行判断.11. （3分）（2015?奉贤区二模）将一个6V、6W”的小灯甲连接在内阻不能忽略的电源上，小灯恰好正常发光，现改将一个6V、3W”的小灯乙连接在这一电源上，则（）A .小灯乙可能正常发光B .小灯乙一定正常发光C.小灯乙可能因电压较低而不能正常发光D .小灯乙可能因电压过高而烧毁【考点】：闭合电路的欧姆定律.【专题】：恒疋电流专题.【分析】：根据公式R^ •可知，乙灯的电阻比甲灯电阻大•根据路端电压与外电阻的关系，P分析乙灯的电压与额定电压的关系，判断乙灯能否正常发光.【解析】：解：根据公式R^ '可知，乙灯的电阻比甲灯电阻大•由题，甲灯接在电源上正常P发光，甲灯的电压为 6V ，乙灯的电阻比甲灯电阻大，当乙灯接到同一电源上时，分得的电压 将大于6V ，所以灯乙可能因电压过高而烧毁，不能正常发光•故 ABC 错误，D 正确.故选：D •【点评】：本题关键抓住路端电压与外电阻的关系，来分析乙灯得到的电压与额定电压的关 系进行判断.12. （ 3分）（2015?奉贤区二模）如图为伽利略设计的一种测温装置示意图，玻璃管的上端与 导热良好的玻璃泡连通，下端插入水中，玻璃泡中封闭有一定量的空气•若玻璃管内水柱上 升，则导致原因可能是外界大气（ ）【考点】：理想气体的状态方程• 【专题】： 理想气体状态方程专题•【分析】：此温度计不是根据液体的热胀冷缩原理制成的，它是靠气体作为膨胀物质，液体 的受热膨胀忽略不计；外界大气的压强变化时，根据理想气体的状态方程也可以判定.【解析】： 解：设玻璃泡中气体压强为 p ,外界大气压强为p',则p'=p+p gh ,且玻璃泡中气体与外界大气温度相同•液柱上升，气体体积减小，根据理想气体的状态方程•可知，若P 不变，则V 减小时，T 减小；若T 不变时，V 减小时，P 变大.即温度T 降低或压强P 变大是 可能的情况•故 BCD 均不符合要求，A 正确. 故选：A【点评】： 本题考查了气体温度计的原理，膨胀物质不是液体，而是气体，依据气体的热胀 冷缩性质制成的13・（3分）（2015?奉贤区二模）如图所示，光滑小球放置在半球面的底端，竖直放置的挡板 水平向右缓慢地推动小球，则在小球运动（始终未脱离球面）的过程中，挡板对小球的推力 F 、 半球面对小球的支持力 F N 的变化情况正确的是（）A • F 增大，F N 减小B • F 减小，F N 增大C • F 减小，F N 减小D • F 增大，F N 增大A •温度降低，压强增大 C ・ 温度升高，压强减小B •温度升高，压强不变 D •温度不变，压强减小【考点】：共点力平衡的条件及其应用；物体的弹性和弹力. 【专题】：共点力作用下物体平衡专题.【分析】：对小球受力分析，受重力、挡板向右的支持力和半球面的支持力，根据平衡条件 列式求解. 【解析】：解：对小球受力分析，受重力、挡板向右的支持力和半球面的支持力，如图F=mgtan 0K― mg rN-C0S e由于0不断增加，故F 增大、F N 增大； 故选：D .【点评】：本题关键是画出受力图，根据平衡条件求解出两个弹力的表达式进行分析讨论.14. （ 3分）（2015?奉贤区二模）如图所示，将小球甲、乙、丙（都可视为质点）分别从 A 、B 、 C 三点由静止同时释放，最后都到达竖直面内圆弧的最低点D ，其中甲是从圆心 A 开始做自由落体运动，乙沿弦轨道从与圆心等咼的 动到D .忽略一切阻力，则正确的是（B 到达D ，丙沿圆弧轨道从C 点（很靠近D 点）运）WA.甲球最先到达D 点，乙球最后到达 D 点B. 甲球最先到达 D 点，丙球最后到达 D 点C. 丙球最先到达 D 点，乙球最后到达 D 点D .甲球最先到达 D 点，无法判断哪个球最后到达 D 点【考点】：牛顿第二定律；自由落体运动. 【专题】：牛顿运动定律综合专题.【分析】：A 为自由落体，运用自由落体的公式求出时间， B 是利用匀变速运动的知识求出所【解析】：解：A 点，AD 距离为r ，加速度为g ，时间 冷二萨；B 点，设/ ADB= 0, BD 距明显t 2>t 3>t 1,知乙球最后到，甲球最先到.故 A 正确，B 、C 、D 错误.用时间,C 是单摆，求出周期，所用时间只是周期.离为2rcos 0,加速度为gcos0,时间丫.-=;C 点，简谐振动，周期T=' —：：_，时间故选：A .【点评】：解得本题的关键是分清三种不同的运动形态，然后分别计算出每条线路所用的时间，比较大小皆可解决.15. （3分）（2015?奉贤区二模）如图，闭合铜环由高处从静止开始下落，穿过一根竖直悬挂的条形磁铁，铜环的中心轴线与条形磁铁的中轴线始终保持重合.若取磁铁中心O为坐标原点，建立竖直向下为正方向的x轴，则下图中最能正确反映环中感应电流i随环心位置坐标x变化的关系图象是（）ft【考点】：楞次定律；闭合电路的欧姆定律.【专题】：电磁感应与电路结合.【分析】：由楞次定律可知，感应线圈中电流方向变化，综合分析两个峰值不可能相等，由排除法可知.【解析】：解：A、圆形闭合铜环由高处从静止开始下落，穿过一根竖直悬挂的条形磁铁，铜环的中心轴线与条形磁铁的中轴始终保持重合，圆环中磁通量变化不均匀，产生的感应电流不是线性变化，故A错误；B、铜环下落到磁铁顶端的速度小于下落到磁铁底端的速度，铜环下落到磁铁顶端产生的感应电流小于下落到磁铁底端产生的感应电流，选项B正确，C错误.D、由楞次定律可知，圆环靠近磁体的过程中向上的磁通量最大，而离开磁体的过程中向上的磁通量减小，磁通量的变化相反，所以感应电流的方向也相反，故D错误.故选：B【点评】：本题考查了对楞次定律的理解和应用，注意重点判断磁通量的大小和方向的变化，其中判断出感应电流的大小与方向不相同是解答的关键.16. （3分）（2015?奉贤区二模）水平路面上行驶的汽车所受到的阻力大小f与汽车行驶的速率成正比.若汽车从静止出发，先做匀加速直线运动，达到额定功率后保持额定功率行驶，则在整个行驶过程中，汽车受到的牵引力大小【考点】：功率、平均功率和瞬时功率. 【专题】：功率的计算专题.【分析】：汽车先做匀加速直线运动，根据牛顿第二定律可知牵引力与阻力的关系式，功率 达到额定功率后，根据 F 宾及f=kv 求出牵引力与阻力的关系式，进而选择图象V【解析】：解：汽车先做匀加速直线运动， 速度增大，f=kv 增大，根据牛顿第二定律得：F=f+ma 可知，牵引力随着f 的增大而均匀增大，图象是一条倾斜的直线，功率达到额定功率后，F=「，f=kv ，则F='，则牵引力与阻力成反比，故 A 正确.v f故选：A【点评】：解决本题的关键会根据汽车的运动情况判断受力情况，能结合牛顿第二定律判断牵引力的变化情况，难度适中 三、（16分）多项选择题.（本大题共4小题，每小题4分.每小题给出的四个答案中，有两 个或两个以上是正确的•每一小题全选对得 4分；选对但不全，得 2分；有选错或不答的, 得0分.）17. （4分）（2015?奉贤区二模）如图所示，用恒力 F l 、F 2分别将同一重物由静止开始沿同一 固定的粗糙斜面由底端推到顶端，F i 沿斜面向上，F 2沿水平方向.已知两次所用时间相等，C.物体克服摩擦力做功相同 D .恒力F i 、F 2对物体做功相同【考点】：功的计算；牛顿第二定律. 【专题】：功的计算专题.【分析】：根据匀变速直线运动的位移时间公式比较加速度的大小.根据末速度的大小关系 得出动能的变化量大小，结合动能定理比较合外力做功情况.根据动能定理，通过摩擦力做 功和重力做功的大小关系得出恒力 F 做功的大小关系 【解析】：解：A 、根据x=—一厂••知，位移相同，运动时间相同，则加速度相同.故A 正确.B 、 根据v=at 知，加速度相同，运动时间相同，则末速度大小相等，根据动能定理知，动能变 化量相等，物体上升的顶端，重力势能的增加量相同，动能的增加量相等，则机械能变化量 相等，故B 正确；C 、 运动过程中，重力做功相等，第二次的摩擦力大于第一次的摩擦力，则第二次克服摩擦力做功大，而动能的变化量相等，根据动能定理知， F i 做的功比F 2做的少，故CD 错误.故选：AB【点评】：本题考查了动能定理的基本运用，关键抓住动能变化量相等，重力做功相等，通 过摩擦力做功不等，比较出恒力F 做功的大小关系18. （ 4分）（2015?奉贤区二模）位于正方形四角上的四个等量点电荷的电场线分布如图所示, ab 、cd 分别是正方形两条边的中垂线，O 点为中垂线的交点，P 、Q 分别为cd 、ab 上的点,则正确的是（）B .物体机械能增量相同A . P、O两点的电势关系为护=如B. P、Q两点电场强度的大小关系为E p> E QC. 若在0点放一正点电荷，则该正点电荷受到的电场力为零D. 若将某负电荷由P点沿着曲线PQ移到Q点，电场力做负功【考点】：电势；电场强度.【专题】：电场力与电势的性质专题.【分析】：根据电场线的方向确定场源电荷的正负•电势的高低看电场线的指向，沿着电场线电势一定降低.电场线的疏密表示场强的大小，根据电势高低判断功的正负.【解析】：解：A、根据电场叠加，由图象可知ab、cd两中垂线上各点的电势都为零，所以P、0两点的电势相等.故A正确.B、电场线的疏密表示场强的大小，根据图象知E P>E Q,故B正确；C、四个点电荷在0点产生的电场相互抵消，场强为零，故在0点放一正点电荷，则该正点电荷受到的电场力为零，故C正确.D、P、Q电势相等，所以a、c两点电势相等，若将某一负电荷由P点沿着图中曲线PQ移到Q 点，电场力做功为零•故D错误.故选：ABC【点评】：本题的关键要掌握电场线的分布情况，能根据曲线的弯曲方向可知粒子的受力方向.通过电场线的指向看电势的高低19. （4分）（2015?奉贤区二模）如图，电源电动势为3V，内阻不可忽略，滑动变阻器的滑片P从a滑向b的过程中，三只理想电压表的示数变化的绝对值分别为汕1、汕2、△U s，下列各值不可能出现的是（）A . ^U i=3V、^U2=2V、△U3=1VB . △U1=1V、^U2=3V、△U3=2VC. △U i=0.5V、^U2=1V、^U3=1.5V D . △U l=0.2V、^U2=1V、△U3=0.8V【考点】：闭合电路的欧姆定律.【专题】：恒定电流专题.【分析】：滑动变阻器的滑片P从a滑向b的过程中，变阻器接入电路的电阻减小，电路中电流增大，电灯两端的电压增大，路端电压减小，分析变阻器两端电压的变化，根据路端电压的变化，判断A U2、汕3的大小，再选择符合题意的选项.【解析】：解：滑动变阻器的滑片 P 从a 滑向b 的过程中，变阻器接入电路的电阻减小，电 路中电流增大，电灯两端的电压U 2增大，电源的内电压增大，则路端电压 U i 减小，则变阻器两端电压 U 3减小.由于 U 仁U 2+U 3, U i 减小，则知 △U 2>A U 3, △U I V A U2,由于E=3V , △U 2< 3V ，所以△U i =0.2V 、^U 2=1V 、△U 3=0.8V 是可能的， ^U i =3V 、^U 2=2V 、△U 3=1V ;△U i =1V 、△U 2=3V 、^U 3=2V ; △△J i =0.5V 、^U 2=1V 、^U 3=1.5V 不可能.故 D 正确，ABC 错误. 本题选不可能的，故选：ABC .【点评】：本题解题的关键是抓住 U 仁U 2+U 3,根据总量法分析三个电压表读数变化量的大小.20. （ 4分）（2015?奉贤区二模）如图所示，由一根绝缘导线绕成半径相同的两个小圆（中央缺口很小）组成的线圈水平放置，匀强磁场 B 垂直通过线圈平面，若将磁场的磁感强度从 B 增大到2B 的过程中通过线圈的电量为 Q ,则下列可使线圈中通过电量为 Q 的过程是（ ）【考点】：带电粒子在匀强磁场中的运动. 【专题】：带电粒子在磁场中的运动专题. 【分析】：由法拉第电磁感应定律和欧姆定律，结合电量表达式，可得出电量的综合表达式. 从而分析判断.【解析】：解：电量公式q=l而法拉第电磁感应定律 E=N 厶一，电流1=：'，因此电量的At F表达式Q=：；上一.由题意可知，当磁感应强度由B 增强至2B 的过程中有电量 Q 通过线圈，要使线圈通过电量仍为 Q ,则有：磁通量变化为 BS ，或电阻变为原来的一半，A 、 保持B 不变，将线圈平面翻转 90°,根据磁通量的变化，可知， △①=BS ，故A 正确； B 、 保持B 不变，将线圈平面翻转 180°,根据磁通量的变化，可知， △①=2BS ，故B 错误；C 、 保持B 不变，将线圈的一个小圆平面翻转180°根据磁通量的变化，可知，△①=BS ，故C正确；D 、 保持B 不变，将线圈拉大成一个大圆，根据周长相同，则有大圆的半径是小圆半径的2倍，因此磁通量的变化，为 △①=B?2S -BS=BS ，故D 正确； 故选：ACD .【点评】：考查电量的综合表达式，注意掌握经过变化后，磁通量变化必须为 BS ,才能满足题意，同时注意线圈拉成大圆时，圆的周长不变，从而确定大圆与小圆的半径关系.四、（20分）填空题.（本大题共6小题，每空格2分.不要求写出演算过程.本大题中第 22题，23题为分叉题，考生可任选一题答题.若两题均做，一律按第一题题计分.）21. （ 4分）（2015?奉贤区二模）禾U 用发波水槽得到的水面波形如a 、b 所示，则图a 显示了波的 衍射 现象.为了能更明显看到此现象， 可以 减小 水波的振源频率（选填 增大” 减小”.x KaoX X X XA . 保持磁场B 不变，将线圈平面翻转 90° B . 保持磁场B 不变，将线圈平面翻转180°C . 保持磁场B 不变，将线圈的一个小圆平面翻转D . 保持磁场B 不变，将线圈拉成一个大圆180°【考点】：波的干涉和衍射现象.【分析】：波绕过障碍物继续传播的现象就是波的衍射现象；当频率相同的两列波相遇时有的地方振动减弱，有的地方振动加强，且加强和减弱的区域交替出现说明发生了干涉现象.【解析】：解：波绕过障碍物继续传播的现象就是波的衍射现象，故图a说明发生了明显的衍射现象.当频率相同的两列波相遇时当波程差为波长的整数倍时振动加强，当波程差为半个波长的奇数倍时振动减弱，使有的地方振动加强有的地方振动减弱，且加强和减弱的区域交替出现，故图b是发生了干涉现象.当波长越长，发生明显干涉和衍射越明显；再由于波速有介质决定，所以增大波长需减小波的频率.故答案为：衍射；减小.【点评】：掌握干涉和衍射的图样的特点和发生条件，是解决此类题目的关键所在•注意干涉现象的条件与明显的衍射现象条件.22. （4分）（2015?奉贤区二模）总质量为M的装砂的小车，正以速度V0在光滑水平面上向右前进，突然车底漏了，不断有砂子漏出来落到地面，则漏出的砂的速度为__ vo_，在漏砂的过程中，小车的速度的变化情况是不变（选填变大” 不变”或变小”.【考点】：动量守恒定律.【专题】：动量定理应用专题.【分析】：车以及沙子在水平方向动量守恒，根据动量守恒列方程求解即可.【解析】：解：设漏掉质量为m的沙子后，砂子从车上漏掉的瞬间由于惯性速度仍然为v o,汽车速度为v',根据水平方向动量守恒可得：Mv o=mv o+ （M - m）v，解得：v'=v o，小车的速度保持不变. 故答案为：v o,不变.【点评】：本题考查了动量守恒定律的应用，应用时注意：正确选取研究对象，明确公式中各物理量含义.23.（2oi5?奉贤区二模）已知一颗人造卫星在某行星表面上空做匀速圆周运动，经时间t,卫星的行程为s,它与行星中心的连线扫过的角度为1rad,则卫星的环绕周期为 2 n ，若减小该卫星的速度，则卫星的轨道半径将变小（选填变大” 不变”或变小”）.【考点】：人造卫星的加速度、周期和轨道的关系.【专题】：人造卫星问题.【分析】：根据圆周运动的规律间的关系解出T ,由万有引力与需要的向心力之间的关系，判定轨道半径的变化.【解析】：解：（1）由圆周运动的规律得：T=<'，。

第二学期普陀区高三数学质量调研考生注意:1. 本试卷共4页，21道试题，满分150分. 考试时间120分钟.2. 本考试分试卷和答题纸. 试卷包括试题与答题要求. 作答必须涂（选择题）或写（非选择题）在答题纸上，在试卷上作答一律不得分.3. 答卷前，务必用钢笔或圆珠笔在答题纸正面清楚地填写姓名、准考证号，并将核对后的条码贴在指定位置上，在答题纸反面清楚地填写姓名.一、填空题（本大题共有12题，满分54分）考生应在答题纸相应编号的空格内直接填写结果，每个空格填对前6题得4分、后6题得5分，否则一律得零分.1. 计算：=⎪⎭⎫⎝⎛+∞→311lim n n .2. 函数⎪⎭⎫⎝⎛-=x y 11log 2的定义域为 . 3. 若παπ<<2，53sin =α，则=2tan α. 4. 若复数()21i i z ⋅+=（i 表示虚数单位），则=z .5. 曲线C ：⎩⎨⎧==θθtan sec y x （θ为参数）的两个顶点之间的距离为 .6. 若从一副52张的扑克牌中随机抽取2张，则在放回抽取的情形下，两张牌都是K 的概率为 （结果用最简分数表示）.7. 若关于x 的方程0cos sin =-+m x x 在区间⎥⎦⎤⎢⎣⎡2,0π上有解，则实数m 的取值范围是 . 8. 若一个圆锥的母线与底面所成的角为6π,体积为π125,则此圆锥的高为 . 9. 若函数1log log )(222+-=x x x f （2≥x ）的反函数为)(1x f-，则)3(1-f= .10. 若三棱锥ABC S -的所有的顶点都在球O 的球面上，⊥SA 平面ABC ，2==AB SA ，4=AC ，3π=∠BAC ，则球O 的表面积为 .11.设0<a ，若不等式01cos )1(sin 22≥-+-+a x a x 对于任意的R ∈x 恒成立，则a 的取值范围是 .12.在△ABC 中，D 、E 分别是AB 、AC 的中点，M 是直线DE 上的动点.若△ABC 的面积为1，则2BC MC MB +⋅的最小值为 .二、选择题（本大题共有4题，满分20分）每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分.13. 动点P 在抛物线122+=x y 上移动，若P 与点()1,0-Q 连线的中点为M ，则动点M 的轨迹方程为……………………………………………………………………………………………………………（ ）)A ( 22x y = ()B 24x y = ()C 26x y = ()D 28x y =14. 若α、β∈R ，则“βα≠”是“βαtan tan ≠”成立的……………………………………（ ）)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15. 设l 、m 是不同的直线，α、β是不同的平面，下列命题中的真命题为…………………………（ ）)A ( 若α//l ，β⊥m ，m l ⊥，则βα⊥ ()B 若α//l ，β⊥m ，m l ⊥，则 βα//()C 若α//l ，β⊥m ，m l //，则βα⊥ ()D 若α//l ，β⊥m ，m l //，则βα//16. 关于函数x y 2sin =的判断，正确的是……………………………………………………………（ ）)A (最小正周期为π2，值域为[]1,1-，在区间⎥⎦⎤⎢⎣⎡-2,2ππ上是单调减函数()B 最小正周期为π，值域为[]1,1-，在区间⎥⎦⎤⎢⎣⎡2,0π上是单调减函数()C 最小正周期为π，值域为[]1,0，在区间⎥⎦⎤⎢⎣⎡2,0π上是单调增函数()D 最小正周期为π2，值域为[]1,0，在区间⎥⎦⎤⎢⎣⎡-2,2ππ上是单调增函数三、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相应编号的规定区域内写出必要的步骤17. （本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分在正方体1111D C B A ABCD -中，E 、F 分别是BC 、11D A 的中点. （1）求证：四边形EDF B 1是菱形；（2）求异面直线C A 1与DE 所成角的大小 (结果用反三角函数值表示) .1A1B1C1DF18.（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分 已知函数x b x a x f cos sin )(+=（a 、b 为常数且0≠a ，R ∈x ）.当4π=x 时，)(x f 取得最大值.（1）计算⎪⎭⎫⎝⎛411πf 的值； （2）设⎪⎭⎫⎝⎛-=x f x g 4)(π，判断函数)(x g 的奇偶性，并说明理由.19.（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分某人上午7时乘船出发，以匀速v 海里/小时（54≤≤v ）从A 港前往相距50海里的B 港，然后乘汽车以匀速ω千米/小时（10030≤≤ω）自B 港前往相距300千米的C 市，计划当天下午4到9时到达C 市.设乘船和汽车的所要的时间分别为x 、y 小时，如果所需要的经费()()y x P -+-+=853100（单位：元） （1）试用含有v 、ω的代数式表示P ；（2）要使得所需经费P 最少，求x 和y 的值，并求出此时的费用.20. （本题满分16分）本题共有3小题，第1小题4分，第2小题6分，第3小题6分. 已知曲线Γ：13422=+y x ，直线l 经过点()0,m P 与Γ相交于A 、B 两点. （1）若()3,0-C 且2=PC ，求证：P 必为Γ的焦点；（2）设0>m ，若点D 在Γ上，且PD 的最大值为3，求m 的值； （3）设O 为坐标原点，若3=m ，直线l 的一个法向量为()k ,1=，求∆AOB 面积的最大值.xyo21.（本题满分18分）本题共有3小题，第1小题4分，第2小题6分，第3小题8分.已知数列{}n a （*N ∈n ），若{}1++n n a a 为等比数列，则称{}n a 具有性质P .（1）若数列{}n a 具有性质P ，且3,1321===a a a ，求4a 、5a 的值； （2）若()nn n b 12-+=，求证：数列{}n b 具有性质P ；（3）设=+++n c c c Λ21n n +2，数列{}n d 具有性质P ，其中11=d ，123c d d =-，232c d d =+，若310>m d ，求正整数m 的取值范围.第二学期普陀区高三数学质量调研一、填空题（本大题共有12题，满分54分）考生应在答题纸相应编号的空格内直接填写结果，每个空格填对前6题得4分、后6题得5分，否则一律得零分.1.12. ()()+∞∞-,10,Y3.34. i +-15.26.1691 7. 21≤≤m . 8. 5 9. 4 10.π20 11. 2-≤a 12. 3二、选择题（本大题共有4题，满分20分）每题有且只有一个正确答案，考生应在答题纸的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分.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相应编号的规定区域内写出必要的步骤17. （本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分 【解】设正方体的棱长为1，建立空间直角坐标系，如图所示：则()1,0,11B ,⎪⎭⎫ ⎝⎛0,21,1E ,()0,1,0D ,⎪⎭⎫⎝⎛1,21,0F ……1分⎪⎭⎫⎝⎛-=0,21,1DE ,⎪⎭⎫ ⎝⎛-=0,21,11FB ……2分所以1FB DE =,即1//FB DE 且1FB DE =,故四边形EDF B 1是平行四边形……3分又因为⎪⎭⎫ ⎝⎛-=1,21,01E B ,25==……5分 故平行四边形EDF B 1是菱形……6分（2）因为()0,1,11=A ()()1,1,101,0--=-,⎪⎭⎫⎝⎛-=0,21,1DE ……8分 设异面直线C A 1与DE 所成的角的大小为θ……9分cos =θ……10分()()15152111110121)1(11222222=+⎪⎭⎫ ⎝⎛-+⋅+-+-⨯+⎪⎭⎫⎝⎛-⨯-+⨯-=……12分 所以1515arccos=θ……13分, 故异面直线C A 1与DE 所成的角的大小为1515arccos ……14分 18.（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分 【解】（1）x b x a x f cos sin )(+=()ϕ++=x b a sin 22，其中abarctan =ϕ……2分根据题设条件可得，224b a f +=⎪⎭⎫⎝⎛π 即()2222b a b a +=+ ……4分 化简得()()2222b a b a +=+，所以0222=+-b ab a即()02=-b a ，故0=-b a ……………5分所以()022411cos 411sin411=-=+=⎪⎭⎫⎝⎛b a b a f πππ……………6分 （2）由（1）可得，b a =，即()⎪⎭⎫ ⎝⎛+=+=4sin 2cos sin )(πx a x x a x f ……8分故x a x a x a x f x g cos 22sin 244sin 24)(=⎪⎭⎫⎝⎛-=⎪⎭⎫ ⎝⎛+-=⎪⎭⎫ ⎝⎛-=ππππ所以x a x g cos 2)(=（R ∈x ）…………10分对于任意的R ∈x ，x a x a x g cos 2)cos(2)(=-=-（0≠a ）……12分即)()(x g x g =-，所以)(x g 是偶函数.…………14分19.（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分 【解】（1）v x 50=，204≤≤v ，得22510≤≤x ……2分 ω300=y ，10030≤≤ω，得103≤≤y ……4分()()y x P -+-+=853100⎪⎭⎫ ⎝⎛-+⎪⎭⎫ ⎝⎛-+=ω30085053100v所以ω300150123--=v P （其中204≤≤v ，10030≤≤ω）……6分 （2）()()y x P -+-+=853100)3(123y x +-=其中⎪⎪⎩⎪⎪⎨⎧≤≤≤≤≤+≤10322510149y x y x ，……9分令目标函数y x k +=3,⎪⎭⎫213，()3,6 …12分则当3,11==y x 时，333max =+=k 所以8736123min =-=P （元）,此时11x 3答：当3,11==y x 时，所需要的费用最少，为87元。

2019届上海市普陀区曹杨第二中学高三上学期9月月考数学试题一、单选题1．已知数列{}n a 是等差数列，则12a a <是数列{}n a 为递增数列的（ ） A ．充分非必要条件 B ．必要非充分条件 C ．充分必要条件 D ．非充分非必要条件 【答案】C【解析】设{}n a 的公差为d ,分析当数列{}n a 为递增数列时d 满足的条件,再根据12a a <分析即可.【详解】设{}n a 的公差为d ,则121200a a a a d ⇒->⇒><.若{}n a 为递增数列则对任意的正整数n 有1100n n n n a a a a d ++>⇒->⇒>. 故12a a <是数列{}n a 为递增数列的充分必要条件. 故选：C 【点睛】本题主要考查了等差数列递增的问题与充分必要条件,属于基础题型.2．已知直线l 、m ，平面α、β，且l α⊥，m β⊂，下列命题中正确的是（ ） A ．若αβ∥，则l m ⊥ B ．若l m ⊥，则αβ∥ C ．若αβ⊥，则l m D ．若l m ，则αβ∥【答案】A【解析】根据线面垂直的判定与性质逐个判断,同时结合长方体举反例即可. 【详解】 画出如图长方体.对A, 若αβ∥,则因为l α⊥,故l β⊥,又m β⊂,所以l m ⊥,故A 正确.对B,当l 为AE ,m 为AB ,面α为ABCD ,β为面ABFE 时,满足l α⊥,m β⊂,l m ⊥, 但αβ∥不成立.故B 错误.对C, 当l 为AE ,m 为AB ,面α为ABCD ,β为面ABFE 时, 满足l α⊥,m β⊂,αβ⊥,但l m 不成立.故C 错误.对D, 当l 为AE ,m 为BF ,面α为ABCD ,β为面ABFE 时, 满足l α⊥,m β⊂,l m ，但αβ∥不成立.故D 错误.故选：A 【点睛】本题主要考查平行垂直的判断,可直接利用线面垂直的方法进行判定,或者在长方体中举出反例即可.属于基础题型. 3．将函数sin 6y x π⎛⎫=-⎪⎝⎭的图像上所有的点向右平移4π个单位长度，再把图形上各点的横坐标扩大到原来的2倍（纵坐标不变），则所得图像的解析式为（ ）A ．5sin 212x y π⎛⎫=- ⎪⎝⎭B ．5sin 212x y π⎛⎫=+ ⎪⎝⎭C ．5sin 212y x π⎛⎫=- ⎪⎝⎭ D ．5sin 224x y π⎛⎫=-⎪⎝⎭【答案】A【解析】根据三角函数平移伸缩的变换求解即可. 【详解】将函数sin 6y x π⎛⎫=-⎪⎝⎭的图像上所有的点向右平移4π个单位长度得到5sin sin 4612y x x πππ⎛⎫⎛⎫=--=- ⎪ ⎪⎝⎭⎝⎭.再把图形上各点的横坐标扩大到原来的2倍（纵坐标不变）则变成15sin 212y x π⎛⎫=- ⎪⎝⎭. 故选：A 【点睛】本题主要考查了三角函数图像的变换,属于基础题型.4．设θ是两个非零向量a ，b 的夹角，若对于任意实数t ，||a b +得最小值为1，则下列判断正确的是（ ） A ．若||a 确定，则θ唯一确定 B ．若||b 确定，则θ唯一确定 C ．若θ确定，则||b 唯一确定 D ．若θ确定，则||a 确定【答案】D【解析】可将||a b +平方,利用二次函数不等式进行求解,也可以利用几何的方法画图求解. 【详解】解法一,一般法：2222()2a b a ta b t b +=+⋅+,则令222()2g t a ta b t b =+⋅+, 可得判别式222222222224()44cos 44sin 0a b a b a b a b a b θθ-∆=⋅-=-=-<, 由二次函数的性质,可得()0g t >恒成立.当且仅当22||cos 2||a b a t b b θ⋅=-=-时,()g t 最小,且最小为1. 即22222||cos ||cos ||||sin 1||a g a a a b θθθ⎛⎫-=-+== ⎪⎝⎭, 故当若θ确定,则||a 唯一确定；解法二,几何法：|||()|a tb a t b +=--,则画出大致向量图像,如下图：令OA a =,OC tb =-,AC a tb =-,OB b =,则由图像可知,因为|1|AC =,则当()AOB θ∠确定时,||OA 确定. 故选：D 【点睛】本题主要考查了向量含参数的范围问题,也考查了向量的几何法,属于基础题型.二、填空题5．已知集合{}2|230A x x x =--≤，{||2|2}B x x =-<，则A B =________；【答案】(0,3]【解析】分别利用二次不等式与绝对值不等式的求解方法解出,A B 再计算A B 即可.【详解】{}{}{}2|230|(3)(1)0|13A x x x x x x x x =--≤=-+≤=-≤≤ {||2|2}{|222}{|04}B x x x x x x =-<=-<-<=<<,所以 130304x x x -≤≤⎧⇒<≤⎨<<⎩,故(0,3]A B =； 故答案为：(0,3] 【点睛】本题主要考查了二次不等式与绝对值不等式的求解方法与交集的运算,属于基础题型. 6．已知复数：z 满足1iz i+=（i 为虚数单位），则z 的虚部为________； 【答案】-1；【解析】求出z 的值再判断即可. 【详解】()211111i i i i z i i i ++-====--,故z 的虚部为1- 故答案为：1- 【点睛】本题主要考查了复数的基本运算与虚部的定义,属于基础题型.7．613x x ⎛⎫- ⎪⎝⎭的二项展开式中的常数项为________；【答案】-540【解析】展开式中的常数项则相乘的6项里面需要选3个3x 与3个1x-,再利用二项式公式求解即可. 【详解】二项展开式中的常数项为33361(3)2027540C x x ⎛⎫-=-⨯=- ⎪⎝⎭, 故答案为：540- 【点睛】本题主要考查了二项式定理求常数项的方法,属于基础题型.8．已知双曲线221x y m -=和椭圆221124x y +=焦点相同，则该双曲线的方程为__________．【答案】2217x y -=【解析】分析：根据题意，求出椭圆的焦点坐标，由双曲线的几何性质可得若双曲线221x y m -=和椭圆221124x y +=焦点相同，则有18m +=，解得m 的值，将m 的值代入双曲线的方程，即可得答案.详解：根据题意，椭圆221124x y +=的焦点在x 轴上，且焦点坐标为()±，若双曲线221x y m -=和椭圆221124x y +=焦点相同，则有18m +=，解得7m =，则双曲线的方程为2217x y -=.故答案为：2217x y -=.点睛：本题考查双曲线的几何性质，关键是掌握双曲线的标准方程的形式.9．如图，正四棱锥P ABCD -的底面一边AB 的长为，侧面积为2，则它的体积为___．【答案】4.【解析】由题设14822h h ⨯=⇒=,则四棱锥的高1H ==,所以该四棱锥的体积2111433V SH ==⨯⨯=，应填答案4。

目录第一套：2019年上海市静安区高考数学二模试卷第二套：2019年上海市虹口高考数学二模试卷2019年上海市静安区高考数学二模试卷一、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）考生应在答题纸的相应位置直接填写结果．1．设f ﹣1（x ）为的反函数，则f ﹣1（1）= ．2．函数y=2sin 2（2x ）﹣1的最小正周期是 ． 3．设i 为虚数单位，复数，则|z|= ．4．= ．5．若圆锥的侧面积是底面积的2倍，则其母线与轴所成角的大小是 ．6．设等差数列{a n }的前n 项和为S n ，若=，则= ．7．直线（t 为参数）与曲线（θ为参数）的公共点的个数是 ．8．已知双曲线C 1与双曲线C 2的焦点重合，C 1的方程为，若C 2的一条渐近线的倾斜角是C 1的一条渐近线的倾斜角的2倍，则C 2的方程为 ． 9．若，则满足f （x ）＞0的x 的取值范围是 ．10．某企业有甲、乙两个研发小组，他们研发新产品成功的概率分别为和．现安排甲组研发新产品A ，乙组研发新产品B ，设甲、乙两组的研发相互独立，则至少有一种新产品研发成功的概率为 ．11．设等差数列{a n }的各项都是正数，前n 项和为S n ，公差为d ．若数列也是公差为d 的等差数列，则{a n }的通项公式为a n = ．12．设x ∈R ，用[x]表示不超过x 的最大整数（如[2.32]=2，[﹣ 4.76]=﹣5），对于给定的n ∈N *，定义C =，其中x ∈[1，+∞），则当时，函数f （x ）=C的值域是 ．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．考生应在答题纸的相应位置，将代表正确选项的小方格涂黑．13．命题“若x=1，则x 2﹣3x+2=0”的逆否命题是（ ） A ．若x ≠1，则x 2﹣3x+2≠0 B ．若x 2﹣3x+2=0，则x=1 C ．若x 2﹣3x+2=0，则x ≠1 D ．若x 2﹣3x+2≠0，则x ≠1 14．如图，在正方体ABCD ﹣A 1B 1C 1D 1中，M 、E 是AB 的三等分点，G 、N 是CD 的三等分点，F 、H 分别是BC 、MN 的中点，则四棱锥A 1﹣EFGH 的左视图是（ ）A ．B ．C ．D ．15．已知△ABC 是边长为4的等边三角形，D 、P 是△ABC 内部两点，且满足，，则△ADP 的面积为（ ） A ．B ．C ．D ．16．已知f （x ）是偶函数，且f （x ）在[0，+∞）上是增函数，若f （ax+1）≤f （x ﹣2）在上恒成立，则实数a 的取值范围是（ ）A ．[﹣2，1]B ．[﹣2，0]C ．[﹣1，1]D ．[﹣1，0]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．在△ABC 中，内角A ，B ，C 的对边分别为a ，b ，c ，已知a ﹣b=2，c=4，sinA=2sinB ． （Ⅰ）求△ABC 的面积； （Ⅱ）求sin （2A ﹣B ）．18．如图，在长方体ABCD ﹣A 1B 1C 1D 1中，AB=8，BC=5，AA 1=4，平面α截长方体得到一个矩形EFGH ，且A 1E=D 1F=2，AH=DG=5．（1）求截面EFGH 把该长方体分成的两部分体积之比； （2）求直线AF 与平面α所成角的正弦值．19．如图，已知椭圆C ：（a ＞b ＞0）过点，两个焦点为F 1（﹣1，0）和F 2（1，0）．圆O 的方程为x 2+y 2=a 2． （1）求椭圆C 的标准方程；（2）过F 1且斜率为k （k ＞0）的动直线l 与椭圆C 交于A 、B 两点，与圆O 交于P 、Q 两点（点A 、P 在x 轴上方），当|AF 2|，|BF 2|，|AB|成等差数列时，求弦PQ 的长．20．如果函数y=f （x ）的定义域为R ，且存在实常数a ，使得对于定义域内任意x ，都有f （x+a ）=f （﹣x ）成立，则称此函数f （x ）具有“P（a ）性质”．（1）判断函数y=cosx 是否具有“P （a ）性质”，若具有“P （a ）性质”，求出所有a 的值的集合；若不具有“P（a ）性质”，请说明理由；（2）已知函数y=f （x ）具有“P（0）性质”，且当x ≤0时，f （x ）=（x+m ）2，求函数y=f （x ）在区间[0，1]上的值域； （3）已知函数y=g （x ）既具有“P （0）性质”，又具有“P （2）性质”，且当﹣1≤x ≤1时，g （x ）=|x|，若函数y=g （x ）的图象与直线y=px 有2019个公共点，求实数p 的值．21．给定数列{a n }，若满足a 1=a （a ＞0且a ≠1），对于任意的n ，m ∈N *，都有a n+m =a n •a m ，则称数列{a n }为指数数列． （1）已知数列{a n }，{b n }的通项公式分别为，，试判断{a n }，{b n }是不是指数数列（需说明理由）；（2）若数列{a n }满足：a 1=2，a 2=4，a n+2=3a n+1﹣2a n ，证明：{a n }是指数数列；（3）若数列{a n }是指数数列，（t ∈N *），证明：数列{a n }中任意三项都不能构成等差数列．2019年上海市静安区高考数学二模试卷参考答案与试题解析一、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）考生应在答题纸的相应位置直接填写结果．1．设f﹣1（x）为的反函数，则f﹣1（1）= 1 ．【考点】4R：反函数．【分析】根据反函数的性质，原函数的值域是反函数的定义域即可求解【解答】解：的反函数，其反函数f﹣1（x），反函数的性质，反函数的定义域是原函数的值域，即．可得：x=1，∴f﹣1（x）=1．故答案为1．2．函数y=2sin2（2x）﹣1的最小正周期是．【考点】H1：三角函数的周期性及其求法．【分析】利用二倍角公式基本公式将函数化为y=Acos（ωx+φ）的形式，再利用周期公式求函数的最小正周期，【解答】解：函数y=2sin2（2x）﹣1，化简可得：y=1﹣cos4x﹣1=﹣cos4x；∴最小正周期T=．故答案为3．设i为虚数单位，复数，则|z|= 1 ．【考点】A8：复数求模．【分析】利用复数的运算法则、模的计算公式即可得出．【解答】解：复数===﹣i，则|z|=1．故答案为：1．4． = 3 ．【考点】8J：数列的极限．【分析】通过分子分母同除3n+1，利用数列极限的运算法则求解即可．【解答】解： ===3．故答案为：3．5．若圆锥的侧面积是底面积的2倍，则其母线与轴所成角的大小是30°．【考点】MI：直线与平面所成的角．【分析】根据圆锥的底面积公式和侧面积公式，结合已知可得l=2R ，进而解母线与底面所成角，然后求解母线与轴所成角即可． 【解答】解：设圆锥的底面半径为R ，母线长为l ，则： 其底面积：S 底面积=πR 2，其侧面积：S 侧面积=2πRl=πRl， ∵圆锥的侧面积是其底面积的2倍， ∴l=2R ，故该圆锥的母线与底面所成的角θ有， cosθ==， ∴θ=60°，母线与轴所成角的大小是：30°． 故答案为：30°．6．设等差数列{a n }的前n 项和为S n ，若=，则=．【考点】85：等差数列的前n 项和． 【分析】=，可得3（a 1+4d ）=5（a 1+2d ），化为：a 1=d ．再利用等差数列的求和公式即可得出． 【解答】解：∵=，∴3（a 1+4d ）=5（a 1+2d ），化为：a 1=d ．则==．故答案为：．7．直线（t为参数）与曲线（θ为参数）的公共点的个数是 1 ．【考点】QK：圆的参数方程；QJ：直线的参数方程．【分析】根据题意，将直线的参数方程变形为普通方程，再将曲线的参数方程变形为普通方程，分析可得该曲线为圆，且圆心坐标为（3，5），半径r=，求出圆心到直线的俄距离，分析可得直线与圆相切，即可得直线与圆有1个公共点，即可得答案．【解答】解：根据题意，直线的参数方程为，则其普通方程为x+y﹣6=0，曲线的参数方程为，则其普通方程为（x﹣3）2+（y ﹣5）2=2，该曲线为圆，且圆心坐标为（3，5），半径r=，圆心到直线x+y﹣6=0的距离d===r，则圆（x﹣3）2+（y﹣5）2=2与直线x+y﹣6=0相切，有1个公共点；故答案为：1．8．已知双曲线C1与双曲线C2的焦点重合，C1的方程为，若C2的一条渐近线的倾斜角是C1的一条渐近线的倾斜角的2倍，则C2的方程为．【考点】KC：双曲线的简单性质．【分析】求出双曲线的焦点坐标，利用渐近线的倾斜角的关系，列出方程，然后求解即可．【解答】解：双曲线C1与双曲线C2的焦点重合，C1的方程为，焦点坐标（±2，0）．双曲线C1的一条渐近线为：y=，倾斜角为30°，C 2的一条渐近线的倾斜角是C1的一条渐近线的倾斜角的2倍，可得C2的渐近线y=．可得，c=2，解得a=1，b=，所求双曲线方程为：．故答案为：．9．若，则满足f（x）＞0的x的取值范围是（1，+∞）．【考点】7E：其他不等式的解法．【分析】由已知得到关于x的不等式，化为根式不等式，然后化为整式不等式解之．【解答】解：由f（x）＞0得到即，所以，解得x＞1；故x的取值范围为（1，+∞）；故答案为：（1，+∞）；10．某企业有甲、乙两个研发小组，他们研发新产品成功的概率分别为和．现安排甲组研发新产品A ，乙组研发新产品B ，设甲、乙两组的研发相互独立，则至少有一种新产品研发成功的概率为．【考点】C9：相互独立事件的概率乘法公式． 【分析】利用对立事件的概率公式，计算即可，【解答】解：设至少有一种新产品研发成功的事件为事件A 且事件B 为事件A 的对立事件，则事件B 为一种新产品都没有成功， 因为甲乙研发新产品成功的概率分别为和． 则P （B ）=（1﹣）（1﹣）=，再根据对立事件的概率之间的公式可得P （A ）=1﹣P （B ）=，故至少有一种新产品研发成功的概率．故答案为．11．设等差数列{a n }的各项都是正数，前n 项和为S n ，公差为d ．若数列也是公差为d 的等差数列，则{a n }的通项公式为a n =．【考点】84：等差数列的通项公式． 【分析】由题意可得：S n =na 1+d ．a n ＞0.=+（n ﹣1）d ，化简n ≠1时可得：a 1=（n ﹣1）d 2+2d ﹣d ．分别令n=2，3，解出即可得出．【解答】解：由题意可得：S n =na 1+d ．a n ＞0．=+（n ﹣1）d ，可得：S n =a 1+（n ﹣1）2d 2+2（n ﹣1）d ．∴na 1+d=a 1+（n ﹣1）2d 2+2（n ﹣1）d ． n ≠1时可得：a 1=（n ﹣1）d 2+2d ﹣d ． 分别令n=2，3，可得：a 1=d 2+2d ﹣d ，a 1=2d 2+2d ﹣d ．解得a 1=，d=． ∴a n =+（n ﹣1）=．故答案为：．12．设x ∈R ，用[x]表示不超过x 的最大整数（如[2.32]=2，[﹣ 4.76]=﹣5），对于给定的n ∈N *，定义C =，其中x ∈[1，+∞），则当时，函数f （x ）=C的值域是．【考点】57：函数与方程的综合运用．【分析】分类讨论，根据定义化简C x n ，求出C x 10的表达式，再利用函数的单调性求出C x 10的值域．【解答】解：当x ∈[，2）时，[x]=1，∴f （x ）=C =， 当x ∈[，2）时，f （x ）是减函数，∴f （x ）∈（5，）；当x ∈[2，3）时，[x]=2，∴f （x ）=C=，当x ∈[2，3）时，f （x ）是减函数，∴f （x ）∈（15，45]； ∴当时，函数f （x ）=C 的值域是，故答案为：．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．考生应在答题纸的相应位置，将代表正确选项的小方格涂黑．13．命题“若x=1，则x 2﹣3x+2=0”的逆否命题是（ ） A ．若x ≠1，则x 2﹣3x+2≠0 B ．若x 2﹣3x+2=0，则x=1 C ．若x 2﹣3x+2=0，则x ≠1 D ．若x 2﹣3x+2≠0，则x ≠1 【考点】25：四种命题间的逆否关系．【分析】根据逆否命题的定义，我们易求出命题的逆否命题 【解答】解：将命题的条件与结论交换，并且否定可得逆否命题：若x 2﹣3x+2≠0，则x ≠1 故选：D14．如图，在正方体ABCD ﹣A 1B 1C 1D 1中，M 、E 是AB 的三等分点，G 、N 是CD 的三等分点，F 、H 分别是BC 、MN 的中点，则四棱锥A 1﹣EFGH 的左视图是（ ）A ．B ．C ．D ．【考点】L7：简单空间图形的三视图．【分析】确定5个顶点在面DCC 1D 1上的投影，即可得出结论． 【解答】解：A 1在面DCC 1D 1上的投影为点D 1，E 在面DCC 1D 1的投影为点G ，F 在面DCC 1D 1上的投影为点C ，H 在面DCC 1D 1上的投影为点N ，因此侧视图为选项C 的图形． 故选C15．已知△ABC 是边长为4的等边三角形，D 、P 是△ABC 内部两点，且满足，，则△ADP 的面积为（ ） A ．B ．C ．D ．【考点】9V ：向量在几何中的应用．【分析】以A 为原点，以BC 的垂直平分线为y 轴，建立直角坐标系．由于等边三角形△的边长为4，可得B ，C 的坐标，再利用向量的坐标运算和数乘运算可得，，利用△APD 的面积公式即可得出．【解答】解：以A 为原点，以BC 的垂直平分线为y 轴，建立直角坐标系．∵等边三角形△的边长为4， ∴B （﹣2，﹣2），C （2，﹣2），由足= [（﹣2，﹣2）+（2，﹣2）]=（0，﹣），=（0，﹣）+（4，0）=（，﹣），∴△ADP的面积为S=||•||=××=，故选：A．16．已知f（x）是偶函数，且f（x）在[0，+∞）上是增函数，若f（ax+1）≤f（x﹣2）在上恒成立，则实数a的取值范围是（）A．[﹣2，1] B．[﹣2，0] C．[﹣1，1] D．[﹣1，0]【考点】3N：奇偶性与单调性的综合．【分析】因为偶函数在对称区间上单调性相反，根据已知中f（x）是偶函数，且f（x）在（0，+∞）上是增函数，易得f（x）在（﹣∞，0）上为减函数，又由若时，不等式f（ax+1）≤f（x﹣2）恒成立，结合函数恒成立的条件，求出时f（x﹣2）的最小值，从而可以构造一个关于a的不等式，解不等式即可得到实数a的取值范围．【解答】解：∵f（x）是偶函数，且f（x）在（0，+∞）上是增函数，∴f（x）在（﹣∞，0）上为减函数，当时，x﹣2∈[﹣，﹣1]，故f（x﹣2）≥f（﹣1）=f（1），若时，不等式f（ax+1）≤f（x﹣2）恒成立，则当时，|ax+1|≤1恒成立，∴﹣1≤ax+1≤1，∴≤a≤0，∴﹣2≤a≤0，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．在△ABC中，内角A，B，C的对边分别为a，b，c，已知a ﹣b=2，c=4，sinA=2sinB．（Ⅰ）求△ABC的面积；（Ⅱ）求sin（2A﹣B）．【考点】GL：三角函数中的恒等变换应用．【分析】解法一：（I）由已知及正弦定理可求a，b的值，由余弦定理可求cosB，从而可求sinB，即可由三角形面积公式求解．（II）由余弦定理可得cosA，从而可求sinA，sin2A，cos2A，由两角差的正弦公式即可求sin（2A﹣B）的值．解法二：（I）由已知及正弦定理可求a，b的值，又c=4，可知△ABC为等腰三角形，作BD⊥AC于D，可求BD==，即可求三角形面积．（II）由余弦定理可得cosB，即可求sinB，由（I）知A=C⇒2A ﹣B=π﹣2B．从而sin（2A﹣B）=sin（π﹣2B）=sin2B，代入即可求值．【解答】解：解法一：（I）由sinA=2sinB⇒a=2b．又∵a﹣b=2，∴a=4，b=2．cosB===．sinB===．=acsinB==．∴S△ABC（II）cosA===．sinA===．sin2A=2sinAcosA=2×．cos2A=cos2A﹣sin2A=﹣．∴sin（2A﹣B）=sin2AcosB﹣cos2AsinB==．解法二：（I）由sinA=2sinB⇒a=2b．又∵a﹣b=2，∴a=4，b=2．又c=4，可知△ABC为等腰三角形．作BD ⊥AC 于D ，则BD===．∴S △ABC ==． （II ）cosB===． sinB===．由（I ）知A=C ⇒2A ﹣B=π﹣2B ． ∴sin （2A ﹣B ）=sin （π﹣2B ）=sin2B =2sinBcosB =2××=．18．如图，在长方体ABCD ﹣A 1B 1C 1D 1中，AB=8，BC=5，AA 1=4，平面α截长方体得到一个矩形EFGH ，且A 1E=D 1F=2，AH=DG=5． （1）求截面EFGH 把该长方体分成的两部分体积之比； （2）求直线AF 与平面α所成角的正弦值．【考点】MI ：直线与平面所成的角；LF ：棱柱、棱锥、棱台的体积．【分析】（1）由题意，平面α把长方体分成两个高为5的直四棱柱，转化求解体积推出结果即可．（2）解法一：作AM ⊥EH ，垂足为M ，证明HG ⊥AM ，推出AM ⊥平面EFGH ．通过计算求出AM=4．AF ，设直线AF 与平面α所成角为θ，求解即可．解法二：以DA 、DC 、DD 1所在直线分别为x 轴、y 轴、z 轴建立空间直角坐标系，求出平面α一个法向量，利用直线AF 与平面α所成角为θ，通过空间向量的数量积求解即可．【解答】（本题满分，第1小题满分，第2小题满分8分） 解：（1）由题意，平面α把长方体分成两个高为5的直四棱柱，，… ，…所以，．…（2）解法一：作AM ⊥EH ，垂足为M ，由题意，HG ⊥平面ABB 1A 1，故HG ⊥AM ，所以AM ⊥平面EFGH ． … 因为，，所以S △AEH =10，）因为EH=5，所以AM=4． … 又，…设直线AF 与平面α所成角为θ，则．… 所以，直线AF 与平面α所成角的正弦值为． …解法二：以DA 、DC 、DD 1所在直线分别为x 轴、y 轴、z 轴建立空间直角坐标系，则A （5，0，0），H （5，5，0），E （5，2，4），F （0，2，4），… 故，，…设平面α一个法向量为，则即所以可取． …设直线AF 与平面α所成角为θ，则． …所以，直线AF 与平面α所成角的正弦值为． …19．如图，已知椭圆C ：（a ＞b ＞0）过点，两个焦点为F 1（﹣1，0）和F 2（1，0）．圆O 的方程为x 2+y 2=a 2． （1）求椭圆C 的标准方程；（2）过F 1且斜率为k （k ＞0）的动直线l 与椭圆C 交于A 、B 两点，与圆O 交于P 、Q 两点（点A 、P 在x 轴上方），当|AF 2|，|BF 2|，|AB|成等差数列时，求弦PQ 的长．【考点】KH：直线与圆锥曲线的综合问题；K3：椭圆的标准方程；KL：直线与椭圆的位置关系．【分析】（1）求出c=1，设椭圆C的方程为，将点代入，解得a2=4，然后求解椭圆C的方程．（2）由椭圆定义，|AF1|+|AF2|=4，|BF1|+|BF2|=4，通过|AF2|，|BF2|，|AB|成等差数列，推出．设B（x，y），通过解得B，然后求解直线方程，推出弦PQ的长即可．【解答】（本题满分，第1小题满分，第2小题满分8分）解：（1）由题意，c=1，…设椭圆C的方程为，将点代入，解得a2=4（舍去），…所以，椭圆C的方程为．…（2）由椭圆定义，|AF1|+|AF2|=4，|BF1|+|BF2|=4，两式相加，得|AB|+|AF 2|+|BF 2|=8，因为|AF 2|，|BF 2|，|AB|成等差数列，所以|AB|+|AF 2|=2|BF 2|， 于是3|BF 2|=8，即． …设B （x 0，y 0），由解得，…（或设，则，解得，，所以）． 所以，，直线l 的方程为，即，… 圆O 的方程为x 2+y 2=4，圆心O 到直线l 的距离，…此时，弦PQ 的长． …20．如果函数y=f （x ）的定义域为R ，且存在实常数a ，使得对于定义域内任意x ，都有f （x+a ）=f （﹣x ）成立，则称此函数f （x ）具有“P（a ）性质”．（1）判断函数y=cosx 是否具有“P （a ）性质”，若具有“P （a ）性质”，求出所有a 的值的集合；若不具有“P（a ）性质”，请说明理由；（2）已知函数y=f （x ）具有“P（0）性质”，且当x ≤0时，f （x ）=（x+m ）2，求函数y=f （x ）在区间[0，1]上的值域； （3）已知函数y=g （x ）既具有“P （0）性质”，又具有“P （2）性质”，且当﹣1≤x ≤1时，g （x ）=|x|，若函数y=g （x ）的图象与直线y=px 有2019个公共点，求实数p 的值．【考点】57：函数与方程的综合运用．【分析】（1）根据题意可知cos（x+a）=cos（﹣x）=cosx，故而a=2kπ，k∈Z；（2）由新定义可推出f（x）为偶函数，从而求出f（x）在[0，1]上的解析式，讨论m与[0，1]的关系判断f（x）的单调性得出f（x）的最值；（3）根据新定义可知g（x）为周期为2的偶函数，作出g（x）的函数图象，根据函数图象得出p的值．【解答】解：（1）假设y=cosx具有“P（a）性质”，则cos（x+a）=cos（﹣x）=cosx恒成立，∵cos（x+2kπ）=cosx，∴函数y=cosx具有“P（a）性质”，且所有a的值的集合为{a|a=2kπ，k∈Z}．（2）因为函数y=f（x）具有“P（0）性质”，所以f（x）=f （﹣x）恒成立，∴y=f（x）是偶函数．设0≤x≤1，则﹣x≤0，∴f（x）=f（﹣x）=（﹣x+m）2=（x﹣m）2．①当m≤0时，函数y=f（x）在[0，1]上递增，值域为[m2，（1﹣m）2]．②当时，函数y=f（x）在[0，m]上递减，在[m，1]上递增，y=f（m）=0，，值域为[0，（1﹣m）2]．min③当时，y=f（m）=0，，值域为[0，m2]．min④m＞1时，函数y=f（x）在[0，1]上递减，值域为[（1﹣m）2，m2]．（3）∵y=g（x）既具有“P（0）性质”，即g（x）=g（﹣x），∴函数y=g（x）偶函数，又y=g（x）既具有“P（2）性质”，即g（x+2）=g（﹣x）=g （x），∴函数y=g（x）是以2为周期的函数．作出函数y=g（x）的图象如图所示：由图象可知，当p=0时，函数y=g（x）与直线y=px交于点（2k，0）（k∈Z），即有无数个交点，不合题意．当p＞0时，在区间[0，2016]上，函数y=g（x）有1008个周期，要使函数y=g（x）的图象与直线y=px有2019个交点，则直线在每个周期内都有2个交点，且第2019个交点恰好为，所以．同理，当p＜0时，．综上，．21．给定数列{a n }，若满足a 1=a （a ＞0且a ≠1），对于任意的n ，m ∈N *，都有a n+m =a n •a m ，则称数列{a n }为指数数列． （1）已知数列{a n }，{b n }的通项公式分别为，，试判断{a n }，{b n }是不是指数数列（需说明理由）；（2）若数列{a n }满足：a 1=2，a 2=4，a n+2=3a n+1﹣2a n ，证明：{a n }是指数数列；（3）若数列{a n }是指数数列，（t ∈N *），证明：数列{a n }中任意三项都不能构成等差数列． 【考点】8B ：数列的应用．【分析】（1）利用指数数列的定义，判断即可； （2）求出{a n }的通项公式为，即可证明：{a n }是指数数列；（3）利用反证法进行证明即可．【解答】（1）解：对于数列{a n }，因为a 3=a 1+2≠a 1•a 2，所以{a n }不是指数数列． …对于数列{b n }，对任意n ，m ∈N *，因为，所以{b n }是指数数列． …（2）证明：由题意，a n+2﹣a n+1=2（a n+1﹣a n ），所以数列{a n+1﹣a n }是首项为a 2﹣a 1=2，公比为2的等比数列． … 所以．所以，=，即{a n }的通项公式为（n ∈N *）． …所以，故{a n }是指数数列． …（3）证明：因为数列{a n }是指数数列，故对于任意的n ，m ∈N *，有a n+m =a n •a m ，令m=1，则，所以{a n }是首项为，公比为的等比数列，所以，． …假设数列{a n }中存在三项a u ，a v ，a w 构成等差数列，不妨设u ＜v ＜w ，则由2a v =a u +a w ，得，所以2（t+4）w ﹣v （t+3）v ﹣u =（t+4）w ﹣u +（t+3）w ﹣u ，… 当t 为偶数时，2（t+4）w ﹣v （t+3）v ﹣u 是偶数，而（t+4）w ﹣u 是偶数，（t+3）w ﹣u 是奇数，故2（t+4）w ﹣v （t+3）v ﹣u =（t+4）w ﹣u +（t+3）w ﹣u 不能成立； … 当t 为奇数时，2（t+4）w ﹣v （t+3）v ﹣u 是偶数，而（t+4）w ﹣u 是奇数，（t+3）w ﹣u 是偶数，故2（t+4）w ﹣v （t+3）v ﹣u =（t+4）w ﹣u +（t+3）w ﹣u 也不能成立．… 所以，对任意t ∈N *，2（t+4）w ﹣v （t+3）v ﹣u =（t+4）w ﹣u +（t+3）w ﹣u不能成立，即数列{a n }的任意三项都不成构成等差数列． …2019年上海市虹口高考数学二模试卷一、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）考生应在答题纸的相应位置直接填写结果．1．设f ﹣1（x ）为的反函数，则f ﹣1（1）= ．2．函数y=2sin 2（2x ）﹣1的最小正周期是 ． 3．设i 为虚数单位，复数，则|z|= ．4．= ．5．若圆锥的侧面积是底面积的2倍，则其母线与轴所成角的大小是 ．6．设等差数列{a n }的前n 项和为S n ，若=，则= ．7．直线（t 为参数）与曲线（θ为参数）的公共点的个数是 ．8．已知双曲线C 1与双曲线C 2的焦点重合，C 1的方程为，若C 2的一条渐近线的倾斜角是C 1的一条渐近线的倾斜角的2倍，则C 2的方程为 ．9．若，则满足f （x ）＞0的x 的取值范围是 ．10．某企业有甲、乙两个研发小组，他们研发新产品成功的概率分别为和．现安排甲组研发新产品A ，乙组研发新产品B ，设甲、乙两组的研发相互独立，则至少有一种新产品研发成功的概率为 ．11．设等差数列{a n }的各项都是正数，前n 项和为S n ，公差为d ．若数列也是公差为d 的等差数列，则{a n }的通项公式为a n = ．12．设x ∈R ，用[x]表示不超过x 的最大整数（如[2.32]=2，[﹣ 4.76]=﹣5），对于给定的n ∈N *，定义C =，其中x ∈[1，+∞），则当时，函数f （x ）=C的值域是 ．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．考生应在答题纸的相应位置，将代表正确选项的小方格涂黑．13．命题“若x=1，则x 2﹣3x+2=0”的逆否命题是（ ） A ．若x ≠1，则x 2﹣3x+2≠0 B ．若x 2﹣3x+2=0，则x=1 C ．若x 2﹣3x+2=0，则x ≠1 D ．若x 2﹣3x+2≠0，则x ≠1 14．如图，在正方体ABCD ﹣A 1B 1C 1D 1中，M 、E 是AB 的三等分点，G 、N 是CD 的三等分点，F 、H 分别是BC 、MN 的中点，则四棱锥A 1﹣EFGH 的左视图是（ ）A．B．C．D．15．已知△ABC是边长为4的等边三角形，D、P是△ABC内部两点，且满足，，则△ADP的面积为（）A．B．C．D．16．已知f（x）是偶函数，且f（x）在[0，+∞）上是增函数，若f（ax+1）≤f（x﹣2）在上恒成立，则实数a的取值范围是（）A．[﹣2，1] B．[﹣2，0] C．[﹣1，1] D．[﹣1，0]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．在△ABC中，内角A，B，C的对边分别为a，b，c，已知a ﹣b=2，c=4，sinA=2sinB．（Ⅰ）求△ABC 的面积； （Ⅱ）求sin （2A ﹣B ）．18．如图，在长方体ABCD ﹣A 1B 1C 1D 1中，AB=8，BC=5，AA 1=4，平面α截长方体得到一个矩形EFGH ，且A 1E=D 1F=2，AH=DG=5． （1）求截面EFGH 把该长方体分成的两部分体积之比； （2）求直线AF 与平面α所成角的正弦值．19．如图，已知椭圆C ：（a ＞b ＞0）过点，两个焦点为F 1（﹣1，0）和F 2（1，0）．圆O 的方程为x 2+y 2=a 2． （1）求椭圆C 的标准方程；（2）过F 1且斜率为k （k ＞0）的动直线l 与椭圆C 交于A 、B 两点，与圆O 交于P 、Q 两点（点A 、P 在x 轴上方），当|AF 2|，|BF 2|，|AB|成等差数列时，求弦PQ 的长．20．如果函数y=f （x ）的定义域为R ，且存在实常数a ，使得对于定义域内任意x ，都有f （x+a ）=f （﹣x ）成立，则称此函数f （x ）具有“P（a ）性质”．（1）判断函数y=cosx 是否具有“P （a ）性质”，若具有“P （a ）性质”，求出所有a 的值的集合；若不具有“P（a ）性质”，请说明理由；（2）已知函数y=f （x ）具有“P（0）性质”，且当x ≤0时，f （x ）=（x+m ）2，求函数y=f （x ）在区间[0，1]上的值域； （3）已知函数y=g （x ）既具有“P （0）性质”，又具有“P （2）性质”，且当﹣1≤x ≤1时，g （x ）=|x|，若函数y=g （x ）的图象与直线y=px 有2019个公共点，求实数p 的值．21．给定数列{a n }，若满足a 1=a （a ＞0且a ≠1），对于任意的n ，m ∈N *，都有a n+m =a n •a m ，则称数列{a n }为指数数列． （1）已知数列{a n }，{b n }的通项公式分别为，，试判断{a n }，{b n }是不是指数数列（需说明理由）；（2）若数列{a n }满足：a 1=2，a 2=4，a n+2=3a n+1﹣2a n ，证明：{a n }是指数数列；（3）若数列{a n }是指数数列，（t ∈N *），证明：数列{a n }中任意三项都不能构成等差数列．2019年上海市虹口高考数学二模试卷参考答案与试题解析一、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）考生应在答题纸的相应位置直接填写结果．1．设f﹣1（x）为的反函数，则f﹣1（1）= 1 ．【考点】4R：反函数．【分析】根据反函数的性质，原函数的值域是反函数的定义域即可求解【解答】解：的反函数，其反函数f﹣1（x），反函数的性质，反函数的定义域是原函数的值域，即．可得：x=1，∴f﹣1（x）=1．故答案为1．2．函数y=2sin2（2x）﹣1的最小正周期是．【考点】H1：三角函数的周期性及其求法．【分析】利用二倍角公式基本公式将函数化为y=Acos（ωx+φ）的形式，再利用周期公式求函数的最小正周期，【解答】解：函数y=2sin2（2x）﹣1，化简可得：y=1﹣cos4x﹣1=﹣cos4x；∴最小正周期T=．故答案为3．设i为虚数单位，复数，则|z|= 1 ．【考点】A8：复数求模．【分析】利用复数的运算法则、模的计算公式即可得出．【解答】解：复数===﹣i，则|z|=1．故答案为：1．4． = 3 ．【考点】8J：数列的极限．【分析】通过分子分母同除3n+1，利用数列极限的运算法则求解即可．【解答】解： ===3．故答案为：3．5．若圆锥的侧面积是底面积的2倍，则其母线与轴所成角的大小是30°．【考点】MI：直线与平面所成的角．【分析】根据圆锥的底面积公式和侧面积公式，结合已知可得l=2R ，进而解母线与底面所成角，然后求解母线与轴所成角即可． 【解答】解：设圆锥的底面半径为R ，母线长为l ，则： 其底面积：S 底面积=πR 2， 其侧面积：S 侧面积=2πRl=πRl， ∵圆锥的侧面积是其底面积的2倍， ∴l=2R ，故该圆锥的母线与底面所成的角θ有， cosθ==， ∴θ=60°，母线与轴所成角的大小是：30°． 故答案为：30°．6．设等差数列{a n }的前n 项和为S n ，若=，则=．【考点】85：等差数列的前n 项和． 【分析】=，可得3（a 1+4d ）=5（a 1+2d ），化为：a 1=d ．再利用等差数列的求和公式即可得出． 【解答】解：∵=，∴3（a 1+4d ）=5（a 1+2d ），化为：a 1=d ．则==．故答案为：．7．直线（t为参数）与曲线（θ为参数）的公共点的个数是 1 ．【考点】QK：圆的参数方程；QJ：直线的参数方程．【分析】根据题意，将直线的参数方程变形为普通方程，再将曲线的参数方程变形为普通方程，分析可得该曲线为圆，且圆心坐标为（3，5），半径r=，求出圆心到直线的俄距离，分析可得直线与圆相切，即可得直线与圆有1个公共点，即可得答案．【解答】解：根据题意，直线的参数方程为，则其普通方程为x+y﹣6=0，曲线的参数方程为，则其普通方程为（x﹣3）2+（y ﹣5）2=2，该曲线为圆，且圆心坐标为（3，5），半径r=，圆心到直线x+y﹣6=0的距离d===r，则圆（x﹣3）2+（y﹣5）2=2与直线x+y﹣6=0相切，有1个公共点；故答案为：1．8．已知双曲线C1与双曲线C2的焦点重合，C1的方程为，若C2的一条渐近线的倾斜角是C1的一条渐近线的倾斜角的2倍，则C2的方程为．【考点】KC：双曲线的简单性质．【分析】求出双曲线的焦点坐标，利用渐近线的倾斜角的关系，列出方程，然后求解即可．【解答】解：双曲线C1与双曲线C2的焦点重合，C1的方程为，焦点坐标（±2，0）．双曲线C1的一条渐近线为：y=，倾斜角为30°，C 2的一条渐近线的倾斜角是C1的一条渐近线的倾斜角的2倍，可得C2的渐近线y=．可得，c=2，解得a=1，b=，所求双曲线方程为：．故答案为：．9．若，则满足f（x）＞0的x的取值范围是（1，+∞）．【考点】7E：其他不等式的解法．【分析】由已知得到关于x的不等式，化为根式不等式，然后化为整式不等式解之．【解答】解：由f（x）＞0得到即，所以，解得x＞1；故x的取值范围为（1，+∞）；故答案为：（1，+∞）；10．某企业有甲、乙两个研发小组，他们研发新产品成功的概率分别为和．现安排甲组研发新产品A ，乙组研发新产品B ，设甲、乙两组的研发相互独立，则至少有一种新产品研发成功的概率为．【考点】C9：相互独立事件的概率乘法公式． 【分析】利用对立事件的概率公式，计算即可，【解答】解：设至少有一种新产品研发成功的事件为事件A 且事件B 为事件A 的对立事件，则事件B 为一种新产品都没有成功， 因为甲乙研发新产品成功的概率分别为和． 则P （B ）=（1﹣）（1﹣）=，再根据对立事件的概率之间的公式可得P （A ）=1﹣P （B ）=，故至少有一种新产品研发成功的概率．故答案为．11．设等差数列{a n }的各项都是正数，前n 项和为S n ，公差为d ．若数列也是公差为d 的等差数列，则{a n }的通项公式为a n =．【考点】84：等差数列的通项公式． 【分析】由题意可得：S n =na 1+d ．a n ＞0.=+（n ﹣1）d ，化简n ≠1时可得：a 1=（n ﹣1）d 2+2d ﹣d ．分别令n=2，3，解出即可得出．【解答】解：由题意可得：S n =na 1+d ．a n ＞0．=+（n ﹣1）d ，可得：S n =a 1+（n ﹣1）2d 2+2（n ﹣1）d ．∴na 1+d=a 1+（n ﹣1）2d 2+2（n ﹣1）d ． n ≠1时可得：a 1=（n ﹣1）d 2+2d ﹣d ． 分别令n=2，3，可得：a 1=d 2+2d ﹣d ，a 1=2d 2+2d ﹣d ．解得a 1=，d=． ∴a n =+（n ﹣1）=．故答案为：．12．设x ∈R ，用[x]表示不超过x 的最大整数（如[2.32]=2，[﹣ 4.76]=﹣5），对于给定的n ∈N *，定义C =，其中x ∈[1，+∞），则当时，函数f （x ）=C的值域是．【考点】57：函数与方程的综合运用．【分析】分类讨论，根据定义化简C x n ，求出C x 10的表达式，再利用函数的单调性求出C x 10的值域．【解答】解：当x ∈[，2）时，[x]=1，∴f （x ）=C =， 当x ∈[，2）时，f （x ）是减函数，∴f （x ）∈（5，）；当x ∈[2，3）时，[x]=2，∴f （x ）=C=，当x ∈[2，3）时，f （x ）是减函数，∴f （x ）∈（15，45]； ∴当时，函数f （x ）=C 的值域是，故答案为：．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．考生应在答题纸的相应位置，将代表正确选项的小方格涂黑．13．命题“若x=1，则x 2﹣3x+2=0”的逆否命题是（ ） A ．若x ≠1，则x 2﹣3x+2≠0 B ．若x 2﹣3x+2=0，则x=1 C ．若x 2﹣3x+2=0，则x ≠1 D ．若x 2﹣3x+2≠0，则x ≠1 【考点】25：四种命题间的逆否关系．【分析】根据逆否命题的定义，我们易求出命题的逆否命题 【解答】解：将命题的条件与结论交换，并且否定可得逆否命题：若x 2﹣3x+2≠0，则x ≠1 故选：D14．如图，在正方体ABCD ﹣A 1B 1C 1D 1中，M 、E 是AB 的三等分点，G 、N 是CD 的三等分点，F 、H 分别是BC 、MN 的中点，则四棱锥A 1﹣EFGH 的左视图是（ ）。

上海市华东师大二附中2019-2020学年高三下学期期中物理试题一、单项选择题：本题共6小题，每小题4分，共24分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1、如图所示，在等量异种点电荷形成的电场中有A 、B 、C 三点，A 点为两点电荷连线的中点，B 点为连线上距A 点距离为d 的一点，C 点为连线中垂线上距A 点距离也为d 的一点。

则下列说法正确的是（ ）A ．A CB E E E =>，AC B ϕϕϕ=>B ．B AC E E E >>，A C B ϕϕϕ=>C ．将正点电荷q 沿AC 方向移动到无穷远处的过程中，电势能逐渐减少D ．将负点电荷q 沿AB 方向移动到负点电荷处的过程中，所受电场力先变小后变大2、将一小球从高处水平抛出，最初1 s 内小球动能E k 随时间t 变化的图象如图所示，不计空气阻力，g 取10 m/s 1．根据图象信息，不能确定的物理量是( )A ．小球的质量B ．小球的初速度C ．最初1 s 内重力对小球做功的平均功率D ．小球抛出时的高度3、 “蹦极”运动中，长弹性绳的一端固定，另一端绑在人身上，人从几十米高处跳下。

将蹦极过程简化为人沿竖直方向的运动。

从绳恰好伸直，到人第一次下降至最低点的过程中，若不计空气阻力，下列分析正确的是（ ）A ．绳对人的冲量始终向上，人的动量一直减小B ．绳对人的拉力始终做负功，人的动能一直减小C ．绳恰好伸直时，人的动能最大D ．人的动量最大时，绳对人的拉力等于人所受的重力4、研究光电效应的实验规律的电路如图所示，加正向电压时，图中光电管的A 极接电源正极，K 极接电源负极时，加反向电压时，反之．当有光照射K 极时，下列说法正确的是A．K极中有无光电子射出与入射光频率无关B．光电子的最大初动能与入射光频率有关C．只有光电管加正向电压时，才会有光电流D．光电管加正向电压越大，光电流强度一定越大5、物体做匀速圆周运动时，在任意相同时间间隔内，速度的变化量（）A．大小相同、方向相同B．大小相同、方向不同C．大小不同、方向不同D．大小不同、方向相同6、如图所示，足够长的传送带与水平面的夹角为θ，传送带以速度v0逆时针匀速转动。

静安区2019届高三物理等级考（二模)试卷考生注意：1．试卷满分100分，考试时间60分钟。

2．本考试分设试卷和答题纸。

试卷包括三部分，第一部分为选择题,第二部分为填空题，第三部分为综合题。

3．答卷前，务必用钢笔或圆珠笔在答题纸正面清楚地填写姓名、准考证号等。

作答必须涂或写在答题纸上，在试卷上作答一律不得分。

第一部分的作答必须涂在答题纸上相应的区域，第二、三部分的作答必须写在答题纸上与试卷题号对应的位置.一、单项选择题（共40分。

第1—8小题，每小题3分，第9—12小题，每小题4分。

每小题只有一个正确答案。

）1。

在真空中传播速度等于光速的射线是( )A. 阴极射线B。

α射线 C. β射线D。

γ射线【答案】D【解析】【详解】阴极射线是低压气体放电管阴极发出的电子在电场加速下形成的电子流，传播速度小于光速,α射线是核原子核，速度为光速的十分之一,β射线是高速电子流，速度为光速的99%,γ射线是光子流,传播速度等于光速，故D正确，ABC错误。

2。

链式反应中，重核裂变时放出的可以使裂变不断进行下去的粒子是( )A. 质子B. 中子C. β粒子D。

α粒子【答案】B【解析】重核裂变时放出中子，中子再轰击其他重核发生新的重核裂变,形成链式反应，B项正确.3。

声波能绕过某一建筑物传播而光波却不能绕过该建筑物，这是因为A。

声波是纵波，光波是横波B。

声波振幅大，光波振幅小C。

声波波长较长,光波波长很短D。

声波波速较小,光波波速很大【答案】C【解析】波发生明显的衍射现象的条件是:孔缝的宽度或障碍物尺寸与波长相比差不多或比波长更短．由于声波的波长比较大（1.7cm～17m）和楼房的高度相近,故可以发生明显的衍射现象,而可见光的波长很小，无法发生明显的衍射现象．故只有C正确．4.布朗运动实验，得到某个观察记录如图，该图反映了（）A. 液体分子的运动轨迹B。

悬浮颗粒的运动轨迹C. 每隔一定时间液体分子的位置D。

每隔一定时间悬浮颗粒的位置【答案】D【解析】【详解】A、布朗运动是悬浮在液体中的固体小颗粒的无规则运动,而非分子的运动，故A错误；B、布朗运动是无规则运动，微粒没有固定的运动轨迹，故B错误；CD、该图记录的是按等时间间隔依次记录的某个运动悬浮微粒位置的连线，故C错误,D正确。

2019-2020学年上海市第二中学高三英语月考试题及答案解析第一部分阅读（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AIn the 1994 filmForrest Gump, there’s a famous saying, “Life is like a box of chocolates; you never know what you’re gonna get.” The surprise is part of the fun. Now blind box toys are bringing the magic of surprise to online shopping.A blind box toy is hidden inside uniform packaging but invisible from the outside. You don’t know what will be inside, although the toys typically come from pop culture, ranging from movies to comics and cartoons.Blind boxes have caught on since they were first introduced fromJapantoChinain 2014. According to a 2019 Tmall report, the mini-series of Labubu blind box, designed byHong Kong-born Kasing Lung, was named Champion of Unit Sales with 55,000 sold in just 9 seconds during the Singles Day shopping event. Most customers for blind boxes are young people aged 18 to 35.According to The Paper, blind box toys are popular in part because of their cute appearances. The typically cute cartoon figurines come in miniature sizes, making them suitable for display almost anywhere.Even if blind boxes are not their top choice for decorations, the mystery and uncertainty of the process also attracts people. It’s the main reason why people buy blind boxes one after another.“Fear of the unknown is always a part of the box-opening process,” said Miss Cao, 24, who lives and works inShenyang. Speaking to Sina News, she said: “Until you open all the boxes, you cannot know what it is inside.”Opening a blind box is a delightful little surprise for our mundane daily lives, something small but fun to wait for each day, week or month. When people open this simple little box, they may be disappointed, but the uncertainty is part of the fun. People will open more blind boxes and hope for a better outcome.When someone re-makesForrest Gump, don't be surprised if he says, “Life is like a blind box...”1. Why is the famous saying in the filmForrest Gumpquoted at the beginning?A. To arouse the readers’ interest.B. To present the writer’s view.C. To introduce the topic.D. To highlight the fun of blind boxes.2. Which of the following is the main feature that makes blind box so popular?A. Miniature sizes.B. Cute appearances.C. Fear of the unknown.D. Mystery and uncertainty.3. What can we learn from the passage?A. Blind box became popular in 2019 after being first introduced fromJapantoChina.B. Blind box toys typically originated in pop culture, varying from movies to cartoons.C. Blind box toys was designed and named by Hong Kong-born Kasing Lung.D. When people open this simple little box, they will feel disappointed.BThere are 195 countries in the world today but almost none of them have purple on their national flag. So what’s wrong with purple? It’s such a popular color1 today. Why would no country use it in their flag? The answer is really quite simple. Purple was just for too expensive.The color1 purple has been associated with royalty power and wealth for centuries. Queen Elizabeth I forbade anyone except close members of the royal family to wear it. Purple’s high status comes from the rarity and cost of the dye (染料)originally used to produce it. Fabric traders got the dye from a small sea snail (海螺)that was only found in the Tyre region of the Mediterranean. More than 10,000 snails were needed to create just one gram of purple; not to mention a lot of work went into producing the dye, which made purple dye so expensive.Since only wealthy rulers could afford to buy and wear the color1 , it became associated with the royal family. Sometimes, however, the dye was too expensive even for royalty. Third century Roman Emperor Aurelio famously wouldn’t allow his wife to buy a scarf made from purple silk because it cost three times its weight in gold. A single pound of dye cost three pounds of gold, which equals 56,000 dollars today. Therefore, even the richest countries couldn’t spend that much having purple on their flags.The dye became more accessible to lower-class about a century and a half ago. In 1856, 18-year-old English chemist William Henry Perkin accidentally created a man-made purple compound (化合物)while attemptingto produce an anti-malaria drug. He noticed that the compound could be used to dye fabrics, so he patented the dye, manufactured it and got rich. Purple dye was then mass-produced so everybody could afford it.Till now, a handful of new national flags have been designed and a few of them have chosen to use purple in their flag. So don’t be making any bets just yet.4. Why was color1 purple expensive in the past?A. Because only royal families were allowed to wear purple.B. Because it took a long time to get purple dye from gold.C. Because purple was worth as much as its weight in gold.D. Because purple dye used to be rare and hard to produce.5. Why did Roman Emperor Aurelio forbid his wife to buy a purple scarf?A. Because of poor quality.B. Because of long tradition.C. Because of bad taste.D. Because of high price.6. What is purple's situation now?A. Purple has been widely used on national flags.B. Purple dye is now affordable to ordinary people.C. Royal family stop using purple because it’s toocommon.D. Fewer snails are used to produce purple dye than before.7. Which of the following would be the best title for the passage?A. No Purple Flags?B. Purple vs GoldC. How to Produce Purple Dye?D. The Birth of Purple ColorCAccording to a survey, the wasteof food on the dining table occupies 10% of the total grain output.Last week, Meituan, a giant online food ordering platform, co-published a proposal with a number of business organizations, calling on restaurants to stop food waste and help develop new eating habits for customers. Following the proposal, merchants are asked to offer guidance for consumers, including reminding them during the ordering process about the taste of the ingredients, portion sizes and other information about the dishes, to helpthem avoid excessive ordering and food waste.Catering（餐饮)associations in more than 18 provinces have also joined the campaign to remove food waste. The Wuhan Catering Association proposed an “NT" ordering code for restaurants in which a group of 10 diners would only order enough for nine people. More food is only brought to the table if required. On Friday, the China Cuisine Association announced that it had teamed up with Ele. me, the Alibaba Group Holding-owned food delivery platform, to launch a "half-dish plan," encouraging restaurants to provide customers with the option to order smaller portions.Tang Zhisong, a professor at Southwest University Education School, said "Evaluating how much you can eat, how much you should buy and how to deal with the leftover is a way for young people to improve their self-management. It's also a means to teach them sharing food, caring about others, and more importantly, developing a mindset of suitability. "8. What's the purpose of the proposal mentioned in the passage?A. To change customers' attitude toward life.B. To promote a new policy on food delivery.C. To spread the idea of healthy eating.D. To encourage restaurants to reduce food waste.9. What does the underlined word “excessive" in Paragraph 2 prolably mean?A. More than enough.B. Less than required.C. Better than ever.D. Worse than before.10. Paragraph 3 is mainly developed by.A. offering analysesB. presenting a surveyC. giving examplesD. making comparisons11. What do Tang's words suggest?A. Sharing food is caring about others.B. Young people should have self-discipline.C. Reducing food waste has all-round benefits.D Saving food contributes to a sustainable society.DMark Bertram lost the tips of two fingers at work in 2018 when his hand became trapped in a fan belt. “It’s life-changing but it’s not life-ending,”he says.After two surgeries and occupational therapy, Bertram decided to ask Eric Catalano, a tattoo artist, to create fingernail tattoos. The idea made everyone in the studio laugh—until they saw the final result. “The mood changed,” Catalano recalls from his Eternal Ink Tattoo Studio in Hecker, Illinois. “Everything turned from funny to wow.”Catalano posted a photo of the tattoos, and it eventually was viewed by millions of people around the world. The viral photo pushed Catalano, 40, further into the world of paramedical tattooing. Now people who want to cover their life-altering scars come from as far away as Ireland to visit his shop.Leslie Pollan, a dog breeder, was bitten on the face by a puppy. She underwent countless surgeries but those gave her no hope. She ultimately traveled six hours for a session with Catalano. HecamouflagedPollan’s lip scar, giving her back confidence.Though he is now known for his talent with intricate fingernail, Catalano uses the techniques he picked up years ago while helping breast cancer survivors. Those tattoos are among the most common paramedical requests. His grandmother had breast cancer, and her battle with the disease is one reason Catalano is so dedicated tohelping those with the diagnosis.Catalano performs up to eight reconstructive tattoos each “Wellness Wednesday”. While he charges $100 per regular tattoo, he doesn’t charge for paramedical tattoos: A GoFundMe page established last year brought in more than $16,000, allowing Catalano to donate his work.“Financially, it doesn’t make sense,” Catalano says. “But every time I see emotions from my customers, I am 100 percent sure this is something that I can’t stop doing.”12. How did people in the studio react to Bertram’s idea at first?A. They took it lightly.B. They found it creative.C. They were confused.D. They were impressed.13. What does the underlined word “camouflaged” in Paragraph 4 probably mean?A. Exposed.B. Hid.C. Ignored.D. Removed.14. What does Catalano say about his work with paramedical tattoos?A. It is flexible.B. It is demanding.C. It is profitable.D. It is rewarding.15. Which of the following can best describe Catalano?A. Humorous and experienced.B. Devoted and generous.C. Cooperative and grateful.D. Professional and tolerant.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。

2019-2020学年上海市二中学高三英语模拟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ANew events and changes of junior golf competition calendarNew eventsNotah BegayⅢJunior Golf National ChampionshipWhat does a junior golfer aim to pursue? One thing is to be noticed, ideally by a college coach. A remarkable opportunity will be offered by the Notah BegayⅢJunior Golf National Championship to its participants: an event broadcast by Golf Channel. Players aged between eight and 18 can compete in the new event; information about where and when it will be held will be released later.Barbasol Junior ChampionshipBeginning the career in the PGA Tour is something that a junior golfer tends to dream of. The Barbasol Junior Championship, which is scheduled to take place between June 29 and July 2 at Keene Trace Golf Club inNicholasville,Kentucky, will provide such an opportunity. Boys under 19 years old will qualify for this new 54-hole event, and the winner will be awarded a spot at the PGA Tour's Barbasol Championship in July.Changed eventsThunderbird International JuniorThe dates of the AJGA's Thunderbird International Junior have to be changed since the NCAA Championships move to Grayhawk Golf Club inScottsdale,Arizona, for the next three years. Generally, the Thunderbird is played at the end of May. However, this year it is scheduled on different dates for the first time, from April 9 to 12, which means, of course, that the finish date is on Masters Sunday.Gator InvitationalJunior golf intends to prepare for the following college golf. If this is the case, then it is crucial to simulate the higher-level experience as much as possible. Because of that, the Gator Invitational, as a junior boys' event, has made a significant decision on becoming a 54-hole event by adding a round this year. The new version will be played from March 13 to 15 at The Country Club of Jackson inJackson,Mississippi.1. Which event can be watched on TV?A. Notah BegayⅢJunior Golf National Championship.B. Barbasol Junior Championship.C. Thunderbird International Junior.D. Gator Invitational.2. When will the Thunderbird International Junior be played?A. At the end of May.B. From April 9 to 12.C. Between June 29and July 2.D. From March 13 to 15.3. What has been changed about the Gator Invitational?A. The award given to the winnerB. The place where it is played.C. The required age of the players.D. The number of rounds it has.BPlease take a few seconds and think of your personal biggest goal. Imagine telling someone you meet today what you’re going to do. Imagine their congratulations and their high image of you. Doesn’t it feel good to say it out loud? Don’t you feel one step closer already? Well, bad news: you should have kept your mouth shut, because that good feeling will make you less likely to do it.Any time you have a goal, there is some work that needs to be done to achieve it. Ideally, you would not be satisfied until you’d actually done the work. But when you tell someone your goal and he acknowledges(认可) it, psychologists have found it’s called a “social reality”. The mind is kind of tricked into feeling that it’s already done. And then, because you’ve felt that satisfaction, you’re less motivated to do the actual hard work necessary. This goes againstthe traditional wisdom that we should tell our friends our goals, right?In 1982, Peter Gollwitzer, a Professor of Psychology, wrote a whole book about this. And in 2009, he did some new tests that were published. It goes like this: 163 people across four separate tests—everyone wrote down their personal goal. Then half of them announced their commitment(许诺) to this goal to the room, and half didn’t. Then everyone was given 45 minutes of work that would directly lead them towards their goal, but they were told that they could stop at any time. Now those who kept their mouths shut worked the entire 45 minutes on average, and when asked afterwards, said they felt they had a long way to go to achieve their goal. But those who had announced it quit after only 33 minutes on average, and when asked afterwards, said that they felt much closer to achieving their goal.4. What do the words “social reality” in Paragraph 2 mean?A. Completion of the goal.B. Necessary hard work.C. People's acknowledgement.D. A sense of satisfaction.5. What does Peter Gollwitzer try to tell us?A. Writing down the goal is very helpful.B. Achieving personal goal needs more time.C. Keeping the goal secret makes people work harder.D. Making the goal public makes people less satisfied.6. How did Peter Gollwitzer prove his idea about people’s goal?A. By giving figures.B. By giving examples.C. By making a survey.D. By making comparison tests.7. What will probably happen if you tell your friends your goal?A. You will be more confident.B. You will not gain satisfaction.C. You are less likely to realize it.D. You’ll be much moremotivated.CYou run into the grocery store to quickly pick up your item. You grab what you need and head to the front of the store. After quickly sizing up the check-out lines, you choose the one that looks fastest. You chose wrong. People getting in other lines long after you have already checked out and headed to the parking lot. Why does this seem to always happen to you?Well, as it turns out, it's just math that is working against you. A grocery store tries to have enough employees at the checkout lines to get all their customers through with minimum delay. But sometimes, like on a Sunday afternoon, they get super busy. Because most grocery stores don't have the physical space to add more checkout lines, their system becomes overburdened. Some small interruption — a price check, a particularly talkative customer — will have downstream effects, holding up the entire line behind them.If there are three lines at the store, these delays will happen randomly at different registers (收银台). Think about the probability. The chances of your line being that fastest one are only one in three, which means you have a two-thirds chance of not being in the fastest line. So it's not just in your mind: Another line is probably moving faster thanyours.Now, mathematicians have come up with a good solution, which they call queuing theory, to this problem: Just make all customers stand in one long snaking line, called a serpentine line, and serve each person at the front with the next available register. With three registers, this method is about three times faster on average than the more traditional approach. This is what they do at most banks, Trader Joe's, and some fast-food places. With a serpentine line, a long delay at one register won't unfairly punish the people who lined up behind it. Instead, it will slow everyone down a little bit.8. What phenomenon is described in the first paragraph?A. Queuing in a line.B. A shopping experience.C. A rush in the morning.D. Cutting in a line.9. According to the article, what may cause delays in checking out?A. The lack of employees in the grocery store.B. Some unexpected delays of certain customers.C. The increasing items bought by customers.D. A worsening shopping system of the store.10. What is the solution given by mathematicians?A. Employing more workers for checking out.B. Limiting the number of queuing people.C. Making only one line available.D. Always standing in the same line.11. What's the principle behind the queuing theory?A. To pursue the maximum benefit.B. To leave success or failure to luck.C. To avoid the minimum loss.D. To spread the risk equally among everyone.DNaomi Cooke was walking with a friend and their dogs through her local park in Burnside, on Tuesday when she heard someone shout to watch out. Cooke turned and hardly had time to react before a flying disc hit her in the face with a "big bang”, leaving her right cheek swollen almost to the size of a golfball.Two men playing disc golf at the course in Jellie Park were about 20 metres fromthe pairwhen one of them threw the disc hard, aiming for a nearby goal.After being hit Cooke immediately went to the emergency department, where two CT scans on her face and cheek found she had escaped any broken bones. "I'm lucky it didn't hit my eye because I think I would have lost it." Cooke said.Cooke often walks her dog at the park and said it was always busy with people playing disc golf, but it was not until after Tuesday that she became concerned about public safety there.There were no signs about the disc golf course in the park, she said, and the area is shared with children and people walking their dogs.“If it had hit one of the kids in the head, it could have killed them.” Cooke did not think she was the only person who had been hit before, and said there would be others who share her concerns.Cooke planned to go to the council, saying it needed to realise how dangerous it was for the space to be shared by everyone and to provide disc golfers with a space where they can play safely. "There should be rules about how it's done, making it safe for everyone.”12. What happened to Cooke on Tuesday?A. She was struck by a golf ball.B. She was hit by a flying disc.C. She was beaten by two men.D. She was frightened by a mad dog.13. What do the underlined words "the pair" in Paragraph 2 refer to?A. Cooke and her friend.B. Cooke and her dog.C. The two disc golfers.D. The two CT scans.14. How did Cooke feel about people playing disc golf in the park?A. Acceptable.B. Shocked.C. Angry.D. Worried.15. Why did Cooke plan to go to the council?A. To get the two men in trouble.B. To call for a ban on disc golf.C. To ask for personal protection.D. To call for safer places for disc golf.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。

2019-2020学年上海市十五中学高三英语二模试卷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APeople in the Middle Ages did eat with their hands. Personal utensils (餐具) were mostly unheard of, especially forks. There were spoons to help serve, but only special guests would receive a knife from the host. Everyone else would be expected to bring their own. Of course, eating with one's hands can be quite a sticky situation, so towels were provided to help diners stay at least somewhat clean as they ate.Still, dining was often a messy affair. At special occasions in the wealthiest households, women tended to dine alone, separate from the men. Women were expected to uphold a quality of grace. Eating greasy meat by hand would certainly not help! Once the men and women had finished their meals, they would come together to socialize.Dietary scholars of the Middle Ages believed that the foods in a meal needed to be served and eaten in order of heaviness. The lightest and most easily digested foods, such as fruits and cheeses, were eaten first to help the digestive (消化) system get started. Once digestion was underway, greens and light meats, such as lettuce, cabbage and chicken, could be eaten. Last came the heavier vegetables and meats, such as carrots, beans, beef, pork, and mutton. This method was considered the most healthful way to eat.The main and largest meal of the day was supper, and it was eaten at midday. Dinner was a light meal, and many of those in nobility (贵族) -the highest levels of the Middle Ages society-skipped breakfast altogether. Breakfast was considered unnecessary for those who did not perform physical work. Snacks and any other eating during the day were viewed the same way. Commoners, or the working class, were allowed to eat breakfast and small meals throughout the day.1. What did people in the Middle Ages usually do at the dinner table?A. They cleaned hands before meals.B. They used personal utensils.C. They had to use knives at dinner.D. They ate mostly with hands.2. What is the third paragraph mainly about?A. The order of eating foods.B. The weight of various foods.C. The principles of digesting foods.D. The list of healthy foods.3. Why did the nobility avoid eating breakfast?A. Breakfast was wasted for the nobility.B. Breakfast was viewed as unnecessary.C. Breakfast was considered as a light meal.D. Breakfast was saved for commoners.BWhen my friend suggested going to the op shop (二手商店)，instantly I thought “I hope no one I know sees me”. It was the same when my cousin commented on my new furniture and Japanese, fine-bone-china bowls and asked where I got them. They were from the local op shop but instead I said “from the antique shop”.Many people in my Greek-Cypriot community would look down on me if I said I shopped at the op shop. They may pity me, consider me poor, a failure. Immigrants sacrificed their families and homes for a better life. Buying a house and having enough money to live comfortably, to educate your children and see them also live comfortably, are a big part of the immigrant dream, But has this dream made us materialistic at the cost of our own planet?Our love for purchasing the latest trendy clothes or furniture, then donating them when we are tired of them has become normal. I was once like this. But after watching the documentaryThe True CostI learned donated clothes that don’t get sold are sent to developing nations, many of them ending up in landfills (垃圾填埋地). In addition, your new dress requires electricity and materials to make. But if you buy a second-hand dress, that’s one less dress in a landfill and one less new dress to be made.A friend introduced me to op shopping only a few years ago. My firstitem was a dress she gifted me. It was lovely and I loved it. Nobody could tell it was second-hand. This opened me up to purchasing more second-hand high quality branded clothes. Once I visited a friend and was impressed by how she decorated her apartment. “It’s all second-hand,” she said. I couldn’t believe it. The truth is a lot of things sold at the op shop are in new or almost new condition. That’s when I made the decision to only buy second-hand things.Selling second-hand things isn’t anything new butwhat the planet needs is more buyers. There is so much excess (过量) production in the world. So stop feeling ashamed, and let’s get shopping.4. What kind of feeling is expressed in Paragraph 1?A. Pride.B. Embarrassment.C. Delight.D. Sympathy.5. Whatare many people in the author’s community like?A. They are probably materialistic.B. They care about the environment.C. They think highly of op shopping.D. They look down upon immigrants.6. What was the author encouraged to do after visiting her friend’s apartment?A. Watch the documentaryThe True Cost.B. Donate more to local charities.C. Avoid shopping too much.D. Stop buying new things.7. What’s the purpose of the text?A. To entertain.B. To advertise.C. To persuade.D. To describe.CA crew（全体成员）of six teenage girls completed a nine-day sailing trip in the US recently, after having seasickness and strong winds.For the past three years, the Sea Cadet teenagers whoset sail were all male. Roger Noakes, who captained（担任队长）the boat, said this was the first time he’d taken out an all-female crew.The girls asked for an all-girls trip in August this year. The crew set sail along with three adults, Noakes and two Sea Cadet representatives. The original plan was for the girls to sail 24 hours a day in rotating shifts（轮流换班）along the coast and then return. Things turned out differently, however. “The first night was difficult because the wind was really hard. The waves were going up and down,” said Abby Fairchild,16. “Everybody got seasick.” Noakes gave the girls the choice of just sailing in the bay and not going into open water. “But they decided they were going.”The teenagers then sailed a long way overnight and slept in shifts. “We’ve learned everything from controlling the boat to putting up the sails while we have rough seas,” said 15-year-old Olivia Wilcox.The teenagers stopped on land in Massachusetts. They didn’t make it to their original destination（目的地）in Maine, where they were supposed to have a celebratory dinner, due to the weather and winds. They said they weren’t disappointed, however, as they’d learned a lot. “They learned about boating, and above all, they built confidence and character,” said Noakes.8. What was special about the Sea Cadet trip this year?A. It was the longest sailing trip ever.B. It was the first all-female-crew sailing trip.C. It was the most dangerous sailing trip ever.D. It was the first sailing trip for teenagers.9. What happened on the crew’s first day of the trip?A. They all felt sick on the boat.B. Some of them were hurt.C. Their boat was out of control.D. They went into open water by mistake.10. Which of the following best describes these young sailors?A. Strong-minded and having a strong sense of teamwork.B. Hard-working and having great leadership skills.C. Understanding and creative.D. Adventurous and skillful.11. According to Noakes, what was the sailors’ greatest benefit from the trip?A. They knew the sea better.B. They made many friends.C. They got excellent sailing skills.D. They developed good personalities.DDengue is a very painful illness spread by mosquitoes. In severe cases, dengue can even be deadly. Dengue is a serious disease affecting people in around 120 countries. It can cause high fevers, headaches, and severe pain. It’s caused by a virus spread by bites from mosquitoes. Therefore, dengue is more common in warm areas. Every year, roughly 390 million people get dengue, and as many as 25,000 die from it.Now scientists seem to have found a way to protect humans from dengue by first protecting mosquitoes. Dengue fever is caused by a virus. Though it may seem strange to think of it this way, the mosquitoes that spread the dengue virus are also infected with it. But the virus doesn’t seem to hurt the mosquitoes.Wolbachia is a kind of bacteria commonly found in many insects. In some insects, Wolbachia can keep some viruses fromduplicatingthemselves, which is how viruses grow inside a body. Wolbachia isn’t naturally found in mosquitoes. But by infecting these mosquitoes with Wolbachia, scientists can keep the mosquitoes from catchingthe dengue virus. Even better, the young mosquitoes coming from the eggs of the infected mosquitoes also carry Wolbachia.Researchers working with the World Mosquito Program (WMP) ran a 27-month study in Yogyakarta, Indonesia. They split a 10-square-mile area up into 24 smaller areas. In half of the areas, the scientists did nothing. In the other half, they set out containers of eggs from mosquitoes that had Wolbachia. They did this every two weeks for just 4 to 6 months.Ten months later, 80% of the mosquitoes in the treated areas carried Wolbachia. The researchers report the number of dengue cases in the treated areas was reduced by 77% and that the number of people needing hospital care for dengue dropped by 86%.Because the results of the experiment were so good, the WHO has placed Wolbachia-infected mosquito eggs in all parts of Yogyakarta and surrounding areas. The WHO says that within a year, their efforts will protect 2.5 million people against dengue and that their efforts will be turned into a program that can be repeated worldwide.12. What kind of disease is dengue?A. It is likely to cause death.B. It causes no pain but fevers.C. It happens less often in hot areas.D. It hurts both people and mosquitoes.13. The underlined word “duplicating” in paragraph 3 most probably means “________”.A. worsening the harm ofB. expanding the size ofC. increasing forces ofD. making copies of14. What can be inferred about the method from the figures listed in paragraph 5?A. Its wide use.B. Its effectiveness.C. Its complexity.D.Its easy operation.15. What’s the WHO’s attitude towards the method?A. Ambiguous.B. Positive.C. Tolerant.D. Skeptical.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。

2019-2020学年上海市二中学高三英语模拟试卷及答案解析第一部分阅读（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AWhen you grow up in Voss, outdoor adventures become a way of living. This is why visitors will find outdoor activities for all ages and levels. Many people get the impression that such activities as river sports, air sports and other extreme sports are only for the experts. Actually, you will find many outdoor adventures for those who simply want a taste of these elements in Voss.◆River KayakingThe river in Voss are great for river kayaking. If you are a beginner, we advise you to try an introduction course of 3 hours. A course of 2 days can be tested out if you really want to learn the sport of kayaking. Get a totally new experience with one of the best kayak centers inNorway.Prices From NOK 850 per person◆RaftingThe most popular summer activity in Voss. Thrilling, fun and suitable for beginners as well as those with experience! Includes transportation, safety instruction, swim test and about8kmof breathtaking rafting starting off in softer steams before getting on to the more exciting streams.Season Daily May—OctoberPrices From NOK 1,120Info All necessary equipment is provided. Please bring your own swimwear and towel.◆Bavallsekspressen Chair LiftExplore the mountain by riding the Bavallsekpressen chair lift all the way to the top to get immediate access to a great variety of hiking trails in beautiful scenery. The lift is also open for those who want to bring their bike or paraglider. Start and end: From Bavallen to Hangurstoppen.Season: Sat/Sun 24 June—06 August 12:00-16:00Prices Single trip: NOK 100 Day pass: NOK 250◆HusdyrparkenAt Husdyrparken, visitors get to experience Norwegian farm animals. You can participate in animal feeding and farm competitions, or simply relax with an organic ice cream in the café. You can also visit a small museumwith old farming equipment.Season: Daily 18 June—21 AugustPrices: Adults NOK 120 Children NOK 60 Senior NOK 60Family Pass NOK 200 (For up to two adults and two kids)1. Who are the intended readers of the passage?A. Local residents.B. Professional athletes.C. Travel experts.D. Common Tourists.2. Which of the following activities provide instructions for beginners?A. River Kayaking and Rafting.B. Husdyrparken and River Kayaking.C. Rafting and Bavallsekspressen Chair Lift.D. Bavallsekspressen Chair Lift and Husdyrparken.3. How much should Jan pay for a farm trip with her little son and her mother in Voss?A. NOK 240.B. NOK 200.C. NOK 180.D. NOK 120.BThis past year, I've found myself returning again and again to lines of poetry by Emily Dickinson. Like many people, I've needed the curing effects of reading more than ever. As scientists and psychologists will tell you, books are good for the brain and their benefits are particularly vital now.Books expand our world, providing an escape and offering novelty, surprise and excitement. They broaden our view and help us connect with others. Books can also distract us and help reduce ourmental chatter.When we hit the “flow state" of reading where we're fully lost in a book, our brain's mode network calms down. It's a network of brain that is active and gets absorbed in thinking and worrying endlessly when we are not doing anything else.There is so much noise in the world right now and the very act of reading is kind of meditation. You disconnect from the chaos around you.You reconnect with yourself when you are reading. And there's no more noise.In 2020, the NPD Group recorded the best year of book sales since 2004. Yet even as people are buying more books,many are reporting they're having a harder time getting through them. It's difficult for your brain to focus on a book when it's constantly scanning for threats to keep you alive.Our fight-or-flight response has been consistently activated.Sometimes I picture my brain as a cartoon brain with little arms and legs, fighting with a book I am holdingand screaming: “Can't you see I'm busy!” Anxiety causes our brain to produce a flood of stress,which consumes our energy and makes it harder to concentrate.Then one day in December sitting on my couch, I remembered how much I like to read"The House of Mirth" every few years around the holidays. The memory inspired me to pick up the familiar book, opened it up and started reading.I just kept going.The comfort and distraction and brain-opening experience gave me peace.So return to something familiar.4. What does the underlined part “mental chatter” in Paragraph 2 mean?A. Getting lost in a book.B. Non-stop inner anxiety.C. Chatting with the author.D. Powerful network of brain.5. What do we know about reading according to the text?A. It can treat our headache.B. It can calm down the noisy people.C. It forces us to concentrate.on thinking.D. It makes us communicate with ourselves.6. Why was it difficult for people to finish reading books in 2020?A. People bought too many books.B. The books were too difficult to understand.C. People just wanted to escape from the threat.D. The life threat disturbed people's focus on books.7. Why is the author's experience mentioned in the last paragraph?A. To rid people of concern for safety.B. To present an effective reading way.C. To wake up memories of an old book.D. To recommend the book he/she reads.CYu Chenrui, 29, is a maker of automata (机关人偶) in Chengdu, Sichuan province. Automata are built to look like humans or animals and give the illusion (错觉) of being able to move ontheir own, “The art form amazes me, because it combines various skills, from storytelling to mechanics, and the pieces are built with a sense of humor,”Yu says. His creations have caught the attention of well-known artists and his fancy pieces of art have attracted collectors worldwide.Interested in handcrafts as a boy, Yu first encountered automata designed by Japanese artist Kazuaki at an exhibition in 2015 when he studied at the Communication University of China in Beijing. “It was like meeting a like-minded friend, ” Yu says, recalling the moment. As an art and design major, he began to learn the craft by himself and, with the support of his tutor, he kept studying and examining automata in school.When he graduated in 2016, Yu landed a job at an advertising agency in Beijing. He stayed on at the company for three years because, at that time, he was not sure that he could make a living out of his hobby. While working as a designer Yu kept exploring and advancing his skills in wood carving and mechanics. Eventually, despite the job’s good salary, it was not enough to make up for not following his true passion. Finally, in 2018, Yu quit his job and returned to Chengdu to open his automata workshop.To keep himself occupied while running his workshop, he planted blueberries, raspberries and cherries. He watered, weeded and added fertilizer (肥料) every day. “Daily routines helped me calm down and inspire my creativity, which resulted in an automaton called To Observe the Autumn,” Yu says.Over time, Yu’s reputation grew and his business flourished (兴旺). Many of Yu’s creations are built with a dash of wisdom, a sprinkle of humor and are inspired by observations of real life. Yu knows that there are many more creative ideas waiting to be expressed. “It feels quite good to be fully devoted to automata creation and I am still searching for myself.”8. What can we learn from the first paragraph?A. Yu is equipped with various skills.B. Yu is a person full of sense of humor.C Yu is now gaining recognition worldwide.D. Yu is following the latest trend in handcraft.9. Why did Yu quit his job in Beijing?A. He couldn't make a living out of it.B. He wanted to pursue his own dream.C. He missed his family in Chengdu.D. He thought he had a lot experience.10. According to Yu, what contributes to his creativity as an automata artist?A. The fruits he grows.B. Success of his business.C. His devotion to the job.D. Observation from daily life.11. Which of the following can best describe Yu’s story?A. There is no end to learning.B. Great hopes make great man.C. Actions speak louder than words.D. Experience is the mother of wisdom.DIt is a question people have been asking for ages. Is there a way to turn back the aging process?For centuries, people have been looking for a “fountain of youth”. The idea is that if you find a magical fountain, and drink from its waters, you will not age.Researchers in New York did not find an actual fountain of youth, but they may have found a way to turn back the aging process. It appears the answer may be hidden right between your eyes, in an area called the hypothalamus (下丘脑). The hypothalamus is part of your brain. It controls important activities within the body.Researchers at New York’s Albert Einstein College of Medicine found that hypothalamus neural (神经的) stem cells also influence how fast aging takes place in the body.What are stem cells(干细胞)? They are simple cells that can develop into specialized cells, like blood or skin cells. Stem cells can also repair damaged tissues and organs.Dongsheng Cai is a professor at the Albert Einstein College of Medicine. He was the lead researcher in a study on aging in mice. He and his team reported their findings in the journal Nature, Cai explained when hypothalamus function is in decline, particularly the loss of hypothalamus stem cells, and this protection against the aging development is lost. it eventually leads to aging.Using this information, the researchers began trying to activate, or energize, the hypothalamus laboratory mice. They did this by injecting the animals with stem cells, Later, the researchers examined tissues and tested for changes in behavior. They looked for changes in the strength and coordination (协调) of the animals muscles. They also studied the social behavior and cognitive ability of the mice. The researchers say the results show that the treatment slowed aging in the animals, Cai says injecting middle-aged mice with stem cells from younger mice helped the older animals live longer.But these results were just from studying mice in a laboratory. If the mice can live longer, does that mean people could have longer lives? The next step is to see if the anti-aging effects also work in.12. In Paragraph 2 a “fountain of youth” is mentioned to ________.A. introduce the main topicB. show a hidden secret.C. describe scientists researchD. recommend a way to stay young13. Aging takes place in the body when _______.A. stem cells develop into specialized cellsB. there are important activities within the bodyC. hypothalamus neural stem cells fail to protect against agingD. the hypothalamus fails to repair damaged tissues and organs14. What do we know about the researchers at Albert Einstein College of Medicine from the text?A. They did experiments to see how stem cells work.B. They studied mice to find their connection with humans.C. They have found a possible way to slow the aging progress.D. They have found no changes in mice s behavior during the experiment15. What will the researchers probably do next?A. They will help some animals live longer.B. They will announce the fountain of youth doesn’t existC. They will develop products to help people live a longer life immediatelyD. They will do research to see if what they have found in mice will apply to humans.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。