2020年5月上海市浦东新区普通高中2020届高三下学期期中教学质量监测(二模)语文试题及答案

浦东新区2019学年度第二学期高中教学质量检测试题高三数学2020.05考生注意：1.本场考试时间120分钟，试卷共4页，满分150分，另有答题纸。

2.作答前，在答题纸正面填写姓名，编号等信息。

3.所有作答务必填涂或书写在答题纸上与试卷题号相对应的区域，不得错位，在试卷上作签一律不得分，4.用2B 铅笔作答选择题，用黑色字迹钢笔、水笔或圆珠笔作答非选择题。

一、填空题（本大题满分54分）本大题共有12题，考生应在答题纸相应编号的空格内直接填写结果，1-6题每题填对得4分，7-12题每题填对得5分，否则一律得零分。

1.设全集{0,1,2}U =，集合{0,1}A =，则u C A = . 【答案】{}22.某次考试，5名同学的成绩分别为：96、100、95、108、115，则这组数据的中位数为 . 【答案】1003.若函数12()f x x =，则1(1)f -= .【答案】14.若1i -是关于x 的方程20x px q ++=的一个根（其中i 为虚数单位，,p q R ∈），则p q += .【答案】05.若两个球的表面积之比为1:4，则这两个球的体积之比 . 【答案】81:6.在平面直角坐标系xOy 中，直线l 的参数方程为1x t y t=-⎧⎨=⎩（t 为参数），圆O 的参数方程为cos sin x y θθ=⎧⎨=⎩（θ为参数），则直线l 与圆O 的位置关系是 .【答案】相交7.若二项式4(12)x +展开式的第4项的值为，则23lim()nn x x x x →∞+++⋅⋅⋅+= .【答案】158.已知双曲线的渐近线方程为y x =±，且右焦点与抛物线24y x =的焦点重合，则这个双曲线的方程是 . 【答案】12222=-y x9.从(,4)m m N m *∈≥且个男生、6个女生中任选2个人发言.假设事件A 表示选出2个人性别相同，事件B 表示选出的2个人性别不同.如果事件A 和事件B 的概率相等，则m = .【答案】1010．已知函数222()log (2)2f x x a x a =+++-的零点有且只有一个，则实数a 的取值集合为 . 【答案】{}111、如图，在ABC V 中，3BACπ∠=，D 为AB 中点，P 为CD 上一点，且满足13AP t AC AB =+u u u r u u u r u u u r ，若ABC V 的面积为2，则AP u u u r 的最小值为 . .【解析】1323AP t AC AB t AC AB =+=+u u u r u u u r u u u r u u u r u u u r Q ，21133t t ∴+=∴=设||,||AC b AB C ==u u u r u u u r11sin 2222ABC S bC A bc ∆==⋅⋅=，6bc ∴= 22222c 91112os 26932AP AC AB AC AB b c π∴⎪⎛⎫⎛⎫=++⋅⋅=++⨯⨯ ⎪ ⎝⎭⎝⎭u u u r u u u r u u u r u u u r u u u r1(269)2bc ≥+=||min AP ∴=u u u r【点评】此题与2019长宁嘉定二模第10题相似在ABC V 中，已知2CD DB =u u u r u u u r，P 为线段AD 上的一点，且满足49CP mCA CB =+u u u r u u u r u u u r ，若ABC V3ACB π∠=，则CP u u u r的最小值为 .【解析】4293CP mCA CB mCA CD =+=+u u u r u u u r u u u r u u u r u u u r ，由共线定理，13m =，由ABC S =V 可得4,2,CA CB CA CB ⋅=∴⋅=u u u r u u u r222214148393927CP CA CB CA CB CA CB ⎛⎫⎛⎫⎛⎫=+=++⋅ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭u u u r u u u r u u u r u u u r u u u r u u u r u u u r14161642,392793CA CB CP ⎛⎫⎛⎫≥⋅⋅+=∴≥ ⎪ ⎪⎝⎭⎝⎭u u ur12.已知数列{}{},n n a b ，满足111a b ==，对任何正整数n均有1n n n a a b +=++，1n n n b a b +=+，设113n n n n c a b ⎛⎫=+ ⎪⎝⎭，则数列{}n c 的前2020项之和为 . 【答案】202133-【解析】()()222112n n n n n n n n a b a b a b a b ++⋅=+-+=，12n n n a b -=()1122,n n n n n n n a b a b a b ++++∴=+= 113323nn n n nn n n n n a b c b b a a ⎛⎫+∴=+==⨯ ⎪⎝⎭2020202120206133313S ⎡⎤-⎣⎦∴==--二、选择题（本大愿满分20分）本大题共有4题，每题有且只有一个正确答案，考生必须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．若x y 、满足01,0x y x y y +≥⎧⎪+≤⎨⎪≥⎩则目标函数2f x y =+的最大值为（ ）.A 1 .B 2 .C 3 .D 4【答案】.B14.如图，正方体1111A B C D ABCD -中，E F 、分别为棱1AA BC 、上的点，在平面11ADD A 内且与平面DEF 平行的直线（ ）.A 有一条 .B 有两条 .C 有无数条 .D 不存在【答案】.C15.已知函数()cos cos ,f x x x =⋅ 给出下列结论： ① ()f x 是周期函数；② 函数()f x 图像的对称中心(),0;2k k Z ππ⎛⎫+∈ ⎪⎝⎭③ 若()()12,f x f x =则()12;x x k k Z π+=∈④ 不等式sin 2sin 2cos2cos2x x x x ππππ⋅>⋅的解集为15,.88x k x k k Z ⎧⎫+<<+∈⎨⎬⎩⎭.A ①② .B ②③④ .C ①③④ .D ①②④【答案】D 16.设集合{1,2,3,,2020}S=⋯，设集合A 是集合S 的非空子集，A 中最大元素和最小元素之差称为集A 合的直径，那么集合S 所有直径为71的子集元素个数之和为（ ）7070.711949.21949.2371949.2721949x A B C D ⋅⋅⋅⋅⋅⋅【答案】C【解析】n 和71n +为最小和最大元素的子集有702个其中1,2,,70n n n +++L 每个元素出现次数是692所以n 和71n +出现次数是702,这些子集元素个数之和为69707070222372⨯+⨯=⨯ -11949n ∴≤≤，所以总的元素个数之和为702371949⋅⋅，故选C .三、解答题（本大题满分76分）本大题共有5题，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7分，第2小题满分7分.如图所示的几何体是圆柱的一部分，它是由边长为2的正方形ABCD （及其内部）AB 边所在直线为旋转轴顺时针旋转120︒得到的. （1）求此几何体的体积；（2）设P 是弧EC 上的一点，且BP BE ⊥，求异面直线FP 与CA 所成角的大小.（结果用反三角函数表示【解析】（1）因为34232212122π=⨯π⨯=θ=r S EBC 扇形． 所以，38234π=⨯π=⋅=h S V ． （2）如图所示，以点B 为坐标原点建立空间直角坐标系．则()200,,A ，()202,,F ，()020,,P ，()031,,C -．所以，()222--=,,FP ，()231--=,,AC设异面直线FP 与CA 所成的角为α=αcos 426+=所以，异面直线FP 与CA 所成角426+=αarccos【点评】考察几何体体积的计算公式，比较常规。

浦东高三数学C 答案20.061．2.2． (]2,∞-. 3． 2. 4.()Z k k k ∈⎥⎦⎤⎢⎣⎡+-432,42ππππ. 5. 1. 6.-20. 7. 12-n . 8.10. 9. 92π.10. 132y 324x 22=--.11.[]72,8-12.1t >-.13.A 14.B 15.C 16.D 17.解：(1) 解法一：如图所示，建立直角坐标系，则有关点的坐标为()001,,B ，()2011,,B ，()2111,,C ，()100,,E ，所以，()01011,,C B =，()1011--=,,E B ．… ……………………(3分)设平面E C B 11的法向量()w ,v ,u n =1， 则由111C B n ⊥且E B n 11⊥得，⎪⎩⎪⎨⎧=⋅=⋅0011111E B n C B n ⎩⎨⎧=+=⇒00w u v ， 于是平面E C B 11的一个法向量为()1011-=,,n ．………… (5分) 且()2001,,BB =，所以，点B 到平面E C B 11的距离为()2101210010222=-++⨯-⨯+⨯==d ．……………… (7分)解法二：用等体积法，参照解法一给分．(2) 因为()2111,,C ，()100,,E ，()011,,C ，所以，()2001,,CC =，()111,,CE --=………………… (10分)设平面E CC 1的法向量()w ,v ,u n =2，则由12CC n ⊥且CEn ⊥2得，⎪⎩⎪⎨⎧=⋅=⋅00212CE n CC n ⎩⎨⎧=+--=⇒002w v u w ⎩⎨⎧==+⇒00w v u ， 于是平面E CC 1的一个法向量为()0112,,n -=．……………………… (12分)设平面E C B 11的一个法向量()1011-=,,n 与平面E CC 1的一个法向量 为()0112,,n -=的夹角为ϕ，则21==ϕcos ，………………… (13分) 所以，23=ϕsin ．……………………… (14分) 所以二面角C EC B --11的正弦值为23．18. 解：（1）由三角函数的图像可知，直线y m =与正弦函数图像相交的三个相邻交点中，第一个点和第三个点之间正好一个周期 ……………………… (3分) 则5()236T πππ=--=………………………(4分) 所以2125T πω== . ………………………（6分） （2）由OA 、OB 、OC 成等差数列得2=OB OA OC + …………………（7分）在同一周期内，不妨设0B x ωϕ+=， πA x ωϕ+=，2πC x ωϕ+=…………（9分） 得π2π,,B A C x x x ϕϕϕωωω--=-==，………………………（11分） 由2=OB OA OC +，得3π22ϕϕωω-=，解得3π4ϕ=. …………………（14分）19.解：（1）有题意可知该商品的利润函数为：()()10()180=-⋅-f x Q x x ，*0100,<≤∈x x N ，……………………………（2分） 则由()*10()18000100,⎧-⋅->⎪⎨<≤∈⎪⎩Q x x x x N解得63≥x ． ………………………（5分）所以至少生产并销售63台这款产品，才能实现盈利． ………………………（6分）（2）法一：由(1)可知，当产量060<≤x ，*x N ∈时，无法实现盈利. ……………（7分）当产量60100<≤x ，*x N ∈时，由题意可知利润函数为()()10()60(60)180=-⋅---f x Q x x …………………（9分）化简得135()18160(1)18011⎡⎤=-⋅++≤-=⎢⎥+⎣⎦f x x x ， 当且仅当89=x 时等号成立 ………………………（13分）所以可以实现盈利，利润最大时，产量为89台．………………………（14分）法二：由(1)可知，当产量060<≤x ，*x N ∈时，无法实现盈利．…………………(7分) 当产量60100<≤x ，*x N ∈时，由题意可知利润函数为()()10()60(60)180=-⋅---f x Q x x ……………………(9分) 则由()*10()60(60)180060100，⎧-⋅--->⎪⎨<≤∈⎪⎩Q x x x x N解得8099<<x ，*x N ∈．……………… (12分) 所以可以实现盈利，比较(81),(82),...,(98)f f f 可知，当产量为89台时，利润最大．………………………(14分)法三：由(1)可知，当产量060<≤x ，*x N ∈时，无法实现盈利．…………………(7分) 当产量60100<≤x ，*x N ∈时，由题意可知利润函数为()()10()60(60)180=-⋅---f x Q x x ……………………(9分) 则由计算器TAB 键功能，列出60100<≤x ，*x N ∈的所有值，发现当产量为89台时，利润最大．………………………(14分)20. 解：（1）B =10.22A B A B p p A AF BF x x x x p +=+++=++= ………… (4分) （2）① 当直线设AB 的斜率存在时，设线段AB 的中点为),(00y x M ，则1212003,22x x y y x y ++===， ………… (5分) 21212221212108488AB y y y y k y y x x y y y --====-+- . ………… (7分)线段AB 的垂直平分线的方程是00(3)4y y y x -=--，即0(7)4y y x =--. ……… (9分) ② 当直线设AB 的斜率不存在时，此时线段AB 的垂直平分线的方程是0y =.所以线段AB 的垂直平分线经过一个定点()7,0C . ………… (10分)（3）设()0,m Q ，过Q 点直线方程为m ty x +=，联立088822=--⇒⎩⎨⎧+==m ty y mty x x y ， 则032642>+=∆m t ，t y y 821=+，m y y 821-=. ………… (12分) 则()()212212121y t y m x AQ +=+-=，()()222222221y t y m x BQ +=+-=，…… (13分) 所以，()()22221222111111y t y t BQ AQ +++=+ ()()()()()()164166412122222122122122122221++=+-+=++=t m m t y y t y y y y y y t y y ，………………(15分) 所以当4=m 时，1611122=+BQ AQ ，故Q 点的坐标为()0,4， 并且满足032642>+=∆m t .……………… (16分)21.解：（1）()x x g cos =是()x x f sin =的关联平方差函数，………………………(2分)()()()()()()sin sin sin cos cos sin sin cos cos sin f x y f x y x y x y x y x y x y x y +-=+-=+-()()2222222222sin cos cos sin cos cos cos cos cos cos x y x y y x y x x y =-=---()()2222cos cos y x g y g x =-=-………………………(4分)（2）()f x 是非常值函数，所以存在(),0a f a ≠，………………………(5分)下证对任意实数b ，()()f b f b -=- 令,22a b a b x y +-==可得()()2222a b a b f a f b g g -+⎛⎫⎛⎫=- ⎪ ⎪⎝⎭⎝⎭； 再令,22a b a b x y -+==可得()()22+22a b a b f a f b g g -⎛⎫⎛⎫-=- ⎪ ⎪⎝⎭⎝⎭……………(8分) 两式相加可得()()()0f a f b f b +-=⎡⎤⎣⎦，()0f a ≠，()()f b f b ∴-=-，所以()f x 为奇函数 ………………………(10分)（3）令0y =可得()()()()222201f x g g x g x =-=-，即()()221f x g x +=，………………………(11分)()()()21,220,g f f =-∴=-= ……………………(12分)令2y x =+，()()()()2222220f x f g x g x +-=+-=，………………………(13分)令2y =，()()()2221f x f x g x +-=-，………………………(14分)用2x +替换x 可得()()()()()2224121f x f x g x g x f x +=-+=-=，[1]若()0f x ≠，那么()()4f x f x =+；[2]若()0f x =，那么()()()()()2222211224f x g x g x f x f x =-=-+=+=+；所以()()40f x f x =+= ………………………(16分) 综上可知4T =满足要求，下证4T =是满足要求的最小正数，用反证法，若存在004T <<也满足要求，令00,2T x y ==可得()2200000222T T T f fg g ⎛⎫⎛⎫⎛⎫-=-< ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，而 000000,,0222222T T T T T T f f f f f f ⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫-=-=-∴-== ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭，矛盾！ 所以4T =是满足要求的最小正数 ………………………(18分)。

浦东新区2019学年度第二学期期中教学质量监测高三数学试卷一、填空题（本大题满分54分）本大题共有12题，1-6题每题4分，7-12题每题5分．考生应在答题纸相应编号的空格内直接填写结果，每个空格填对得4分或5分，否则一律得零分．1.设全集{}0,1,2U =，集合{}0,1A =，则U C A =________． 【答案】{}2 【解析】 【分析】由补集的运算法则可得解.【详解】{}{}0,1,2,0,1U A ==Q{}2U C A ∴=故答案为：{}2【点睛】本题考查了补集的运算，属于基础题.2.某次考试，5名同学的成绩分别为：96,100,95,108,115，则这组数据的中位数为___． 【答案】100 【解析】 【分析】数据个数为奇数时，中位数为从小到大排列后中间的那一个数字. 【详解】5名同学的成绩由小到大排序为：95,96,100,108,115，∴这组数据的中位数为100.故答案为：100【点睛】本题考查了一组数据中中位数的求法，属于基础题. 3.若函数()12f x x =，则()11f -=__________．【答案】1 【解析】 【分析】由()12f x x=可得：()12,0f x x x -=≥，问题得解.【详解】由()12f x x =可得：()12,0f x x x -=≥()12111f -∴==故答案为：1【点睛】本题考查了反函数的求法，属于基础题.4.若1i -是关于x 的方程20x px q ++=的一个根（其中i 为虚数单位，,p q R ∈），则p q +=__________． 【答案】0 【解析】 【分析】直接利用实系数一元二次方程的虚根成对原理及根与系数关系求解.【详解】1i -Q 是关于x 的实系数方程20x px q ++=的一个根，1i ∴+是关于x 的实系数方程20x px q ++=的另一个根，则(1)(1)2p i i -=-++=，即2p =-，2(1)(1)12q i i i =-+=-=，0p q ∴+=.故答案为：0【点睛】本题考查了一元二次方程的虚根特征和虚数的运算,考查了计算能力，属于中档题. 5.若两个球的表面积之比为1:4，则这两个球的体积之比为 ． 【答案】1:8 【解析】试题分析：由求得表面积公式24S R π=得半径比为1:2，由体积公式343V R π=可知体积比为1:8 考点：球体的表面积体积6.在平面直角坐标系xOy 中，直线l 的参数方程为1x t y t =-⎧⎨=⎩（t 为参数），圆O 的参数方程为cos sin x y θθ=⎧⎨=⎩（θ为参数），则直线l 与圆O 的位置关系是________． 【答案】相交 【解析】【分析】由已知可得：直线l 的标准方程为10x y -+=，圆O 的标准方程为221x y +=，再计算出圆心到直线的距离d r =<，问题得解. 【详解】由直线l 的参数方程1x t y t =-⎧⎨=⎩，可得：直线l 的标准方程为：10x y -+=，由圆O 的参数方程cos sin x y θθ=⎧⎨=⎩，可得：圆O 的标准方程为：221x y +=，圆心为(0,0)，半径1r = 圆心为(0,0)到直线l的距离12d ==< 则直线l 与圆O 的位置关系是相交. 故答案为：相交【点睛】本题考查了参数方程与普通方程的转化，考查了直线与圆的位置关系，属于中档题. 7.若二项式()412x +展开式的第4项的值为，则()23lim nn x x x x →∞++++=L __．【答案】15【解析】 【分析】利用二项展开式的通项公式，得：3344(2)x T C ==，解得16x =，再由等比数列求和公式，得：2311156nnx x x x ⎡⎤⎛⎫=⨯-++⎢⎥ ⎪⎝⎭⎢⎥⎣+⎦+L ，从而极限可求.【详解】由已知可得：3344(2)x T C ==，即33(2)2x x ==16x =， 2311166(1)111115616nn n nx x x x x x x ⎡⎤⎛⎫-⎢⎥ ⎪∴+++⎡⎤⎝⎭-⎢⎥⎛⎫⎣⎦===⨯-⎢⎥ ⎪-⎝⎭⎢⎥⎣⎦-+L ，()231111565lim lim nnn n x x x x →∞→∞+++⎡⎤⎛⎫∴⨯-=⎢⎥ ⎪⎝⎭⎢⎥⎣⎦+=L . 故答案为：15【点睛】本题考查了二项式定理，等比数列求和公式以及求极限，考查了计算能力，属于中档题. 8.已知双曲线的渐近线方程为y x =±，且右焦点与抛物线24y x =的焦点重合，则这个双曲线的方程是____________. 【答案】22221x y -= 【解析】 【分析】由已知可得双曲线的右焦点为(1,0)，即1c =，由双曲线的渐近线方程为y x =±，可设其方程为：22,0x y λλ-=>，再由222+=a b c 可得：1λλ+=，求出λ，问题得解.【详解】Q 抛物线24y x =的焦点为：(1,0)∴双曲线的右焦点为：(1,0)，即1c = Q 双曲线的渐近线方程为y x =±，∴双曲线的方程可设为：22,0x y λλ-=>，即221x y λλ-=，22a b λ∴==由222+=a b c 可得：1λλ+=，12λ∴=， 双曲线的方程是22221x y -=. 故答案为：22221x y -=【点睛】本题考查了双曲线的标准方程和其渐近线方程，关键是掌握共渐近线的曲双线方程的设法，属于中档题.9.从m （N m *∈且4m ≥）个男生、6个女生中任选2个人当发言人，假设事件A 表示选出2个人性别相同，事件B 表示选出的2个人性别不同．如果A 的概率和B 的概率相等，则m =_____________． 【答案】10 【解析】【分析】从m 个男生、6个女生中任选2个人当发言人,共有26m C +种情况，事件A 表示选出的2个人性别相同，共有226m C C +情况，事件B 表示选出的2个人性别不同，共有116m C C情况，由已知可得：2211662266m m m m C C C C C C +++=，即221166m m C C C C +=，解之即可.【详解】从m 个男生、6个女生中任选2个人当发言人,共有26m C +种情况，事件A 表示选出的2个人性别相同，共有226m C C +情况， 事件B 表示选出的2个人性别不同，116m C C 情况()()P A P B =Q ，2211662266m m m m C C C C C C +++∴= 221166m m C C C C ∴+=，即(1)65622m m m -⨯+= 整理，得：213300m m -+=，即(3)(10)0m m --=N m *∈Q 且4m ≥，10m ∴=故答案为：10【点睛】本题考查了概率计算和组合数及其计算，考查了计算能力和分析能力，属于中档题.10.已知函数()()222log 22f x x a x a =+++-的零点有且只有一个，则实数a 的取值集合为________．【答案】{}1 【解析】 【分析】由已知可得：()f x 为R 上的偶函数，又函数()f x 的有且只有一个零点，所以()00f =，由此可得：2log 220a a +-=，解得1a =【详解】显然，由()()222log 22f x x a x a =+++-，可得：()()f x f x =-， ()f x \为R 上的偶函数.函数()f x 的有且只有一个零点， ()0=0f ∴ 由此可得：2log 220a a +-=，解得1a =故答案为：{}1【点睛】本题考查了偶函数的对称性，属于中档题.11.如图，在ABC V 中，3BAC π∠=，D 为AB 中点，P 为CD 上一点，且满足13t AC AB AP =+u u u r u u u r u u u r，若ABC V 的面积为332，则AP u u u r 的最小值为__________.2 【解析】 【分析】设,AB AC m n ==u u u r u u u r ，由133sin 2BA AB A C C ⋅⋅∠=u u u r u u u r ，可得：6mn =再由1233t AC AB t AC A AP D =++=u u u r u u u r u u u r u u u r u u u r ，可得：13t =，则2221123393m n AP AC AB +⎛⎫=+=+ ⎪⎝⎭u u u r u u u r u u u r 后由222m n mn +≥可得解.【详解】设,AB AC m n ==u u u r u u u rABC QV 33， 1sin 2AB AC S BAC =⋅⋅∠u u u r u u u r 13332mn ==6mn ∴=Q D 为AB 中点，2AB AD ∴=u u u r u u u r1233t AC AB t AC AD AP +==+∴u u u r u u u u r u u r u u u r u u u r又C 、P 、Q 三点共线，213t ∴+=，即13t =1133AP AC AB ∴=+u u u r u u u r u u u r则()2222911112=3399APAC AB AC AB AC AB ⎛⎫=+++⋅ ⎪⎝⎭u u u r u u ur u u u r u u u r u u u r u u u r u u u r22112=cos 999AC AB AC AB BAC ++⋅⋅∠u u ur u u u r u u u r u u u r 222211212=992993m n m n m n +++⋅⋅=+AP ∴=≥=u u u r当且仅当m n ==时取得最小值.【点睛】本题考查了向量的模的运算和数量积运算及三角形的面积公式，考查了计算能力，属于中档题.12.已知数列{}{},n n a b 满足111a b ==，对任何正整数n均有1n n n a a b +=+1n n n b a b +=+，设113nn n n c a b ⎛⎫=+⎪⎝⎭，则数列{}n c 的前2020项之和为_____________. 【答案】202133- 【解析】 【分析】由已知得：()112+n n n n a b a b +++=，2,n n n a b n N *∴+=∈；11n n a b ++=2n n a b ，12,n n n a b n N -*∴=∈，由此可得：12333n n n n c +=⋅=-，再由等比数列求和公式可得解.【详解】1n n n a a b +=++Q ①，1n n n b a b +=+②两式相加可得：()112+n n n n n n n n a b a b a b a b ++++=+=，{}n n a b ∴+是公比为2的等比数列，首项112a b +=2,n n n a b n N *∴+=∈两式相乘可得：(11n n n n nna b a b a b ++=++()22n n n n a b a b =+=-{}n n a b ∴是公比为2的等比数列，首项111a b =113323n n n n n n n n n n a bc a b a b ⎛⎫+=+=⋅=⋅ ⎪⎝⎭，由等比数列求和公式，得：()2020202120206133313S -==--故答案为：202133-【点睛】本题考查了等比数列的通项公式和求和公式，考查了转化能力和计算能力，属于中档题.二、选择题（本大题满分20分）本大题共有4题，每题有且只有一个正确答案．考生必须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.若x 、y 满足 010x y x y y -≥⎧⎪+≤⎨⎪≥⎩， 则目标函数2z x y =+的最大值为（ ）A. 1B. 2C. 3D. 4【答案】B 【解析】 【分析】作出可行域和目标函数，找到目标函数取最大值的最优解即可. 【详解】由已知，可作出满足条件的可行域和目标函数如下：由图可知目标函数2y x z =-+中z 取最大值的最优解为：(1,0)故选：B【点睛】本题考查了线性规划求线性目标函数的最值问题，考查了数形结合思想，属于中档题.14.如图，正方体1111A B C D ABCD -中，E 、F 分别为棱1A A 、BC 上的点，在平面11ADD A 内且与平面DEF 平行的直线（ ）A. 有一条B. 有二条C. 有无数条D. 不存在【答案】C 【解析】 【分析】易知当//l DE 时即可满足要求，所以存在无数条. 【详解】若l ∃⊂平面11ADD A ，使得//l DE ， 又DE ⊂平面DEF ，l ⊄平面DEF ，//l ∴平面DEF ，显然满足要求的直线l 有无数条. 故选：C【点睛】本题考查了线面平行的判定，属于基础题. 15.已知函数()cos cos f x x x =⋅.给出下列结论： ①()f x 是周期函数；② 函数()f x 图像的对称中心+,0)()2（ππ∈k k Z ；③ 若()()12f x f x =，则()12x x k k Z π+=∈；④不等式sin 2sin 2cos2cos2x x x x ππππ⋅>⋅的解集为15,88x k x k k Z ⎧⎫+<<+∈⎨⎬⎩⎭.则正确结论的序号是（ ）A. ①②B. ②③④C. ①③④D. ①②④【答案】D 【解析】 【分析】由()()2f x f x π+=，可知()f x 是周期为2π的函数， 当22x ππ-≤≤时，()11cos 222f x x =+；当322x ππ<≤时，()11cos 222f x x =--，画出()f x 在一个周期3,22ππ⎛⎫- ⎪⎝⎭内的函数图象，通过图象去研究问题.【详解】()()()()2cos 2cos 2cos cos f x x x x x f x πππ+=+⋅+=⋅=()f x ∴是周期为2π的函数，①正确；当22x ππ-≤≤时，cos 0x ≥，()211cos cos 222f x x x ==+当322x ππ<≤时，cos 0x <，()211cos cos 222f x x x =-=--可以画出()f x 在一个周期3,22ππ⎛⎫- ⎪⎝⎭内的函数图象，如下由图可知：函数()f x 的对称中心为+,0)()2（ππ∈k k Z ，②正确；函数()f x 的对称轴为,x k k Z π=∈ 若()()12f x f x =，则122x x k π+=，即()122x x k k Z π+=∈，③错误； sin 2sin 2cos 2cos 2cos 2cos 22222x x x x x x ππππππππππ⎛⎫⎛⎫⎛⎫⎛⎫⋅=-⋅-=-⋅- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭不等式sin 2sin 2cos2cos2x x x x ππππ⋅>⋅等价于：()222f x f x πππ⎛⎫-> ⎪⎝⎭由图可知：52+2,+2,44x k k k Z πππππ⎛⎫∈∈⎪⎝⎭解得15,,88x k k k Z ⎛⎫∈++∈ ⎪⎝⎭，④正确. 故选：D.【点睛】本题考查了诱导公式，降幂公式及三角函数的性质，考查了数形结合思想，属于难题.16.设集合{}1,2,3,...,2020S =，设集合A 是集合S 的非空子集，A 中的最大元素和最小元素之差称为集合A 的直径. 那么集合S 所有直径为71的子集的元素个数之和为（ ）A. 711949⋅B. 7021949⋅C. 702371949⋅⋅D. 702721949⋅⋅【答案】C 【解析】 【分析】先考虑最小元素为1，最大元素为72的情况：{}1,72只有1种情况；{}1,,72,271a a ≤≤且a Z ∈，共有170C 种情况；{}1,,,72,2,71b c b c ≤≤且,b c Z ∈，共有种270C 情况；以此类推……{}1,2,3,,71,72L ，有1（7070C ）种情况.所以，此类满足要求的子集元素个数之和012697070707070702347172M C C C C C =+++++L ，计算可得：70372M =⨯.再思考可以分为{}{}{}{}{}1,,72,2,,73,3,,74,4,,75,1949,,2020L L L L L L 等1949类，问题可得解.【详解】当最小元素为1，最大元素为72时，集合有如下情况： 集合只含2个元素：{}1,72只有1种情况；集合含有3个元素：{}1,,72,271a a ≤≤且a Z ∈，共有170C 种情况；集合含有4个元素：{}1,,,72,2,71b c b c ≤≤且,b c Z ∈，共有270C 种情况；以此类推……集合含有72个元素：{}1,2,3,,71,72L ，有（7070C ）种情况.所以，此类满足要求的子集元素个数之和M 为：012697070707070702347172,M C C C C C =+++++L L ① 70696810707070707072717032,M C C C C C ∴=+++++L L ② 707070,070,r r C C r r Z -=≤≤∈Q①②两式对应项相加，得：()0126970707070707070274742M C C C C C =+++++=⨯L70372M ∴=⨯同理可得：{}{}{}{}2,,73,3,,74,4,,75,1949,,2020,L L L L L 所有子集元素个数之和都是70372⨯，所以集合S 所有直径为71的子集的元素个数之和为702371949⋅⋅. 故选：C【点睛】本题考查了集合的子集个数和组合数及其计算，考查了分类讨论思想，属于难题.三、解答题（本大题满分76分）本大题共有5题，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.如图所示的几何体是圆柱的一部分，它是由边长为2的正方形ABCD (及其内部)以AB 边所在直线为旋转轴顺时针旋转120o 得到的．（1）求此几何体的体积；（2）设P 是弧EC 上的一点，且BP BE ⊥，求异面直线FP 与CA 所成角的大小．（结果用反三角函数值表示） 【答案】（1）83π（2）62arccos 4【解析】 【分析】（1）先算底面积212EBC S r θ=扇形，再由V S h =⋅算出体积； （2）以点B 为坐标原点建立空间直角坐标系，用空间向量法算出cos FP AC FP ACα⋅=⋅u u u r u u u r u u ur u u u r ，即可得解. 【详解】（1）由已知可得：22112422233EBC S r ππθ==⨯⨯=扇形．48233V S h ππ∴=⋅=⨯=．（2）如图所示，以点B 为坐标原点建立空间直角坐标系B xyz -，则()0,0,2A ，()2,0,2F ，()0,2,0P ，()3C -，所以，()2,2,2FP =--u u u r，()32AC =--u u u r .设异面直线FP 与CA 所成的角为α，则cos FP ACFP ACα⋅=⋅u u u r u u u r u u u r u u u r()()()()()()()()()222222212322222132-⨯-+⨯+-⨯-=-++-⋅-++-624=所以，异面直线FP 与CA 所成角为62arccos4α=. 【点睛】本题考查了柱体体积计算和空间向量法计算异面直线的夹角，考查了计算能力，属于中档题.18.已知锐角αβ、的顶点与坐标原点重合，始边与x 轴正方向重合，终边与单位圆分别交于P 、Q 两点，若P 、Q 31025、（1）求()cos αβ+的大小；（2） 在ABC ∆中，a b c 、、为三个内角、、A B C 对应的边长，若已知角C αβ=+，3tan 4A =，且22a bc c λ=+，求λ的值．【答案】（1）2（2）1=2λ-【解析】 【分析】（1）由已知得：cos αβ==，故而sin α=，sin β=，再由cos(+)cos cos sin sin αβαβαβ=-可得解.（2）由（1）得：4C παβ=+=，所以cos C C =，由3tan 4A =可得34sin ,cos 55A A ==，再由sin sin()B A C =+可得sin B =，最后由正弦定理可得：2222sin sin =sin sin a c A C bc B C λ--=，问题得解.【详解】（1）由三角函数定义，得：cos αβ==αβQ 、为锐角，sin α∴==，sin β== cos(+)cos cos sin sin αβαβαβ∴=-2-=（2）由cos(+)2αβ=，αβQ 、为锐角，得：4C παβ=+=，cos C C ∴=由3tan 4A =，得sin 3cos 4A A =，又22sin cos 1A A +=， 解得34sin ,cos 55A A == []sin sin ()sin()B A C A C π=-+=+sin cos cos sin A C A C =+3455=+=由正弦定理可得：222291sin sin 1=sin sin 5a c A C bc B C λ---===- 【点睛】本题考查了三家函数定义及正余弦和的展开公式，考查了正弦定理边化角的技巧，考查了计算能力，属于中档题.19.疫情后，为了支持企业复工复产，某地政府决定向当地企业发放补助款，其中对纳税额在3万元至6万元（包括3万元和6万元）的小微企业做统一方案．方案要求同时具备下列两个条件：①补助款()f x （万元）随企业原纳税额x （万元）的增加而增加；②补助款不低于原纳税额x （万元）的50%．经测算政府决定采用函数模型()44x bf x x=-+（其中b 为参数）作为补助款发放方案． （1）判断使用参数12b =是否满足条件，并说明理由； （2）求同时满足条件①、②的参数b 的取值范围． 【答案】（1）当12b =时不满足条件②，见解析（2）939,44⎡⎤-⎢⎥⎣⎦【解析】 【分析】（1）因为当12b =时，()33342f =<，所以不满足条件② ； （2）求导得：()2221444b x bf x x x+'=+=，当0b ≥时，满足条件①；当0b <时，()f x在)⎡+∞⎣上单调递增，所以3≤.由条件②可知，()2x f x ≥，即44x bx+≤，等价于()2211481644b x x x ≤-+=--+在[]3,6上恒成立，问题得解.【详解】（1）因为当12b =时，()33342f =<，所以当12b =时不满足条件② .（2）由条件①可知，()44x bf x x=-+在[]3,6上单调递增，()2221444b x bf x x x+'=+=Q 所以当0b ≥时，()0f x ¢³满足条件；当0b <时，由()0f x ¢=可得x =当)x ⎡∈+∞⎣时()0f x ¢³，()f x 单调递增，3∴≤，解得904b -≤<， 所以94b ≥-由条件②可知，()2x f x ≥，即不等式44x bx+≤在[]3,6上恒成立， 等价于()2211481644b x x x ≤-+=--+当3x =时，()218164y x =--+取最小值394394b ∴≤综上，参数b 的取值范围是939,44⎡⎤-⎢⎥⎣⎦． 【点睛】本题考查了导数求函数单调性以及恒成立问题，考查了转化思想，属于中档题.20.在平面直角坐标系xOy 中，1F ，2F 分别是椭圆()222 10x y a aΓ+=>：的左、右焦点，直线l 与椭圆交于不同的两点A 、B ，且12AF AF +=（1）求椭圆Γ的方程；（2）已知直线l 经过椭圆的右焦点2F ，,P Q 是椭圆上两点，四边形ABPQ 是菱形，求直线l 的方程； （3）已知直线l 不经过椭圆的右焦点2F ，直线2AF ，l ，2BF 的斜率依次成等差数列，求直线l 在y 轴上截距的取值范围.【答案】（1）2212x y +=（20y ±-=（3）(,)-∞+∞U【解析】 【分析】（1）由已知得：2a =，问题得解；（2）由已知可得：OA OB ⊥，设直线l 方程为：1x my -=，()11,A x y ,()22,B x y ，与椭圆方程2212x y +=联立可得：22(2)210m y my ++-=，由韦达定理，得：12222m y y m +=-+，12212y y m =-+，最后由0OA OB ⋅=u u u r u u u r ，可得：1212x x y y +21212(1)()10m y y m y y =++++=，代入解方程即可；（3）设直线l 方程为：y kx b =+，由已知可得：1212211y y k x x +=--，即1212211kx b kx bk x x +++=--，化简得：12()(2)0b k x x ++-=，有已知可得:122x x +=，联立直线与椭圆方程得：222(21)4(22)0k x kbx b +++-=，由228(21)0k b ∆=-+>，和1224221kbx x k +=-=+可求b 的取值范围.【详解】（1）由12+AF AF =2a =，从而a =2212x y +=.（2）由于四边形ABPQ 是菱形，因此//AB PQ 且||||AB PQ =. 由对称性，1F 在线段PQ 上. 因此，,AP BQ 分别关于原点对称； 并且由于菱形的对角线相互垂直，可得AP BQ ⊥，即OA OB ⊥. 设直线l 方程为：1x my -=，且()11,A x y ,()22,B x y与椭圆方程2212x y +=联立可得：22(2)210m y my ++-=，12222m y y m ∴+=-+，12212y y m =-+， 由0OA OB ⋅=u u u r u u u r，可得：12121212(1)(1)x x y y my my y y +=+++ 21212(1)()1m y y m y y =++++2222121022m m m m +=--+=++解得m =0y ±=. （3）设直线l 方程为：y kx b =+，()()()11222,,,,1,0A x y B x y F Q ，由已知可得： 1212211y y k x x +=--，即1212211kx b kx bk x x +++=--. 1212122()()22(1)(1)kx x b k x x b k x x ∴+-+-=--，化简得：12()(2)0b k x x ++-=.若0b k +=，则:l y kx k =-经过2F ，不符合条件， 因此122x x +=.联立直线与椭圆方程得：222(21)4(22)0k x kbx b +++-=. 因为228(21)0k b ∆=-+>，即22210k b -+>L ①由1224221kb x x k +=-=+得：2212k b k+=-L ②将②代入①得：222212102k k k ⎛⎫+-+> ⎪⎝⎭，解得：212k >令()12f k k k =--，则()222112122k f k k k-'=-+= 当212k >时，()0f k '<，()12f k k k∴=--,2⎛-∞- ⎝⎭或2⎛⎫+∞⎪ ⎪⎝⎭上单调递减，()2f k f ⎛∴>-= ⎝⎭()2f k f ⎛⎫<= ⎪ ⎪⎝⎭所以b 的取值范围为：(,)-∞+∞U .【点睛】本题考查了椭圆与直线的综合性问题，关键是联立方程组，用韦达定理进行求解，考查了分析能力和计算能力，属于难题.21.若数列{}n a 对任意连续三项12,,i i i a a a ++，均有()()2210i i i i a a a a +++-->，则称该数列为“跳跃数列”. （1）判断下列两个数列是否是跳跃数列： ①等差数列：1,2,3,4,5,L ； ②等比数列：11111,,,,24816--L ； （2）若数列{}n a 满足对任何正整数n ，均有11na n a a +=()10a >.证明：数列{}n a 是跳跃数列的充分必要条件是101a <<.（3）跳跃数列{}n a 满足对任意正整数n 均有21195nn a a +-=，求首项1a 的取值范围.【答案】（1）① 等差数列：1,2,3,4,5,...不是跳跃数列；② 等比数列：11111,,,,, (24816)--是跳跃数列.（2）证明见解析（3）()(12,2a ∈-U【解析】 【分析】（1）①数列通项公式为n a n =，计算可得：()()22120i i i i a a a a +++--=-<，所以它不是跳跃数列；②数列通项公式为：112n n a -⎛⎫=- ⎪⎝⎭，计算可得：()()222191042ii i i i a a a a +++⎛⎫--=⨯-> ⎪⎝⎭，所以它是跳跃数列；（2）必要性：若11a >，则{}n a 是单调递增数列，若11a =，{}n a 是常数列，均不是跳跃数列；充分性：用数学归纳法证明证明，1n =命题成立，若n k =时2121222221,k k k k k k a a a a a a -+++<<>>，可得：222423k k k a a a +++>>，所以当1n k =+时命题也成立；（3）有已知可得：21n n a a ++-()()221519195125n n n n a a a a =----，2n n a a +-()()()2123195125n n n n a a a a =----，若1n n a a +>，则12n n n a a a ++>>，解得522n a ⎛⎫-∈ ⎪ ⎪⎝⎭；若1n n a a +<，则12n n n a a a ++<<，解得53,2n a ⎛+∈ ⎝⎭，由2n a ⎫∈⎪⎪⎝⎭，则1n a +⎛∈ ⎝⎭，得()2,2n a ∈-；当n a ⎛∈ ⎝⎭，则()12,2n a +∈-，得(n a ∈，问题得解.【详解】（1）①等差数列：1,2,3,4,5,L 通项公式为：n a n =()()[][]221(2)2(1)20i i i i a a a a i i i i +++--=-++-+=-<Q所以此数列不是跳跃数列；②等比数列：11111,,,,,24816--L 通项公式为：112n n a -⎛⎫=- ⎪⎝⎭()()11122211111910222242i i i i ii i i i a a a a -+++++⎡⎤⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫⎛⎫--=------=⨯->⎢⎥⎢⎥ ⎪ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎝⎭⎢⎥⎢⎥⎣⎦⎣⎦Q所以此数列是跳跃数列 （2）必要性：若11a >，则{}n a 是单调递增数列，不是跳跃数列； 若11a =，{}n a 是常数列，不是跳跃数列. 充分性：（下面用数学归纳法证明）若101a <<，则对任何正整数n ，均有2121222221,n n n n n n a a a a a a -+++<<>>成立.①当1n =时，112111a a a a a =>=， 213112a a a a a a =<=，1212131111,a a a a a a a a =<∴=>=Q ，231a a a ∴>> 321231111342,,a a a a a a a a a a a a >>∴<<<<Q ，所以1n =命题成立②若n k =时，2121222221,k k k k k k a a a a a a -+++<<>>，则22221212322,kk k a a a k k k aa a a a a +++++<<∴<<，212322222423,k k k a a a k k k a a a a a a ++++++>>∴>>，所以当1n k =+时命题也成立，根据数学归纳法，可知命题成立，数列满足()()2210i i i i a a a a +++-->， 故{}n a 是跳跃数列.（3）21195n n a a +-=Q()222212191919251919555125n n n n a a a a ++-⎛⎫- ⎪⨯---⎝⎭∴===()22221192519191255n n n n a a a a ++⨯---∴-=-()()221519195125n n n n a a a a =----()222192519125nn n n a a a a +⨯---=-()()()2123195125n n n n a a a a =---- ①若1n n a a +>，则12n n n a a a ++>>，()()()()()222151919501251231950125n n n n n n n n a a a a a a a a ⎧----<⎪⎪∴⎨⎪---->⎪⎩解得2n a ⎫∈⎪⎪⎝⎭； ②若1n n a a +<，则12n n n a a a ++<<，()()()()()222151919501251231950125n n n n n n n n a a a a a a a a ⎧---->⎪⎪∴⎨⎪----<⎪⎩解得53,2n a ⎛+∈ ⎝⎭；若522n a ⎛⎫-∈ ⎪ ⎪⎝⎭，则211953,52n n a a +⎛⎫-=∈ ⎪ ⎪⎝⎭，所以()2,2n a ∈-，若n a ⎛∈ ⎝⎭，则()21192,25n n a a +-=∈-，所以(n a ∈， 所以()(12,2a ∈-U ，此时对任何正整数n ，均有()(2,2n a ∈-U【点睛】本题考查了与数列相关的不等式证明，考查了数学归纳法，考查了分类与整合思想，属于难题.。

浦东新区2019 学年度第二学期高中教学质量检测试题高三英语2020.05I.Listening ComprehensionSection ADirections: 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. At the butcher's. B. In a restaurant. C. On the farm. D. In a supermarket.2. A. Boss and secretary. B. Operator and caller.C. Librarian and student.D. Customer and repairman.3. A. He must attend a class. B. He must meet his teacher.C. He must finish his homework.D. He must go out with his roommate.4. A. It’s not as good as it was. B. It’s better than it used to be.C. It’s better than people expect.D. It’s even worse than people say.5. A. The woman has a practical goal.B.The woman can surely live a long life.C.The woman has taken the right steps to stay healthy.D.The woman should give up cheeseburgers to live longer.6. A. An attractive hut. B. A sunny day. C. raincoat. D. A lovely hat.7. A. He's not going to cook his own dinner. B. He plans to do the same as his brother.C. He loves the dinner his brother cooks.D. He wants to take on his own responsibility.8. A. Applying to Harvard will be fun. B. He is confident of getting into Harvard.C. He has no choice but to apply to Harvard.D. The woman can get the man into Harvard.9. A. The woman is teaching the man how to cook.B.There is nothing the man can do to cook the dish.C.The cookbook contains difficult instructions to follow.D.The man is good at following what is said in the cookbook.10. A. The woman is too busy to go to the dinner.B.The woman will definitely go to the dinner.C.The woman will probably decline the invitation.D.The woman is asking about the time for the dinner.Section BDirections: In Section B, you will hear two short passages and one longer conversation, and you will be asked several questions on each of the passages and the conversation. The passages and the conversation 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. Boston Cooking School. B. Toll House Inn.C. A chocolate company.D. Nestle's branch.12. A. Mix smashed chocolate with other ingredients and baked it.B.Cover the surface of the cookies with melted chocolate.C.Spread butter on semi-sweet chocolate desserts.D.Shape melted chocolate into thick pieces.13. A. She kept it as a secret. B. She sold it to Nestle.C. She applied for a patent.D. She shared it publicly.Questions 14 through 16 are based on the following passage.14. A. They support various living creatures. B.They reduce greenhouse gas emissions.C.They bring about huge economic benefits.D.They protect the coast against melting ice.15. A. Australia. B. Canada. C. America. D. China.16. A. Tourism will face strong decline.B.Beach losses are causing climate change.C.Half of the world's sandy beaches could disappear.D.Beaches play an important part in the ecosystem.Questions 17 through 20 are based on the following conversation.17. A. London. B. Barcelona. C. Madrid. D. Iceland.18. A. She was scheduled to meet more customers in other cities.B.Her ship was delayed by the wind blowing southwards.C.Clouds of volcano ash threatened passengers’ health.D.Volcanic eruption caused her flight to be cancelled.19. A. She tried various means of transport except the coach.B.She had a tough journey back home with many transfers.C.She enjoyed the lovely scenery in various cities in Spain.D.She managed to book a ticket with the British airline at last.20. A. He paid little attention to the news media.B.He didn’t care about meaningless pastimes.C.He was out of employment for too long.D.He was too busy to make preparations for it.II.Grammar and VocabularySection 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.Green Spring Renews Life’s PromiseFor me, two of the loveliest words in th e English language are “Life persists”. I came across them years ago as a college freshman, sitting in the library on a beautiful spring day, bored, working on a history paper, I don’t recall (21) what I was researching into. Out of nowhere, those two words came (22) dancing (dance) off the page in a quote by Gandhi, “In the midst of death life persists, in the midst of untruth truth persists, in the midst of darkness light persists.”After those words (23) were read (read) again a dozen times, suddenly I was no longer bored. Outside in the sunshine, I kicked off my shoes and danced barefoot across a spring-green lawn.I love spring. And this year, I was especially hungry to see it. Flying home last weekend to Las Vegas, after 10 days in California, I looked down on hills that were so green that I (24) could almost taste them. When I approached Vegas, the green turned a dull desert brown. We landed after sunset, and the only green to be seen was neon（霓虹灯）.But the next morning, to my surprise, I (25) was awake (awake) to find signs of spring all over my yard. (26)With my absence, all sorts of things had leafed and bloomed. Three days later, I drove to Arizona to visit a friend and get yet another taste of spring seeing the Giants play the A’s in spring training. The drive across the desert was completely great, a variety of wildflowers and blooming cactuses.Sometimes we need the chance (27) to remind (remind) that we’re still alive. After my husband died, a friend sent me a card which read: “Just (28) you think you will never smile again, life comes back.”Life persists, and so do (29) we in the green of spring and the dead of winter, in the birth of a child and the passing of a loved one; in the words we leave behind and the hearts of those (30) who will remember us. Spring reminds us that we’re alive forever.Section BDirections: After reading the passage below, fill in each blank with a proper word given in the box. Each word can be used only once. Note that there is one more word than you need.Curiosity and Globalization are Driving a New Approach to TravelToday’s political climate and negative headlines seem to point towards a more inward-looking global population —minds narrowing, borders going up. But with more people living and working overseas and becoming exposed to influences from different cultures, many of us are seeking a(n) 31 , connected world.According to the recently published study from Culture Trip, 60% of people in the US and UK say that their outlook on life is shaped by the 32 from different cultures. As a society, we not only want to discover and experience other cultures, we want to learn from them, too. This is one of the many positive side effects of globalization. At the same time, the economic landscape of the last decade has resulted in a shift in values away from 33 , with younger generations more interested in collecting experiences than possessions.Welcome to the “new culture economy”.The collision（碰撞）of two trends — globalization and the experience economy — has 34 a new attitude to travel, with cultural curiosity at its heart. This is the “new culture economy”. The phenomenon is having a powerful impact on people’s interactions and definitions of 35 exploration, and presents an incredible commercial opportunity.While globalization is usually talked about in the context of the 36 of trade and capital between countries, we shouldn’t forget that the37 force behind it all it people. Education, travel, exposure to other customs and geographies and the cultural integration ( 融合)are the more influential social effects of globalization. People are increasingly living or working in countries other than the ones in which they were born - more than half of respondents from the study have friends living overseas, all of which has 38 in more interaction with global cultures.Also, student debt and unaffordable housing have created a(n) 39 in spending patterns, and so a new set of values has emerged in which experiences matter more than ownership. Travel is absolutely necessary to most people’s lives - in fact, nearly half of all respondents cut down on their daily expenses so they can save money to travel more. For “generation rent” in particular, no matter how expensive an experience or a trip, it is still more 40 than a house.III.Reading Comprehension Section ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Communication, One Major Part of the Scientific MethodScientists may feel it their duty to share their guesses, methods, and findings with the rest of the scientific community .This sharing serves two 41 . First, it supports the basic deal of skepticism （怀疑论）by making it possible for others to say, “Oh, yeah? Let me check that.” It tells others where to see what the scientist saw, and what techniques and tools to use. Second, it gets the word out so that others can use what has been discovered. This is essential because science is a(n) 42 efforts. People who work thousands of miles apart build with and upon each other’s discoveries.The communication of science begins with “peer review”, a process of43 an author’s scholarly work, research or ideas to the inspection of other experts. It typically has three stages. The first occurs when a scientists seeks funding —— from government agencies, foundations, or other 44 —— to carry out a research program. He or she must prepare a report describing the intended work,laying out background,hypothese（s假设）,planned experiments, expected results, and even the 45 impacts on other fields. Committees of other scientists then 46 the report to see whether the scientist knows his or her area, has the necessary abilities, and is realistic in his or her plans.Once the scientist has the needed 47 , has done the work, and has written a report of the results, that reportswill go to a scientific journal. Before publishing the report, the journal’s editors will show it to other workers in the same or 48 fields and ask whether the work was done adequately, the conclusion are justified, and the report should be published.The third stage of peer review happens are publication, when the broader scientific community gets to see and49 the work.This three-stage quality-control process can, of course, be faulty. Any scientist with independent wealth can50 the first stage quite easily but such scientists are much, much rarer today than they were a century or so ago. Those who remain are the object of envy. 51 , it is fair to say that they are not disapproved as were those who avoid the latter two stages of the “peer review” mechanisms by using press conferences.On the other hand, it is certainly possible for the standard peer review mechanisms to 52 . By their nature, these mechanisms are more likely to 53 ideas that are not different from what the reviewers think they already know. Yet the un-traditional or unconventional ideas are not 54 wrong, as Alfred Wegener proved when he tried to gain 55 for the idea of continental drift in the early twentieth century. It was not until the 1960s that most geologists accepted his ideas as genuine insights.41. A. purposes B. duties C. interests D. needs42. A. innovative B. prospective C. cooperative D. plain43. A. accustoming B. addicting C. restricting D. subjecting44. A. projects B. sources C. unions D. departments45. A. stronger B. more limited C. more dramatic D. broader46. A. look up B. go over C. long for D. call for47. A. funds B. fields C. impacts D. experiments48. A. different B. chosen C. related D. academic49. A. substitute B. create C. judge D. undertake50. A. reach B. mark C. hold D. skip51. A. Similarly B. Contrarily C. Surely D. Therefore52. A. fail B. function C. evolve D. work53. A. convey B. overlook C. reject D. approve54. A. necessarily B. particularly C. dramatically D. terribly55. A. confidence B. acceptance C. strength D. weightSection BDirections: Read the following three passage. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have read.(A)To Be a Deaf DJI was born in England with perfect hearing. In 1990, when I was five, my family moved to the United States. I started getting ear infections every three months or so. We didn’t have h ealth insurance at the time, and when I got a third infection, my parents couldn’t afford the treatment. I went deaf in my right ear and was left with 50 percent hearing in my left. Over time, my remaining hearing dropped to 20 percent, where it is today. My doctors predicted that I would be thoroughly deaf by now, so I think I’m doing pretty well.There was always music on in my house in my childhood. I loved listening to Metallica, Led Zeppelin, Michael Jackson. My dad was a DJ, so he played disco, folk, rock, and music from other countries. For my 18th birthday, my dad asked me to deejay at the restaurant be owned. After doing that for a few weeks, I was hooded. I desired to learn more. I e-mailed DJ Shiftee, a distinguished New York City DJ, when I was 25: “I know you like a challenge. How about teaching a deaf person to deejay?” He wrote back the next day; “Challenge accepted.” He tutored me twice a week for two years, helping me develop correct technique. I practiced four hours a day.Now when I’m performing, muscle memory takes over. When I started, I wouldn’t tell the club managers that I was deaf. I would just show up, introduce myself, and start playing music. At the end of the night, someone would say, “Oh, here’s the check.” And I’d say, “What? Oh, I can’t hear.” They were always so astonished. Sometimes I would bring doctor’s notes because they wouldn’t believe me. It was reassurance that they were giving me opportunities to perform because I was brilliant, no out of sympathy. Eventually peopl e started calling me “that deaf DJ,” and the name stuck.What fascinates me about deejaying is the creativity. I use software that turns the music into lines of color on a computer screen. I’m visually hearing the music. The next time you go dancing, cover your ears, and you’ll start seeing that you’re able to hear the music in a different way. Music is not all about hearing. I pay all sorts of get-togethers now, from college parties to corporate events. I also go to elementary schools for the deaf and talk to the students about motivation and believing in themselves. I’m big on talking to the parents. I tell them, “My advice to you is let your kids chase their dreams. I’m a deaf DJ, so why not?”56.Which of the following might result in the author’s hearing l oss?A. Monthly ear infection.B. Moving to the U. S.C. Family financial hardshipD. The doctors’ prediction.57.How did DJ Shiftee help the author during his youth?A. He taught him correct skills.B. He discovered his talent for DJ.C. He played at the restaurant for him.D. He cultivated his taste for foreign music.58.The underlined expression in Paragraph 3 “the name stuck” probably means that .A. the author was in low spiritsB. the author impressed people deeplyC. the audience felt disappointed by the playerD. the audience looked down upon the player59.We can conclude from the passage that the author loves deejaying because .A.working as a DJ involves innovationB. music helps him to see the world virtuallyC. he motivates the kids to realize their dreamD. he desires to challenge something impossibleFREE HomeschoolingIf you are reading this page you are looking into homeschooling. YAHOO! So proud of you for taking the road less traveled for your kids. It may not always be easy, but it is rewarding!123 Homeschool 4 Me is here to help you on your journey! We’ve got lots of tips, resources,and over a million pages of FREE Home-school worksheets, games and lesson plans to helpyou provide a solid, fun, and affordable education for your kids!Let me walk you through some homeschooling basics and how 123 Homeschool 4 Me can help you home-school!Why HomeschoolingFor some it is a better education, impact of being socialized at school, passing on your faith, spending more time with your kids, helping your child with a special need, makinglearning fun, or any number of other reasons. Keep reminding yourself WHY you choose tohome-school and make that your primary focus.How to Home-school●Decided to Home-school after much careful research and thought25 Reasons Why We Love Homeschooling●Find out the legal requirements to home-school in your state —— every state has different requirements that youmust follow to home-school legally —— Homeschooling Laws in your State●Follow any and all legal requirements to home-school legally (see above)●Pick a curriculum - you have tons of choices to fit your family and childrenHow to Choose a Home-school CurriculumOur Curriculum Choices●Plan your school year —— with any state regulations in mind, pick when your school will start end, take breaks,what days of the week you will meet and for how long, and what pace you need to go through your curriculum to finish in a year.Use these Free Home-school Forms to organize your yearHow to Home-school in 15 Hours a Week●Don’t Home-school alone! Just because you aren’t sending your kids to public or private school doesn’t mean theywon’t be with others from outside your family.Getting Social in Your Home-school●Start teaching your child —— Your taught your child how to use the bathroom and put on their shoes. You can dothis! Just dive in!Okay, so that was super simplified, I know! But really that is all you need to start with. Make sure to read the links above for more information on each point.How to Start HomeschoolingMake Home-school FUN and Affordable! This is where comes in! Mom leaves little time to think of fun, creative educational activities that make concepts stick. Plus the cost to buy cool games and additional worksheets for every little skill can be unacceptable!We’ve got you covered! This site is filled with thousands of creative ideas and 1,000,000+ FREE educational print-ables to make learning fun!60.According to the above material, 123 Homeschool 4 Me is probably .A. a websiteB. a counselorC. a magazineD. an advertisement61.Which of the following might be a reason for parents to choose homeschooling?A. Restoring the child’s faith.B. Getting social in the home-schoolC. Challenging the road less traveled.D. Tailoring the courses to kids’ needs.62.123 Homeschool 4 Me is likely to be quite appealing to the readers due to .A. simplified lesson plans and fun activitiesB. interesting games and affordable worksheetsC. free teaching resources and practical suggestionsD. detailed curriculum plans and free homeschoolingChanging the GameOn a warm September evening in London, The Arch climbing wall, just south of the River Thames, is packed. Scores of people wander around on the thick crash pads, chatting, waiting their turn and offering the odd shout of encouragement to those clinging on to the colourful climbing walls.Rock climbing was once classified as an “extreme sport”. But indoor centres like The Arch, which offer c limbing without the need for rocks, are bringing it into the mainstream. The British Mountaineering Council estimates there are at least 248 public climbing walls in Britain, a number that has risen by 30% since 2010. In 2020 the sport’s governing bodies are hoping to see an even bigger increase in interest.Along with skateboarding,surfing and karate（空手道）,rock climbing will be making its first appearance as an Olympic sport at the summer games in Tokyo.The International Olympic Committee (IOC) is frank about the ambition to appeal to a younger crowd who may be less familiar with longer-standing sports such as athletics and weightlifting. The crowd at the Arch is exactly what the IOC has in mind: mostly young professional letting off steam after work, who see climbing as a more engaging ans sociable alternative to jogging on running machines or pumping iron in a gym. Between them, the new sports will mean another 18 events and 474 athletes at the Tokyo games.Officially, all four sports are delighted with their new status. But with the exception of karate, all of them have counter-cultural, anti-establishment roots. Some stars have wondered whether accepting the Olympic torch means going against their beliefs. Owen Wright, a famous surfer, has said that surfing is more art form than sport, and therefore not suitable for the games — though he has since gone back on his word, and hopes to represent Australia in Tokyo.Adam Ondra, a Czech who is one of the world’s climbers, said he might steer clear of the gam es because of the format. The eventual Olympic champion will have to master all the three disciplines including bouldering (climbing without a rope, low to the ground, with a focus on hard, gymnastic moves), lead climbing (roped climbing up a tall wall of increasing difficulty) and speed climbing. Bouldering and lead climbing feature new routes in each stage of a competition, in an effort to imitate the variety of real rock. But speed climbing takes place on a standard, unvarying course. Because of this, said Mr. Ondra, “speed is a kind of artificial discipline ... and this is against the soul of climbing.”Skateboarders, also notably rebellious, can be strikingly young. Sky Brown is set to become Britain’s youngest Olympian and has settled down to training. By the time of the Tokyo games, she will have turned 12.63.Which of the following statements is true about rock climbing？A.It originated in The Arch, a sports centre on the River Thames.B.It has evolved from a mainstream sport into an extreme sport.C.Spectator’s encouragement contributes to its rapid expansion.D.The increase in climbing walls reflects a growing interest in it.64.IOC introduced rock climbing into the Olympics in order to .A.familiarize the global population with the new sportB.attract young people who lack interest in traditional sportsC.enable the young to let off their energy after workD.challenge the dominant status of traditional sports65.What can you infer from the star athletes’ responses according to the passage?A.Surfers are expected to strike a balance between art and sport in the Olympics.B.Rock climbers must be self-disciplined if they are to win the championship.C.Adam believes that the soul of climbing consists in its harmony with nature.D.Strikingly young skateboarders have an advantage over other opponents.66.What is the passage mainly about?A.With the addition of new Olympics sports, stars are divided on whether to participate.B.Rock climbing, skateboard, surfing and karate are accepted as Olympics sports.C.Extreme sports athletes rebel against traditions while training for the Olympics.D.The appeal of a new sport event consists is changing for format of this game.Section CDirections: Read the following passage. Fill in each blank with a proper sentence given in the box. Each sentence canbe used only once. Note that there are two more sentences than you need.A.It all goes back to each country’s distinct cultural heritage.B.American stories are rooted in realism; even our fantasies are rooted in realism.C.Both boys are characterized by their unique roles, thus breathing life into the fancy stories.D.Meanwhile, the United States, also a major player in children’s classics, deals much less in magic.E.Britain’s time-honored countryside, with ancient castles and restful farms, lends itself to fairy-tale i nvention.F.B oth orphans took over the world of children’s literature, but their stories unfold in noticeable different ways.How the British and American Tell Children’s StoriesIf Harry Potter and Huckleberry Finn were each to represent British versus American children’s literature, a curious situation would emerge: In a literary competition for the hearts and minds of children, one is a wizard（巫师) in-training at a boarding school in the Scottish Highlands, while the other is a barefoot boy drifting down the Mississippi, bothered by cheats, slave hunters, and thieves. One defeats evil with a magic stick, the other takes to a raft（筏）to right a social wrong. 67The small island of Great Britain is an unque stionably powerhouse of children’s bestsellers: Alice in Wonderland, Harry Potter, and The Lion, the Witch, and the Wardrobe. Significantly, all are fantasies. 68 Stories like The Call of the Wild. Charlotte’s Web, Little Women, and the Adventures of Tom Sawyer are more notable for their realistic portraits of day-to-day life in the towns and farmlands on the growing frontier. If British children gathered in the dim light of the kitchen fireplace to hear stories about magic swords and talking bears, American children sat at their mother’s knee listening tales with moral messages about a world where life was hard, obedience emphasized, and Christian morality valued. Each style has its virtues, but the British approach undoubtedly creates the kinds of stories that appeal to the furthest reaches of children’s imagination.69 For one, the British have always been in touch with their pagan （异教徒的）folk traditions and stories, says Maria Tatar, a Harvard professor of children’s literature. After all, the country’s very origin story is about a young king tutored by a wizard. Legends have always been accepted as history, from Merlin to Macbeth. “Even as the British were digging into these magical worlds, Americans, much more realistic, always viewed their soil as something to exploit,” says Tatar.American write fantasies too, but nothing like the British, says Jerry Griswold, a San Diego State University professor of children’s literature. He said, “ 70 ” To prove it, he mentioned Dorothy, the heroine of Wizard of Oz （绿野仙踪）who unmasks the great and powerful Wizard as a cheat. Meanwhile, American fantasies differ in another way: They usually end with a moral lesson learned — for example, in Oz, Dorothy’s journey ends with the realization: “There’s no place like home.”IV.S ummary WritingDirections: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your own words as far as possible.71.Britain’s Buses are Getting EmptierIn Britain, buses account for more public - transport trips than trains, tubes and trams put together. People love them, in theory: one poll by Transport Focus, a consumer group, found that 74% of young people think they are a good way of getting around and 85% believe it is important for a place to have a good bus service. There is just one problem. In practice, Britions are taking buses less and less.Why are London buses emptier? One thing that has cha nged is young people’s behavior. The young are heavy bus users when they travel. But, increasingly, they do not travel. According to Transport for London, the average 17 - to 24-year-old took 2.3 transport trips per day in the year 2011- 12 but only 1.7 in 2018 - 19. The National Travel Survey confirms that no group has cut back harder on travel since the early 2000s than teenagers. Young people are more diligent these days, and stay in school for longer. They can do the things that young people love to do on their phones, without going out.The other big bus users are the poor and the old, especially outside London, but both are turning away from buses to cars. Lower lending standards have made cars easier to acquire; a fuel-tax freeze and fuel-saving engines make them cheaper to run. Cars are ever more comfortable and easier to operate, with parking-assist technology and lane-drifting alerts to help starters. Outside London, the average free bus pass was used 90 times in the year 2010-11 but only 74 times in 2018-19, according to the Department for Transport.Finally there is the gig economy (零工经济). Online shopping and Uber probably substitute of bus trips as well as private car journeys. And they put new vehicles on the roads, which slows everything down. The number of light-goods vehicles in London has risen by 28% since 2012. Tony Travers of the London School of Economics points out that bus speeds have fallen slightly in the capital, even though private cars have almost been cleared up from the city center. The average London bus now travels at 9.3 miles per hour. Just as people become less inclined to run after buses, they are becoming easier to catch.。

浦东新区2019~2020学年第二学期高三年级质量调研语文试卷（时间150分钟，满分150分）2020.05一、积累应用（10分）1.按要求填空。

（5分）（1）＿＿＿＿＿＿＿＿＿，水面清圆，一一风荷举。

（＿＿＿＿＿《苏幕遮》）（2）如使人之所欲莫甚于生者，＿＿＿＿＿＿＿＿＿，使人之所恶莫甚于死者，则凡可以辟患者何不为也？（孟子《鱼我所欲也》）（3）《琵琶行》中描写琵琶女初次出场时羞于见人之态的诗句是：“＿＿＿＿＿＿＿＿＿，＿＿＿＿＿＿＿＿＿。

”2.按要求选择。

（5分）（1）下列选项中，成语使用恰当．．的一项是（）A.小明看到上观新闻上发布了本地的疫情数据，立即向同学通风报信。

B.这凹凸二字历来用的人最少，如今作茶楼餐厅之名，更觉不落窠臼。

C.疫情期间，每个上海人都自觉在家，减少出行，整座城市万人空巷。

D.小明平时学习粗心，总是目无全牛，以致很简单的题目也常常出错。

（2）将下列编号的语句依次填入语段空白处，语意连贯的一项是（）○1构成快乐生活的不是无休止的狂欢、美色、鱼肉以及其他餐桌上的佳肴○2就像某些由于无知、偏见或蓄意曲解我们意见的人所认为的那样○3而是清晰的推理、寻求选择和避免的原因、排除那些使灵魂不得安宁的观念○4我们认为，快乐就是身体的无痛苦和灵魂的不受干扰A.○2○4○1○3B.○1○3○2○4C.○4○2○1○3D.○3○2○1○4二、阅读70分（一）阅读下文，完成第3-7题。

（16分）①道德记忆是人类对其道德生活经历的记忆。

作为个人，我们可以凭借天生的记忆能力刻写个体道德记忆；家庭、企业、政党、民族、国家等社会集体则具有集体道德记忆能力，但它们刻写集体道德记忆的方式远比个人刻写个体道德记忆的情况复杂。

需要强调的是，无论道德记忆是以何种形式存在，它都具有选择性特征。

对此，我们需要从三个方面予以解析：②道德记忆的选择性具有一定的客观性。

人类的道德生活经历是复杂的，而我们的记忆容量是有限的，加上我们的记忆还会受到“遗忘”的阻挠，因此，要记住过去的所有道德生活经历客观上是不可能的，人类积累的道德记忆在内容上往往少于我们的实际道德生活经历。

2020届上海市浦东新区三林中学高三语文下学期期中考试试卷及参考答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。

社会是由众多家庭组成的，家庭和谐关乎社会和谐。

要在家庭中建立一种和谐的关系，就需要有家庭伦理。

中国自古以来就有维护家庭关系的种种伦理规范，它们往往体现在各种“礼”之中。

从《礼记》中可以看到各种礼制的记载，如婚丧嫁娶，这些都包含着各种家庭伦理规范，而要使这些规范成为一种社会遵守的伦理，就要使“礼”制度化。

在中国古代，“孝”无疑是家庭伦理中最重要的观念。

《孝经》中有孔子的一段话：“夫孝，天之经也，地之义也，民之行也。

”这是说“孝”是“天道”常规，是“地道”通则，是人们遵之而行的规矩。

为什么“孝”有这样大的意义？这与中国古代宗法制有关。

中国古代社会基本上是宗法性的农耕社会，家庭不仅是生活单位，而且是生产单位。

要较好地维护家庭中长幼尊卑的秩序，使家族得以顺利延续，必须有一套维护当时社会稳定的家庭伦理规范。

这种伦理规范又必须是一套自天子至庶人都遵守的伦理规范，这样社会才得以稳定。

“孝”成为一种家庭伦理规范，并进而成为社会的伦理制度，必有其哲理上的根据。

《郭店楚简·成之闻之》中说：“天登大常，以理人伦，制为君臣之义，作为父子之亲，分为夫妇之辩。

”理顺君臣、父子、夫妇的关系是“天道”的要求。

君子以“天道”常规处理君臣、父子、夫妇伦理关系，社会才能治理好。

所以，“人道”与“天道”是息息相关的。

“孝”作为一种家庭伦理的哲理根据就是孔子的“仁学”。

以“亲亲”（爱自己的亲人）为基点，扩大到“仁民”，以及于“爱物”。

基于孔子的“仁学”，把“孝”看成是“天之经”“地之义”“人之行”是可以理解的。

一方面，它体现了孔子“爱人”（“泛爱众”）的精义；另一方面，在孔子儒家思想中，“孝”在社会生活实践中有一个不断扩大的过程。

因此，“孝”不是凝固教条，而是基于“仁学”的“爱”不断释放的过程，只有在家庭实践和社会实践中，以“仁学”为基础的“孝”的意义才能真正显现出来。

绝密★启用前上海市浦东新区普通高中2020届高三年级下学期期中教学质量监测(二模) 英语试题参考答案及听力材料2020年5月英语试卷听力材料及答案I. Listening ComprehensionSection ADirections: 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 a 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. M: I’ll have the steak. And a bottle of red wine.W: Yes，and I’ll have fish with boiled potatoes. And please see that it isn't overcooked.Q: Where does the conversation probably take place? （B）2. W: How long will it take you to fix my smartphone?M: I’ll call you when it’s ready. But it shouldn’t take longer than a week.Q: What is the probable relationship between the speakers? (D)3. W: John，do you want to work out in the gym with me today?M: Sure，but I can’t leave now. I have an app ointment with my professor at 4 o’clock.Q:Why can’t John go to the gym now? (B)1。

2023届上海市浦东新区高三下学期等级考模拟质量调研（二模）物理试题学校:___________姓名：___________班级：___________考号：___________一、单选题10．如图，将磁铁插入或拔出线圈过程中检流计指针都发生偏转，则（）A．拔出过程磁场力对磁铁做负功，将其他形式的能转化为电能B．拔出过程磁场力对磁铁做负功，将电能转化为其他形式的能C．插入过程磁场力对磁铁做正功，将其他形式的能转化为电能D．插入过程磁场力对磁铁做正功，将电能转化为其他形式的能11．如图，正六边形线框abcdef，各边电阻相同。

线框垂直于匀强磁场放置，b、c点与直流电源相接。

若闭合电键后fe边受到的安培力大小为F，不考虑各边之间的相互作用，则整个线框受到的安培力大小为（）A．0B．2F C．5F D．6F12．如图，一根足够长的均质粗糙杆水平固定，一个圆环套在杆上。

现给环一个向右的初速度0v，同时对环施加一个竖直方向的力，力的大小与环的速度大小成正比。

则环所受摩擦力的瞬时功率P与速度v的图像不可能为（）A．B．C．D．二、填空题13．发生光电效应的条件是入射光频率______（选填“大于”或“小于”）金属的极限频率；当发生光电效应时，从金属表面逸出的粒子，其本质与β粒子一样都是______。

14．利用图示装置中的激光照射细铜丝，在抽丝过程中对铜丝的粗细实施自动控制。

这一技术利用了光的______现象，如果发现光屏上的条纹变宽，表明此时抽出的铜丝变______（选填“粗”或“细”）。

15．图为采用动力学方法测量空间站质量的示意图，若已知飞船质量为33.010kg ⨯，其推进器的平均推力F 为900N ，在飞船与空间站对接后，推进器工作5s 内，测出飞船和空间站的速度变化量是0.05m/s ，则对接后，飞船的加速度大小=a ______2m/s ，空间站的质量为______kg 。

16．图示为控制中心大屏幕上显示的“神舟”十四号飞船在轨运行图，屏幕上的曲线表示它一段时间内先后两次在同一轨道绕地球做匀速圆周运动的“轨迹”。

2020年高三教学质量检测试卷（二）【二模】文综政治试题姓名:________ 班级:________ 成绩:________一、单选题 (共12题；共24分)1. （2分） (2019高二下·黑龙江月考) 若不考虑其他因素，用D表示某商品的需求曲线，如图的经济现象与图象一致的是（）①加大新能源汽车购置税优惠对新能源汽车需求量的影响符合图（I）②高铁提高运行速度对航空客运需求的影响符合图（I）③煤炭限产能后煤炭价格上涨对天然气需求量的影响符合图（Ⅱ）④旅游景区的门票价格上调对游客需求量的影响符合图（Ⅱ）A . ①③B . ①④C . ②③D . ②④2. （2分）当前我国财政支出预算执行具有显著的“前低后高”特点，往往40%左右的预算资金要留到四季度支出，如无法正常支出，则在年末以结余、结转等方式留在国库、财政专户或者部门账户上，累积而成财政存量资金。

对于财政存量资金，下列看法正确的是（）①出现过多的财政存量资金意味着财政资金得到了合理的利用②过多的财政存量资金会削弱积极财政政策的实际效果③财政存量资金的增加是财政收入新常态的重要表现④应当深化预算管理制度改革，盘活财政存量资金A . ①②B . ①③C . ②④D . ③④3. （2分） (2016高一上·梅河口期末) 下列能够反映企业承担社会责任的是（）①依法维护劳动者的合法权益②遵守法律，遵守社会公德③积极参加社会公益事业④依靠技术进步和管理，提高企业的经济效益A . ①②③B . ①②④C . ①③④D . ②③④4. （2分） 2015年，我国对外非金融类直接投资创下ll80.2亿美元的历史最高值，同比增长14.7％，超额完成全年l0％的增长目标，实现中国对外直接投资连续13年增长。

下列对我国对外直接投资快速增长的原因分析正确的是（）①企业“走出去”的动力增强②我国外商投资环境不断优化③境外投资风险防控意识增强④“一带一路”引领作用突出A . ①②B . ①④D . ③④5. （2分） (2018高一下·汕头期末) 我国首个“村务监督委员会”——浙江省武义县白洋街道后陈村村务监督委员会成功换届选举。

浦东新区2019学年度第二学期教学质量检测高三数学C 卷一、填空题（本大题满分54分）本大题共有12题，1-6题每题4分，7-12题每题5分．考生应在答题纸相应编号的空格内直接填写结果，每个空格填对得4分或5分，否则一律得零分．1. 在行列式12011213a-中，元素a 的代数余子式的值是____________．【答案】2 【解析】 【分析】根据代数余子式定义列式求解,即得结果.【详解】在行列式12011213a-中，元素a 的代数余子式为401(1)0(2)221--=--=故答案为：2【点睛】本题考查代数余子式，考查基本求解能力，属基础题. 2.函数y =____________．【答案】(]2-∞，【解析】 【分析】根据偶次根式下被开方数非负列不等式，解得指数不等式得结果. 【详解】2933032x x x ≥∴≤∴-≤,故定义域为(]2-∞， 故答案为：(]2-∞，【点睛】本题考查定义域，考查基本分析求解能力，属基础题.3. 已知x 、R y ∈，i 为虚数单位，且()2i 1i x y -+=-+，则x y +=____________． 【答案】2 【解析】 【分析】根据复数相等列方程组，解得,x y ，即得结果.【详解】()212i 1i 1x x y y -=-⎧-+=-+∴⎨=⎩ 121x x y y =⎧∴∴+=⎨=⎩故答案为：2【点睛】本题考查复数相等，考查基本分析求解能力，属基础题. 4. 函数()sin cos R y x x x =-∈的单调递增区间为____________． 【答案】()32,244k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦【解析】 【分析】先根据辅助角公式化简函数解析式，再根据正弦函数性质求单调区间.【详解】sin cos )4y x x x π=-=-22()242k x k k Z πππππ∴-≤-≤+∈322()44k x k k Z ππππ∴-≤≤+∈ 故答案为：()32,244k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦【点睛】本题考查辅助角公式、正弦函数性质，考查基本分析求解能力，属基础题.5. 已知02x <<_______ 【答案】1 【解析】 【分析】配方后利用二次函数求最值可得结果.==,又因为02x <<所以1x =时 1. 故答案为:1【点睛】本题考查了二次函数求最大值,属于基础题. 6. 61()x x-的展开式中的常数项是： ．(请用数字作答) 【答案】-20 【解析】621661()(1)r n r r rr r r T C x C x x--+=-=-，令620r -=，则3r =，所以常数项为3620C -=-．7. .数列{}n a 的前n 项和为n S ，若点(,)n n S （*n N ∈）在函数的反函数的图像上，则n a =________.【答案】12n - 【解析】 解：因为221log (1)log (1)12212nn n n n n n y x n S S S a -=+∴=+∴+=∴=-∴=8. 一支田径队有男运动员40人，女运动员36人，若用分层抽样的方法从该队的全体运动员中抽取一个容量为19的样本，则抽取男运动员的人数为____________． 【答案】10 【解析】 【分析】由样本容量与总体容量的比值相等计算．【详解】设抽取的男运动员人数为n ，则由分层抽样定义得40194036x =+，解得10x =． 故答案为：10．【点睛】本题考查分层抽样，利用分层抽样中样本容量与总体容量的比值相等求解即可． 9. 36的正六棱柱的所有顶点都在一个平面上，则此球的体积为 ． 【答案】92π 【解析】 【分析】作出六棱柱的最大对角面与外截球的截面,设正六棱柱的上下底面中心分别为12,O O ,球心为O ,一个顶点为A ,可根据题中数据结合勾股定理算出球的半径OA ,再用球的体积公式即可得到外接球的体积.【详解】作出六棱柱的最大对角面与外截球的截面,如右图,则该截面矩形分别以底面外接圆直径和六棱柱高为两边， 设球心为O,正六棱柱的上下底面中心分别为12,O O,则球心O 是12O O 的中点.∴正六棱柱底面边长为3261Rt AO O ∴中,113622AO O O ==,可得221132AO AO O O =+=因此,该球的体积为3439322V ππ⎛⎫=⋅=⎪⎝⎭ 故答案为92π. 【点睛】本题给出一个正六棱柱,求它的外接球的体积,着重考查了球的内接多面体和球体积公式等知识点,属于基础题.10. 如图，已知椭圆1C 和双曲线2C 交于1P 、2P 、3P 、4P 四个点，1F 和2F 分别是1C 的左右焦点，也是2C 的左右焦点，并且六边形121342PP F P P F 是正六边形．若椭圆1C 22142323+=+，则双曲线2C 的方程为____________．22142323=- 【解析】 【分析】先根据椭圆1C 的方程确定半焦距，再根据正六边形性质确定双曲线中,,.a b c 【详解】222142334242323c c +=∴=+=∴=+设22222:1,(0,0)x y C a b a b-=>>22212||||23231a P F P F a =-=∴=2222431)23b c a ∴=-=-=因此2222(31)123C -=22142323=- 22142323=- 【点睛】本题考查求双曲线方程，考查基本分析求解能力，属基础题.11. 已知A 、B 、C 是半径为5的圆M 上的点，若6BC =，则AB AC ⋅的取值范围是____________． 【答案】[]8,72- 【解析】 【分析】由正弦定理求出sin A ，由平方关系得cos A ，然后利用余弦定理和基本不等式求出bc 的范围，最后由数量积的定义可得结论． 【详解】记,,A B C 所对边长分别,,a b c ，由正弦定理得2sin a R A=，即63sin 2255a A R ===⨯，所以4cos 5A =±， 由余弦定理2222cos a b c bc A =+-，4cos 5A =时，228836255b c bc bc bc =+-≥-，90bc ≤，b c =时等号成立，所以4cos 90725AB AC bc A ⋅=≤⨯=，4cos 5A =-时，228836255b c bc bc bc =++≥+，10bc ≤，b c =时等号成立，所以4cos 1085AB AC bc A ⎛⎫⋅=≥⨯-=- ⎪⎝⎭，综上[8,72]AB AC ⋅∈-． 故答案为：[8,72]-．【点睛】本题考查求平面向量数量积的取值范围，考查正弦定理，余弦定理，基本不等式，属于中档题． 12. 对数列{}n a ，{}()*n b n N∈，如果存在正整数k ，使得1kk ab >+，则称数列{}n a 是数列{}n b 的“优数列”，若32222n a n n tn t =+-+，3241n b n n n =+++，并且{}n a 是{}n b 的“优数列”，{}n b 也是{}n a 的“优数列”，则t 的取值范围是____________． 【答案】1t >-． 【解析】 【分析】根据“优数列”列不等式，再根据二次不等式有解求参数范围. 【详解】因为{}n a 是{}n b 的“优数列”， 所以存在正整数k ，1k k a b >+即3223222411k k tk t k k k +-+>++++，22(24)20k t k t --++> 显然成立，所以t R ∈； 因为{}n b 是{}n a 的“优数列”， 所以存在正整数m ，1m m b a >+即3232241221m m m m m tm t +++>+-++，22(24)0m t m t -++<22(24)401t t t ∴∆=+->∴>-当1t >-时，由于对称轴21m t =+>，所以必存在正整数m ，使得22(24)0m t m t -++< 综上，1t >- 故答案为：1t >-【点睛】本题考查数列新定义、不等式有解问题，考查综合分析求解能力，属中档题.二、选择题（本大题满分20分）本大题共有4题，每题有且只有一个正确答案．考生必须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. “a b =”是“a b =”的（ ） A. 充分不必要条件 B. 必要不充分条件 C. 充要条件 D. 既不充分又不必要条件【答案】A 【解析】 【分析】根据充要关系定义进行判断选择.【详解】若a b =，则a b =，所以充分性成立；若a b =，则a b =不一定成立，例如互为相反向量时就不成立，所以必要性不成立； 故选：A【点睛】本题考查充要关系判断，考查基本分析判断能力，属基础题.14. 已知等比数列{}n a 中，各项都是正数，且1a ，312a ，22a 成等差数列，则8978a a a a +=+（ ）A. 1B. 1C. 3+D. 3-【答案】B 【解析】 【分析】根据等差数列列式求得公比，再代入所求式,解得结果.【详解】因为1a ，312a ，22a 成等差数列，所以3121222a a a ⨯=+设{}n a公比为21201q q qq q ∴=+>∴=+从而89781a a q a a +==++故选：B【点睛】本题考查等比数列与等差数列综合，考查基本分析求解能力，属基础题. 15. 对于实数a 、b 、m ，下列说法： ①若a b >，则22am bm >； ②若a b >，则a ab b ；③若0b a >>，0m >，则a m ab m b+>+； ④若 0a b >>，且ln ln a b =，则(2,)a b +∈+∞，其中正确的命题的个数（ ） A. 1 B. 2C. 3D. 4【答案】C 【解析】 【分析】举例说明①错误；根据不等式性质证明②成立；利用作差法证明③成立；根据对勾函数性质说明④成立. 【详解】若,0a b m >=，则22am bm =,所以①错误； 若a b >，则当0a b >≥时a a b a b b a a b b当0a b ≥>时0a a b b a ab b 当0a b >>时0aba ab a b ba ab b ，因此②成立；若0b a >>，0m >，则()0()a m a b a m a m ab m b b b m b m b+-+-=>∴>+++，所以③成立； 若 0a b >>，且ln ln a b =，则ln ln ln()0,11ab b a a b a ∴==∴=>-，1a b a a∴+=+在(1,)+∞上单调递增，即12a b a a+=+>，因此④成立. 故选：C【点睛】本题考查根据不等式性质比较大小、作差法比较大小、对勾函数单调性，考查基本分析判断能力，属基础题.16. 数学试卷的填空题由12道题组成，其中前6道题，每道题4分；后6道题，每道题5分．下面4个数字是某教师给出的一位学生填空题的得分，这个得分不可能是（ ）A. 17B. 29C. 38D. 43【答案】D 【解析】 【分析】根据得分情况可说明ABC 成立，再说明D 一定不成立.【详解】因为17=34+15,296415,382465⨯⨯=⨯+⨯=⨯+⨯，所以得分可能是ABC;因为432475=⨯+⨯，而满足个位数为3的只有这一种，但每道题5分的只有6道题，因此D 得分是不可能的， 故选：D【点睛】本题考查简单推理，考查基本分析判断能力，属基础题.三、解答题（本大题满分76分）本大题共有5题，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. 如图，长方体1111ABCD A B C D -的底面ABCD 是正方形，点E 为棱1AA 的中点，1AB =，12AA =．（1）求点B 到平面11B C E 的距离； （2）求二面角11B EC C --的正弦值． 【答案】（12；（2）32． 【解析】 【分析】（1）建立空间直角坐标系，用向量法求点到平面的距离； （2）用空间向量法求二面角的余弦值，再求正弦值．【详解】解：（1）如图所示，建立直角坐标系，则有关点的坐标为()1,0,0B ，()11,0,2B ，()11,1,2C ，()0,0,1E，所以，()110,1,0B C=，()11,0,1B E=--．设平面11B C E的法向量()1,,n u v w=，则由111n B C⊥且11n B E⊥得，11111vn B Cu wn B E⎧=⋅=⎧⎪⇒⎨⎨+=⋅=⎪⎩⎩．取1u=，于是平面11B C E的一个法向量为()11,0,1n=-．且()10,0,2BB=，所以，点B到平面11B C E的距离为()112220100122101n BBdn⋅⨯+⨯-⨯===++-（2）因为()11,1,2C，()0,0,1E，()11,1,0C，所以，()10,0,2CC=，()1,1,1CE=--设平面1CC E的法向量()1112,,n u v w=，则由21n CC⊥且2n CE⊥得，112111111220000w wn CCu v w u vn CE⎧⎧==⎧⋅=⎪⎪⇒⇒⎨⎨⎨--+=+=⋅=⎪⎩⎪⎩⎩．取11u=于是平面1CC E的一个法向量为()21,1,0n=-．设平面11B C E的一个法向量()11,0,1n=-与平面1CC E的一个法向量为()21,1,0n=-的夹角为ϕ，则12121cos2n nn nϕ⋅==，所以，3sin2ϕ=．所以二面角11B EC C--的正弦值为32．【点睛】本题考查用空间向量法求点到平面的距离，求二面角．在图形中已有两两相互垂直的三条直线时，如长方体，可建立空间直角坐标系，用空间向量法求空间角，距离，研究（或证明）空间线面位置关系．18. 在平面直角坐标系xOy中，已知函数()()()sin0,0f x xωϕωϕπ=+><<．（1）如图所示，函数()f x的图象与直线()1 1y m m=-<<三个相邻交点的横坐标为3π-、6π、2π，求ω的值；（2）函数()()sin0,0y xωϕωϕπ=+><<的图象与x轴的交点A、B、C，且满足OA、OB、OC 成等差数列，求ϕ的值．【答案】（1）125ω=；（2）34πϕ=．【解析】【分析】（1）先确定周期，再求ω的值；（2）根据等差数列性质得2OB OA OC=+，再利用A、B、C坐标表示，解得ϕ的值．【详解】（1）由三角函数图象可知，直线y m=与正弦函数图象相交的三个相邻交点中，第一个点和第三个点之间正好一个周期则5236Tπππ⎛⎫=--=⎪⎝⎭所以2125T πω==． （2）由OA 、OB 、OC 成等差数列得2OB OA OC =+ 同一周期内，不妨设0B x ωϕ+=，A x ωϕπ+=， 2C x ωϕπ+= 得B x ϕω=-，A x πϕω-=，2C x πϕω-=， 由2OB OA OC =+，得322πϕϕωω-=，解得34πϕ=． 【点睛】本题考查根据三角函数图象与性质求参数、等差数列应用，考查基本分析求解能力，属基础题. 19. 某企业准备投产一款产品，在前期的市场调研中发现： ①需花费180万元用于引进一条生产流水线；②每台生产成本Q （x ）（万元）和产量x （台）之间近似满足Q （x ）＝51351x ++，x ∈N *；（注每台生产成本Q （x ）不包括引进生产流水线的费用） ③每台产品的市场售价为10万元； ④每年产量最高可达到100台；（1）若要保证投产这款产品后，一年内实现盈利，求至少需要生产多少台（而且可全部售出）这款产品； （2）进一步的调查后发现，由于疫情，这款产品第一年只能销售出60台，而生产出来的产品如果没有在当年销售出去，造成积压，则积压的产品每台将亏损1万元，试判断该企业能否在投产第一年实现盈利．如果可以实现盈利，则求出当利润最大时的产量；若不能实现盈利，则说明理由．【答案】（1）至少生产并销售63台这款产品，才能实现盈利；（2）可以实现盈利，利润最大时，产量为89台． 【解析】 【分析】（1）由题意可得利润函数为f （x ）＝[10﹣Q （x ）]⋅x ﹣180，0＜x ≤100，x ∈N *，由f （x ）＞0求解不等式得答案；（2）把利润函数f （x ）变形，再由基本不等式求最值． 【详解】（1）由题意可知该商品的利润函数为： f （x ）＝[10﹣Q （x ）]⋅x ﹣180，0＜x ≤100，x ∈N *，则由()1018000100*Q x x x x N ⎧⎡⎤-⋅-⎪⎣⎦⎨≤∈⎪⎩＞＜，，解得x ≥63．∴至少生产并销售63台这款产品，才能实现盈利；（2）由（1）可知，当产量0＜x ≤60，x ∈N *时，无法实现盈利； 当产量60＜x ≤100，x ∈N *时，由题意可知利润函数为f （x ）＝[10﹣Q （x ）]⋅60﹣（x ﹣60）﹣180．化简得f （x ）＝181﹣[()1356011x x ⋅+++]1801≤-=． 当且仅当x ＝89时等号成立．∴可以实现盈利，利润最大时，产量为89台．【点睛】本题考查分式函数模型的实际应用，涉及利用基本不等式求最值，属综合基础题. 20. 已知点F 是抛物线2:8C y x =上的焦点，()11,A x y 、()22,B x y 是抛物线上的两个动点． （1）若直线AB 经过点F ，且126x x +=，求AB ；（2）若126x x +=，求证:线段AB 的垂直平分线经过一个定点C ，并求出C 点的坐标； （3）若线段AB 与x 轴交于Q 点，是否存在这样的点Q ，使得2211AQBQ+为定值，若存在，求出这个定值和Q 点的坐标；若不存在，请说明理由．【答案】（1）10；（2）证明见解析，经过一个定点()70C ，；（3）存在Q 点满足题意，坐标为()4,0，2211116AQBQ+=． 【解析】 【分析】（1）根据抛物线定义求焦点弦弦长；（2）先考虑直线AB 的斜率存在情况，根据点斜式得线段AB 的垂直平分线的方程，确定定点，再验证直线AB 的斜率不存在情况也过此定点；（3）设()0Q m ，，过Q 点直线方程为x ty m =+，与抛物线联立方程组，结合韦达定理化简2211AQBQ+，确定定值取法，即可确定定点与定值. 【详解】（1）1022A B A B p pAB AF BF x x x x p =+=+++=++=． （2）①当直线AB 的斜率存在时，设线段AB 的中点为()00,M x y ，则12032x x x +==，1202y yy +=，21212221212108488AB y y y y k y y x x y y y --====-+-．线段AB 的垂直平分线的方程是()0034y y y x -=--，即()074yy x =--． ②当直线设AB 的斜率不存在时，此时线段AB 的垂直平分线的方程是0y =．所以线段AB 的垂直平分线经过一个定点()70C ，． （3）设()0Q m ，，过Q 点直线方程为x ty m =+，联立228880y xy ty m x ty m⎧=⇒--=⎨=+⎩， 则264320t m ∆=+>，128y y t +=，128y y m =-．则()()222221111AQ x m y t y =-+=+，()()222222221BQ x m y t y =-+=+，所以，()()22222212111111t y t y AQBQ+=+++()()()()()()222212121222222212122641664111y y y y y y t mm t t y y t y y +-++===+++， 所以当4m =时，2211116AQBQ+=，故Q 点的坐标为()4,0， 并且满足264320t m ∆=+>．【点睛】本题考查抛物线焦点弦、抛物线中定点与定值问题，考查综合分析论证与求解能力，属较难题. 21. 定义在R 上的非常值函数()f x 、()g x （()f x 、()g x 均为实数），若对任意实数x 、y ，均有()()()()22f x y f x y g y g x +⋅-=-，则称()g x 为()f x 的关联平方差函数．（1）判断()cos g x x =是否是()sin f x x =的关联平方差函数，并说明理由； （2）若()g x 为()f x 的关联平方差函数，证明：()f x 为奇函数；（3）在（2）的条件下，如果()01g =，()21g =-，当02x <<时()11g x -<<，且f x T f x对所有实数x 均成立，求满足要求的最小正数T 并说明理由． 【答案】（1）是；理由见解析；（2）证明见解析；（3）4T =是满足要求的最小正数，理由见解析．【解析】【分析】（1）根据关联平方差函数定义直接化简判断；（2）结合关联平方差函数定义，证明()()f b f b -=-恒成立； （3）结合关联平方差函数定义先探求4T=，再用反证法证4T =是满足要求的最小正数.【详解】（1）()cos g x x =是()sin f x x =的关联平方差函数，()()()()()()sin sin sin cos cos sin sin cos cos sin f x y f x y x y x y x y x y x y x y +-=+-=+-()()2222222222sin cos cos sin cos cos cos cos cos cos x y x y y x y x x y =-=---()()2222cos cos y x g y g x =-=-（2）()f x 是非常值函数，所以存在a ，()0f a ≠， 下证对任意实数b ，()()f b f b -=- 令2a b x +=，2a b y -=可得()()2222a b a b f a f b g g -+⎛⎫⎛⎫=- ⎪ ⎪⎝⎭⎝⎭；再令2a b x -=，2a b y +=可得()()2222a b a b f a f b g g +-⎛⎫⎛⎫-=- ⎪ ⎪⎝⎭⎝⎭两式相加可得()()()0f a f b f b +-=⎡⎤⎣⎦，()0f a ≠，()()f b f b ∴-=-，所以()f x 为奇函数（3）令0y =可得()()()()222201f x g g x g x =-=-，即()()221fx g x +=，()21g =-，()()220f f ∴=-=，令2y x =+，()()()()2222220f x f gx g x +-=+-=， 令2y =，()()()2221f x f x gx +-=-，用2x +替换x 可得()()()()()2224121f x f x g x g x f x +=-+=-=，[1]若()0f x ≠，那么()()+4f x f x =； [2]若()=0f x ，那么()()()()()2222211+2+2+4fx g x g x f x f x =-=-==；所以()()+40f x f x ==综上可知4T =满足要求，下证4T =是满足要求的最小正数，用反证法，若存在004T <<也满足要求，令0x =，02T y =可得()2200000222T T T f f g g ⎛⎫⎛⎫⎛⎫-=-< ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭，而0022T T f f ⎛⎫⎛⎫-= ⎪ ⎪⎝⎭⎝⎭，0022T T f f ⎛⎫⎛⎫-=- ⎪ ⎪⎝⎭⎝⎭，00022T T f f ⎛⎫⎛⎫∴-== ⎪ ⎪⎝⎭⎝⎭，矛盾！ 所以4T=是满足要求的最小正数.【点睛】本题考查函数新定义、证明奇函数、函数周期、反证法，考查综合分析论证与求解能力，属较难题.。