江苏省盐城中学2019—2020学年度第一学期期中考试【含解析】
江苏省盐城市盐城中学2019_2020学年高二数学上学期期中试题(含解析)
江苏省盐城市盐城中学2019-2020学年高二数学上学期期中试题（含解析）一、选择题：本大题共12小题，每小题5分，计60分．请把答案涂写在答题卡相应位置． 1.已知命题:p ：“2,0x R x ∀∈>”，则p ￢是（ ）A. 2,0x R x ∀∈≤B. 2,0x R x ∃∈>C. 2,0x R x ∃∈<D.2,0x R x ∃∈≤【答案】D 【解析】试题分析：全称命题的否定是特称命题，所以命题p ⌝是2,0x R x ∃∈<，故选D.考点：命题的否定.2.抛物线24y x =的准线方程为（ ） A. 1x =- B. 1y =- C. 1x = D. 1y =【答案】A 【解析】 【分析】利用22y px =的准线方程为2p x =-，能求出抛物线24y x =的准线方程. 【详解】24,24,2y x p p =∴==，∴抛物线24y x =的准线方程为2p x =-， 即1x =-，故选A .【点睛】本题主要考查抛物线的标准方程与简单性质，意在考查对基础知识的掌握与应用，是基础题.3.两个数4和16的等比中项为（ ） A. 8 B. ±8 C. 4 D. ±4【答案】B 【解析】 【分析】由等比中项的定义，即可求出结果.【详解】4和16的等比中项为8=±. 故选:B【点睛】本题考查等比中项的定义，属于基础题4.已知双曲线22194x y -=的渐近线方程为（ ）A. 23y x =±B. 94y x =±C. 32y x =±D.49y x =±【答案】A 【解析】 【分析】根据双曲线的的方程，可直接得出结果.【详解】令22094x y -=，得23y x =±，即双曲线双曲线22194x y -=的渐近线方程为23y x =±. 故选A【点睛】本题主要考查求双曲线的渐近线方程，熟记双曲线的性质即可，属于基础题型. 5.设x ，y 均为正数，且x +4y ＝4，则xy 的最大值为（ ） A. 1 B. 2C. 4D. 16【答案】A 【解析】 【分析】x +4y 为定值，由基本不等式，即可求出积xy 的最大值.【详解】44x y =+≥=1,1xy ≤≤,当且仅当12,2x y ==时，等号成立.故选:A【点睛】本题考查基本不等式求最值，属于基础题.6.ac 2＞bc 2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. 既不充分也不必要条件【答案】A 【解析】 【分析】根据不等式的性质，利用充分条件和必要条件的定义进行判断. 【详解】若22ac bc >成立，则20,c a b >∴>成立；若a b >成立，而2c ≥0，则有22ac bc ≥，故22ac bc >不成立；22ac bc ∴>是a b >的充分不必要条件.故选:A【点睛】本题考查充分条件和必要条件的判断，根据不等式的关系是解决问题的关键，属于基础题.7.求值：1﹣3+5﹣7+9﹣11+…+2017﹣2019＝（ ） A. ﹣2020 B. ﹣1010C. ﹣505D. 1010【答案】B 【解析】 【分析】分组求和，奇数项和相邻的偶数和均为-2，即可求出结果. 【详解】135720172019-+-++-(13)(57)(20172019)(2)5051010=-+-+-=-⨯=-.故选:B【点睛】本题考查分组并项求和，考查计算能力，属于基础题.8.若∃x ∈[0，3]，使得不等式x 2﹣2x +a ≥0成立，则实数a 的取值范围是（ ） A. ﹣3≤a ≤0 B. a ≥0C. a ≥1D. a ≥﹣3【答案】D 【解析】 【分析】等价于二次函数2()2,[0,3]f x x x a x =-+∈的最大值不小于零，即可求出答案. 【详解】设2()2,[0,3]f x x x a x =-+∈，[0,3]x ∃∈，使得不等式220x x a -+≥成立，须max ()0f x ≥，即(0)0f a =≥，或(3)30f a =+≥， 解得3a ≥-. 故选:D【点睛】本题考查特称命题成立求参数的问题，等价转化是解题的关键，属于基础题. 9.已知等差数列{a n }的前n 项和为S n ，若a 6+a 7＞0，a 6+a 8＜0，则S n 最大时n 的值为（ ） A. 4 B. 5C. 6D. 7【答案】C 【解析】 【分析】当10,0n n a a +≥≤时，则有n S 最大，即可求出答案. 【详解】687720,0a a a a +=<∴<,6760,0a a a +>∴>,6n ∴=,n S 最大.故选:C 【点睛】本题考查等差数列的前n 和的最值，以及等差数列的性质，属于基础题.10.若点P 是以F 为焦点的抛物线y 2＝4x 上的一个动点，B （3，2），则|PB|+|PF|的最小值为（ ） A. 3B. 4C. 5D. 6【答案】B 【解析】 【分析】利用抛物线的定义||PF 等于P 到准线的距离，数形结合即可求出答案.【详解】抛物线24y x =的准线l 方程为1x =-，过点P 做PD l ⊥，垂直为D ，||||||||||4PB PF PB PD BD +=+≥=，当且仅当，,,P B D 三点共线时，等号成立. 故选:B【点睛】本题考查抛物线定义的应用，考查数学转化思想方法与数形结合的解题思想方法，是基础题．11.已知正数x ，y 满足log 2（x +y +3）＝log 2x +log 2y +1，则x +y 的取值范围是（ ） A. [6，+∞）B. （0，6]C. )17⎡+∞⎣，D.(017+，【答案】C 【解析】 【分析】根据已知等式，确定出,x y 的等量关系，再用基本不等式，即可求出结果. 【详解】2222log (3)log log 1log 2x y x y xy ++=++=,232,0,0,()2x y x y xy x y xy +∴++=>>∴≤, 2232(),()2()602x y x y x y x y +∴++≤⨯+-+-≥, 解得17,17x x ≥≤（舍去）.故选:C【点睛】本题考查对数的运算性质及基本不等式的应用，考查计算能力，属于中档题. 12.在数列{a n }中，若对任意的n 均有a n +a n +1+a n +2为定值（n ∈N *），且a 2＝4，a 3＝3，a 7＝2，则此数列{a n }的前100项的和S 100＝（ ） A. 296B. 297C. 298D. 299【解析】 【分析】根据递推公式可得{}n a 是周期为3的周期数列，有472a a ==，求得一个周期和，进而可求出前100项和.【详解】因为对任意的n 均有12n n n a a a ++++为定值（n ∈N *）， 所以12n n n a a a ++++1233,n n n n n a a a a a ++++=++∴=,472a a ∴==,122349n n n a a a a a a ++++=++=， 100123456979899100()()()S a a a a a a a a a a =++++++++++3392299=⨯+=.故选:D【点睛】本题考查用分组并项求和，解题的关键是递推公式的灵活应用，属于中档题. 二、填空题：本大题共4小题，每小题5分，共20分．请把答案填写在答题卡相应位置．13.已知方程22112x y m m +=--表示焦点在x 轴上的椭圆，则实数m 的取值范围是____．【答案】（322，） 【解析】 【分析】根据焦点在x 轴上的椭圆标准方程的特征，可得到关于m 的不等式，即可求得结果.【详解】根据题意，方程22112x y m m+=--表示焦点在x 轴上的椭圆， 则必有1220m m m --⎧⎨-⎩＞＞，解可得：32＜m ＜2，即m 的取值范围是3(2)2，. 故答案为:3(2)2，） 【点睛】本题考查椭圆的标准方程，属于基础题. 14.已知正实数a ，b 满足a +b ＝1，则4a bab+的最小值为_____．【解析】 【分析】化简4a bab+，由已知等式，结合基本不等式，即可求出最小值. 【详解】根据题意，4441a b a b ab ab ab b a+=+=+， 又由正实数a ，b 满足a +b ＝1， 则4a b ab +=（41b a +）（a +b ）＝54a bb a ++，又由4a b b a +≥2=4， 当且仅当b ＝2a 23=时等号成立， 则有4a b ab +=54a bb a ++≥9， 即4a b ab+的最小值为9.【点睛】本题考查基本不等式求最值，解题的关键是“1”的代换，属于中档题. 15.已知数列{a n }满足a 1＝21，a n +1﹣a n ＝2n ，则na n的最小值为__． 【答案】415【解析】 【分析】根据已知条件用累加法求出{}n a 的通项，再构造函数，利用函数单调性，求出数列{}na n的单调性，即可求na n的最小值. 【详解】12,2n n a a n n +-=∴≥时12(1)n n a a n --=-，2n ∴≥时，112211()()()n n n n n a a a a a a a a ---=-+-+-+22[(1)(2)1]21(1)2121,n n n n n n =-+-+++=-+=-+121a =，也满足上式，∴n a n =n 211n+-， ∵()21f x x x=+在（0+∞）上单调递增，∴n an在（0，4]上单调递减，在[5，+∞）上单调递增，且n ∈N +， ∴n ＝4或5时最小，n ＝4时，334n a n =；n ＝5时，413354n a n =＜,n a n ∴的最小值为415. 故答案为:415【点睛】本题考查由递推公式求通项公式，考查数列的单调性，解题的关键熟练掌握常考的递推公式求通项的方法，常用的类型有：(1)公式法：①等差数列通项公式；②等比数列通项公式；(2)已知n S （即12()n a a a f n +++=）求n a ，作差法：11,(1),(2)n nn S n a S S n -=⎧=⎨-≥⎩；(3)已知123()n a a a a f n =求n a ，作商法：(1),(1)(),(2)(1)n f n a f n n f n =⎧⎪=⎨≥⎪-⎩；(4)若1()n n a a f n +-=求n a 累加法：11221()()()n n n n n a a a a a a a ---=-+-++-1a +(2)n ≥；(5)已知1()n na f n a +=求n a ，累乘法：121121n n n n n a aa a a a a a ---=⋅⋅⋅⋅(2)n ≥； (6)形如111n n n a a ka --=+或110n n n n a a ka a ----=，倒数成等差；(7)形如1(1,0)n n a ka b k b +=+≠≠用待定系数法转化为等比数列.16.已知椭圆2222x y a b+=1（a ＞b ＞0）的左、右焦点分别为F 1、F 2，半焦距为c ，且在该椭圆上存在异于左、右顶点的一点P ，满足2a •sin∠PF 1F 2＝3c •sin∠PF 2F 1，则椭圆离心率的取值范围为_____．【答案】0＜e 13＜ 【解析】 【分析】根据已知条件等式，结合正弦定理，得出21,,PF PF e 的关系，利用椭圆定义和2PF 的范围，即可求出e 的取值范围.【详解】在△PF 1F 2中，由正弦定理知21221132PF sin PF F csin PF F PF a∠==∠，又∵P 在椭圆上，∴|PF 1|+|PF 2|＝2a ，12,,P F F 三点不共线，所以14(,)23aPF a c a c e =∈-++， 即41123e e e -++＜＜， 解得103e <<.【点睛】本题考查正弦定理，椭圆的定义和性质，考查计算能力，属于中档题.三、解答题：本大题共6小题，共计70分．请在答题卡指定区域内作答，解答时应写出文字说明、证明过程或计算步骤．17.已知p ：∀x ∈R ，x 2+2x ≥a ，q ：x 2﹣4x +3≤0，r ：（x ﹣m ）[x ﹣（m +1）]≤0． （1）若命题p 的否定是假命题，求实数a 的取值范围； （2）若q 是r 的必要条件，求实数m 的取值范围． 【答案】(1) （﹣∞，﹣1]，(2) [1，2]． 【解析】 【分析】（1）由命题间的关系，即求命题p 为真时，a 的取值范围，利用二次函数的性质，可求得结果；（1）求出命题,q r 为真时，x 的集合，q 是r 的必要条件，转化为集合间关系，即可求出m 的取值范围.【详解】p ：∀x ∈R ，x 2+2x ≥a ，q ：x 2﹣4x +3≤0，r ：（x ﹣m ）[x ﹣（m +1）]≤0， ∴根据二次函数的性质可知，x 2+2x 的最小值﹣1， 故P ：a ≤﹣1，由x 2﹣4x +3≤0可得1≤x ≤3，由（x﹣m）[x﹣（m+1）]≤0，可得m≤x≤m+1，故q：A＝[1，3]，r：B＝[m，m+1]，（1）若命题p的否定是假命题，即p为真命题，故a的范围（﹣∞，﹣1]，（2）若q是r的必要条件，则r⇒q，从而有B⊆A，∴113 mm≥⎧⎨+≤⎩，解可得，1≤m≤2，故m的范围[1，2]．【点睛】本题考查命题与命题的否定的真假关系，考查必要条件与集合集合间的关系，属于基础题.18.已知双曲线2213yx-=的右顶点为A，抛物线的焦点与点A重合．（1）求抛物线的标准方程；（2）若直线l过点A且斜率为双曲线的离心率，求直线l被抛物线截得的弦长．【答案】(1) y2＝4x；(2)5【解析】【分析】（1）由双曲线的标准方程得右顶点坐标，即抛物线焦点坐标，可求抛物线标准方程；（2）根据已知条件写出直线l方程，与抛物线方程联立，结合抛物线的定义,即可求出过抛物线焦点的相交弦长.【详解】（1）由双曲线2213yx-=，得a＝1，∴抛物线的焦点即双曲线的右顶点A为（1，0），则抛物线的标准方程为y2＝4x；（2）由双曲线方程可得，a＝1，2c=，则直线l的斜率为2．∴直线l的方程为y＝2（x﹣1），即y＝2x﹣2．联立2224y x y x=-⎧⎨=⎩，得x 2﹣3x +1＝0，9450∆=-=>， 设两交点横坐标分别为12,x x ，则123x x +=，∴直线l 被抛物线截得的弦长为x 1+x 2+p ＝3+2＝5．【点睛】本题考查双曲线简单几何性质、抛物线的标准方程以及直线与抛物线相交弦长，考查计算能力，属于中档题.19.已知等比数列{a n }满足a 1+a 4＝18，a 2+a 5＝36．（1）求数列{a n }的通项公式a n ；（2）若数列{b n }满足b n ＝a n +log 2a n ，求数列{b n }的前n 项和S n ．【答案】(1) a n ＝2n ．(2) S n =2n +1﹣2()12n n ++．【解析】【分析】（1）根据已知条件求出等比数列公比和首项，即可求出通项公式a n ；（2）先求{b n }的通项公式，转化为求等差数列和等比数列的前n 项和，可求出S n ．【详解】（1）设等比数列{a n }的公比为q ，∵a 1+a 4＝18，a 2+a 5＝36．∴a 1（1+q 3）＝18，q （a 1+a 4）＝18q ＝36，解得q ＝2＝a 1，∴a n ＝2n ．（2）由（1）可得：b n ＝a n +log 2a n ＝2n +n ．∴数列{b n }的前n 项和 S n ()()2121122nn n -+=+=-2n +1﹣2()12n n ++．【点睛】本题考查等比数列的通项公式，以及等差、等比数列的前n 项和，属于基础题.20.如图，GH 是东西方向的公路北侧的边缘线，某公司准备在GH 上的一点B 的正北方向的A 处建一仓库，设AB = y km ，并在公路同侧建造边长为x km 的正方形无顶中转站CDEF （其中边EF 在GH 上），现从仓库A 向GH 和中转站分别修两条道路AB ，AC ，已知AB = AC 1，且∠ABC = 60o ．（1）求y 关于x 的函数解析式；（2）如果中转站四周围墙造价为1万元/km ，两条道路造价为3万元/km ，问：x 取何值时，该公司建中转站围墙和两条道路总造价M 最低？【答案】（1）2412(1)x y x -=-（x ＞ 1）；（2）74x =时，该公司建中转站围墙和道路总造价M 最低．【解析】试题分析：(1)利用题意结合余弦定理可得函数的解析式24122x y x -=-（，其定义域是(1,)+∞. (2)结合(1)的结论求得利润函数，由均值不等式的结论即可求得当km 时，公司建中转站围墙和两条道路最低总造价为490万元. 试题解析：（1）在BCF ∆中,,30,CF x FBC CF BF =∠=︒⊥,所以2BC x =.在ABC ∆中，,1,60AB y AC y ABC ==-∠=︒，由余弦定理，得2222cos AC BA BC BA BC ABC =+-⋅∠， 即 222(1)(2)22cos 60y y x y x （-=+-⋅⋅︒， 所以 24122x y x -=-（. 由AB AC BC -<， 得121,2x x >>. 又因为241022x y x -=>-（，所以1x >.所以函数24122x y x -=-（的定义域是(1,)+∞. (2)30(21)40y x =⋅-+ .因为24122x y x -=-（（1x >）， 所以24130(21)4022x M x x -=⋅⋅-+-（ 即 212310(4-1)1x M x x -=⋅+-（）. 令1,t x =-（则. 于是 ，由基本不等式得， 当且仅当34t =（，即74x （=时取等号. 答：当km 时，公司建中转站围墙和两条道路最低总造价为490万元. 21.如图，已知过点222D ⎛⎫- ⎪ ⎪⎭，的椭圆()222210x y a b a b +=＞＞的离心率为3，左顶点和上顶点分别为A ，B ．（1）求椭圆的标准方程；（2）若P 为线段OD 延长线上一点，直线PA 交椭圆于另一点E ，直线PB 交椭圆于另一点Q ． ①求直线PA 与PB 的斜率之积；②判断直线AB 与EQ 是否平行？并说明理由．【答案】(1)22 4x y +=1．(2) ① 14．②平行．理由见解析 【解析】【分析】（1）离心率值转化为,a b 关系，再把点D 坐标代入方程，即可求出椭圆标准方程；（2）①求出OD 方程，设出P 点坐标，可求出直线PA 与PB 的斜率之积；②求出直线,AP BP 方程，分别与椭圆方程联立，求出,E Q 两点坐标，代入斜率公式，求出直线EQ 的斜率，然后再判断与直线AB 是否平行.【详解】（1）∵椭圆过点D，2-）,且离心率为22222222222223411,,44c a b b b e a b a a a a -===-=∴=∴= ∴22222212111,1,2242b a a b b b b +=+==∴==， ∴椭圆的方程为224+=x y 1．（2）①由（1）知A （﹣2，0），B （0，1），直线OD 方程为y 12x =-，点P 在直线OD 上，设P （﹣2y 0，y 0），k PA •k PB 0000112224y y y y -=⋅=-+-．②设E （x 1，y 1），Q （x 2，y 2），联立直线AP ：y ()00222y x y =+-+与椭圆的方程得，（2y 02﹣2y 0+1）x 2+4y 02x +8y 0﹣4＝0，∴﹣2+x 1202004221y y y =--+，∴x 1020042221y y y -+=-+，y 120020022221y y y y -+=-+，联立直线BP ：y 0112y x y -=+与椭圆的方程得，22000200221440y y y x x y y -+-+=，∴x 220020044221y y y y -=-+，y 2020021221y y y -=-+， ∴20122120211422EQ y y y k x x y -+-===--+ 又因为k AB 12=，∴k AB ＝k EQ ， ∴直线AB 与EQ 是平行．【点睛】本题考查椭圆的标准方程，考查直线与圆锥曲线相交点坐标的求法，以及斜率间的关系，考查计算能力，属于难题.22.已知数列{a n }的前n 项和24n n n S +=． （1）求数列{a n }的通项公式a n ；（2）设数列{b n }的前n 项和为T n ，满足b 1＝1，1111122 21n nk n k k k b T T +++=+-=-∑． ①求数列{b n }的通项公式b n ；②若存在p ，q ，k ∈N *，p ＜q ＜k ，使得a m b q ，a m a n b p ，a n b k 成等差数列，求m +n 的最小值．【答案】(1) a n 2n =．(2) ①b n ＝2n ﹣1；②7 【解析】【分析】（1）根据前n 项和与通项的关系，即可求出通项公式；（2）①将11n k n T b T ++=-代入递推公式中，用裂项相消求出n T ，再由前n 项和求出通项n b ； ②由等差数列的中项性质，求出,m n 的不等量关系，结合基本不等式，即可得到m n +最小值． 【详解】（1）∵数列{a n }的前n 项和24n n n S +=． ∴当n ＝1时，a 1＝S 112=， 当n ≥2时，a n ＝S n ﹣S n ﹣1()22(1)1442n n n n n -+-+=-=， 当1n =时，a 112=，满足上式，∴a n 2n =． （2）①∵111111 n n k k k k k k k k k b T T T T T T ++==++-=∑∑1111 n k kk T T =+⎛⎫=- ⎪⎝⎭∑ =（1211T T -）+（2311T T -）+（3411T T -）+…+（111n n T T +-） 1111n T T +=-=111n T +-． ∴11111122112121n n n n T ++++--==---， ∴T n +1＝2n +1﹣1，T n ＝2n ﹣1，把上面两式相减得，b n +1＝2n ，∴2n ≥时，12n n b -=，当1n =时，11b =满足上式，12n n b -∴=②由a m b q ，a m a n b p ，a n b k 成等差数列，有2a m a n b p ＝a m b q +a n b k ，即222m n •12p -2m =•12q -2n +•12k -， 由于p ＜q ＜k ，且为正整数，所以q ﹣p ≥1，k ﹣p ≥2，所以mn ＝m •2q p -+n •2k p -≥2m +4n ，可得 mn ≥2m +4n ，24n m+≤1，2424()()66m n m n m n n m n m+≥++=++≥+,,m n N m n ∈∴+的最小值为12，此时8,4m n ==或7,5m n ==或6,6m n ==，m n ∴+的最小值为12.【点睛】本题考查了求数列的前n 项和以及通项公式，考查了推理能力与计算能力，属于难题．。
江苏省盐城市2019-2020学年高三上学期期中物理试卷 (含解析)
江苏省盐城市2019-2020学年高三上学期期中物理试卷一、单选题（本大题共5小题，共15.0分）1.如图所示，汽车以恒定的速率运动，分别通过凸形桥面的最高点a、水平桥b及凹形桥面最低点c，桥面粗糙情况是相同的，关于汽车发动机功率大小的下述说法中正确的是()A. 汽车通过a点时功率最大，通过c点时最小B. 汽车通过b点时功率最大C. 汽车通过a点时功率最小，通过c点时功率最大D. 汽车功率不变2.如图所示，B为竖直圆轨道的左端点，它和圆心O的连线与竖直方向的夹角为α。
一小球在圆轨道左侧的A点以速度v0平抛，恰好沿B点的切线方向进入圆轨道。
已知重力加速度为g，则AB之间的水平距离为()A. v02tanαg B. 2v02tanαgC. v02gtanαD. 2v02gtanα3.对于做匀速圆周运动的质点，下列说法正确的是()A. 根据公式a=v2r，可知其向心加速度a与半径r成反比B. 根据公式a=ω2r，可知其向心加速度a与半径r成正比C. 根据公式ω=2πn，可知其角速度ω与转数n成正比D. 根据公式ω=vr，可知其角速度ω与半径r成反比4.如图所示，PQ为某电场中的一条电场线，下列说法正确的是()A. 该电场一定是匀强电场B. 该电场一定是点电荷产生的电场C. P点的电场强度一定比Q点的大D. 正电荷在P点受到的电场力方向沿着电场线方向5.质量为1kg的物体在水平粗糙的地面上受到一水平外力F作用运动，如图甲所示，外力F做功和物体克服摩擦力做功与物体位移的关系如图乙所示，重力加速度g为10m/s2.下列分析错误的是()A. s=9m时，物体速度为3m/sB. 物体运动位移为13.5mC. 前3m运动过程中物体的加速度为3m/s2D. 物体与地面之间的动摩擦因数为0.2二、多选题（本大题共5小题，共20.0分）6.如图所示，蹦床运动员从最高点由静止下落，到接触床面，直至下降到最低点的过程中，不计空气阻力，则()A. 从最高点到接触床面前，运动员做自由落体运动B. 从最高点到接触床面前，运动员处于完全失重状态C. 当刚接触蹦床时，运动员的速度最大D. 从接触床面到下降到最低点的过程中，运动员一直处于超重状态7.根据开普勒关于行星运动的规律和圆周运动知识知：太阳对行星的引力F∝mr2，行星对太阳的引力F′∝Mr2，其中M、m分别为太阳和行星的质量，r为太阳与行星间的距离。
2019-2020学年江苏省盐城中学高三英语上学期期中试题及答案解析
2019-2020学年江苏省盐城中学高三英语上学期期中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AThe Costa Book Awards consistently pick winners that are both of the moment and subsequently endure. It's our pleasure to confirm this year’s Category Winners.First Novel Award WinnerBook: Eleanor Oliphant is Completely FineAuthor: Gail HoneymanEleanor is 31 years old; work finishes on a Friday and begins again on a Monday. Between, her only company will be two bottles of vodka and her own solitary, unique wit (机智). It is contentment, of a kind, but an unexpected shared experience suddenly opens the door to possibility. Challenging reader expectations with a living, breathing character, Gail Honeyman’s debut (初次登台、开张)is a funny and moving diamond.Biography Award WinnerBook: In the Days of RainAuthor: Rebecca StottThe Exclusive Brethren were aclosed community who believed the world is ruled by Satan. Into this is born Rebecca. Her father had been an influential Brethren Minister. As her father lay dying, he begged her to help him write the memoir. He wanted to tell the story of their family who for generations had all been members of a fundamentalist Christian sect.Poetry Award WinnerBook: Inside the WaveAuthor: Helen DunmoreTo be alive is to be inside the wave, always travelling until it breaks and is gone. These poems are concerned with the borderline between the living and the dead — the underworld and the human living world – and the acutely intense being of both.Children's Award WinnerBook:The ExplorerAuthor: Katherine Rundell, Hannah HornFour children survive their aircraft plunging into the Amazon jungle, but for Fred and his friends it’s only the beginning of a cruel battle for survival. Brimming with adventure and a real command of character and incident, Rundell has few peers in superb children's fiction.1.What kind of life does Eleanor lead?A.boring and lonely.B.funny and touching.C.exciting and complex.D.ordinary and happy.2.Why did Rebecca Stott writeIn the Days of Rain?A.To introduce beliefs of the Exclusive Brethren.B.To help her father fulfill his last wish.CTo share the life of fundamentalist Christians.D.To pass on her family traditions.3.For a young adventurous soul, which book seems more appealing?A.Eleanor Oliphant is Completely FineB.In the Days of RainC.Inside the WaveD.The ExplorerBImagine that youare a superhero. Your superpowers are activated by a special suit. The suit communicates with your brain. It allows you to do amazing things with only a thought. By concentrating on strength, for example, you can kick a soccer ball across a field. By focusing on swift actions, you can jump to the top of a tree.Such a connection between mind and machine may sound like a fantasy. To scientists, though, it is a very real goal. They are creating machines that let disabled monkeys walk. These machines may soon help disabled humans do the same. Unlike other bionic devices, these robotic “super suits” do not communicate with muscles and nerves. Instead, they have a direct line to the brain.In 2005, doctors drilled a hole in the skull of Hutchinson, who had lost her right arm in an accident. Then they inserted a sensor onto her motor cortex (大脑皮层运动区). Wires connected the sensor to a receiver on her head. After she recovered, researchers pluggedHutchinson’s receiver into a cable that relayed signals from her brain to computers. Then they connected a robotic arm to the computers. The computerscould interpretHutchinson’s brain signals to move the arm.Soon,Hutchinson, the computer, and the robotic arm became a team.Hutchinsonwas even able to lift her hand and drink from a cup. “She smiled when she put down that drink—that’s everything.” says Donoghue, a brain scientist.Today other scientists are building on that success. One of those scientists is Dr. Miguel Nicolelis, who designed a whole-body bionic equipment. In 2014, a disabled former athlete kicked the first ball of the World Cup Games wearing one of Miguel’s full-body exoskeletons.The exoskeleton was connected to brain signal sensors in the man’s cap. By thinking about kicking, he sent signals to a computer on his back. The computer then translated the signal into an exoskeleton-aided kick. Such designs may become common as scientists keep merging mind and machine.4.Which can best describe the tone of paragraph 1?A. Narrative and serious.B. Persuasive and critical.C. Informative and objective.D. Descriptive and imaginative.5. What is paragraph 3 mainly about?A. Reason of the experiment.B. Results of the experiment.C. Process of the experiment.D. Significance of the experiment.6. Why is Dr. Miguel’s exoskeleton special?A. It can be used on animals.B. It can move the whole body.C. It was supported by computers.D. It was connected directly to the brain.7. What can be inferred from the last paragraph?A. Robotic suits may be widely used to help disabled people.B. Exoskeleton is more common thanHutchinson’s bionic arm.C.Scientific experiments are commonly carried out in football fields.D. Disabled athletes can now play football again with the help of computer.CAn anti-obesity program for Australian girls didn’t lead to any improvements in their diet, physical activities or body weight a year later, according to a new report.Findings from the school-based intervention (介入), which involved exercise sessions and nutrition workshops for lower-income girls, are the latest disappointment in a lot of research attempting tohead offadult obesity and the disease risks that come with it.Especially during the middle-and high-school years, girls’ physical activity reduces obviously, according to lead researcher David Lubans, from theUniversityofNewcastleinNew South Wales,Australia. He said, “In the future we need to make the programs more interesting and exciting and present information in a way that is meaningful to adolescent girl.”Lubans and his workmates conducted their study in 12 schools in low-income areas ofNew South Wales. At the start of the study, girls in both groups weighed an average of close to 130pounds, with about four in ten considered overweight. Over the next year, adolescents in the intervention group were given pedometers (计步器) to encourage walking and running and invited to nutrition workshops and regular exercise sessions during the schoolday and at lunchtime. Participation in some of those activities were less than ideal. For example, the girls went to only one-quarter of lunchtime exercise sessions, and less than one in ten completed at-home physical activity or nutrition challenges, the researchers reported. At the end of the year, girls in both groups had gained a similar amount of weight and there was no difference in their average body fat.Preventive medicine researcher Robert Klesges said that although some anti-obesity programs have helped adults lose weight, the teen population has always been a source of failure for researchers. “The common belief is: nothing works,” he said. “And we have got to get beyond that.”“We need to think outside the box,” said Klesges, who wasn’t involved in the new study. “That could include learning from what has worked in adult studies, such as giving meal replacement drinks or prepared foods to teens who have trouble making changes to their diet. Or, it could mean using a “step-care” method — rather than researchers or their doctor telling them to keep doing the same thing.” Klesges said.8. The underlined words “head off” in Paragraph 2 can best be replaced by “________”.A. damageB. defendC. preventD. affect9. The methods used in the program to stop obesity don’t include ________.A. walking and runningB. inviting them to nutrition workshopsC. joining exercise sessions regularlyD. giving meal replacement drinks10. The main reason for the failure of the anti-obesity program is probably that ________.A. the participants didn’t take an active part in itB. the program was not interesting and exciting to participantsC. the participants didn’t get extra nutrition or exercise helpD. the program didn’t pay attention to healthy exercise11. What can be inferred from the last paragraph?A. As researchers, it is important to have creative research methods.B. Researchers need to give meals or prepare foods to participants.C. Teen girls have no difficulty in making changes to their diet.D. Some ant-obesity programs have not helped adults lose weight.DThe history of the flying car is almost as old as that of powered flight itself. It started with the Curtiss Autoplane of 1917, an awkward-looking machine with removable wings. It never left the ground. Later machines made it into the skies but failed to take off commercially. Money is now pouring into flying taxis. On March 30th Lilium, a German company that develops them, announced a merger with SPAC, an acquisition company that values it at $3. 3 bn -- a sign that investors think the business will fly.Thanks to better batteries and lightweight materials, some of them are ready to carry passengers. Up to 300 firms are working on short-range battery-powered craft that take off and land vertically. Carmakers, tech companies and others are investing money into the field. The government isoffering a glide pathto certification.America's Federal Aviation Administration is engaged in the process with around 30 firms, says Natasha Santha of LEK, a consulting company.Midway between a cab and a helicopter, flying taxis have distinct advantages over both. Quiet electric motors allow them to operate frequent services. They require only a patch of concrete to land, unlike noisy helicopters, which face severe operating restrictions in most cities. They can fly four or five times faster than a cab can drive and do not get stuck in traffic. Prices can be kept low by ride-sharing. Joby, based inCalifornia, says its five-seater machine will enter commercial service in 2024. The firm calculates the initial cost of around $4 per person per mile may soon fall by 25%. A trip fromManhattanto JFK airport would then cost $30-40 per passenger.The real revolution will come when full autonomy takes out the cost of a pilot. Archer hopes to run such aircraft by 2028. They face fewer obstacles in the air than earth-bound cars do on the road; airliners mostly fly on autopilot as it is. Still, as one industry insider puts it, it is probably best to accustom passengers and regulators to airborne taxis before getting rid of the driver.12. What can we learn from Paragraph 1?A. The flying car can date back to the 1920s.B. Investors see the potential of the business of the flying car.C. The flying car never left the ground successfully in history.D. A German company has launched a new flying car into the market.13. What does "offering a glide path" in Paragraph 2 probably refer to?A. Giving the green light.B. Providing timely assistance.C. Presenting legal guidance.D. Conducting strict management.14. Which of the following is the strength of flying taxis?A. Costing as little as cabs.B. Saving passengers from the traffic jam.C. Reducing air pollution.D. Having no operating restrictions.15. What can be inferred about the flying taxis from the last paragraph?A. They will develop faster than cars.B. Passengers will quickly get used to taking them.C. The regulators will take measures to promote them.D. Autopilot flying taxis will probably replace those with pilots.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学年江苏省盐城中学高二(上)期中数学试卷 (含答案解析)
2019-2020学年江苏省盐城中学高二（上）期中数学试卷一、选择题（本大题共12小题，共60.0分）1.已知命题p：∀x∈N∗，2x>x2，则￢p是()A. ∃x∈N∗，2x>x2B. ∀x∈N∗，2x≤x2C. ∃x∈N∗，2x≤x2D. ∀x∈N∗，2x<x22.已知抛物线y2=2px的准线方程是x=−2，则p的值为()A. 2B. 4C. −2D. −43.2+√3和2−√3的等比中项是()A. 1B. −1C. ±1D. 24.双曲线x24−y216=1的渐近线方程为()A. y=±2xB. y=±12x C. y=±√5x D. y=±√52x5.已知x、y均为正数，2x +8y=1，则xy有()A. 最大值64B. 最大值164C. 最小值64 D. 最小值1646.若a，b∈R，则1a2>1b2成立的一个充分不必要的条件是()A. b>a>0B. a>b>0C. b<aD. a<b7.在等差数列{a n}中，a3+a8=8，则S10=()A. 20B. 40C. 60D. 808.已知函数f(x)=x+4x ，g(x)=2x+a，若∀x1∈[12,3]，∃x2∈[2,3]，使得f(x1)≥g(x2)，则实数a的取值范围是()A. a≤1B. a≥1C. a≤0D. a≥09.已知等差数列{a n}的前n项和为S n，且−11<S11≤22，则a6的取值范围是()A. (−1,2]B. (1,2]C. [−1,2]D. [1,2]10.已知M为抛物线y2=4x上一动点，F为抛物线的焦点，定点P(3,1)，则|MP|+|MF|的最小值为()A. 3B. 4C. 5D. 611.若log2x+log2y=2，则x+2y的最小值为()A. 2B. 2√2C. 4D. 4√212.已知数列{a n}的前n项和为S n，a1=1，a2=2，且对于任意n>1，n∈N∗，满足S n+1+S n−1=2(S n+1)，则S10的值为()A. 90B. 91C. 96D. 100二、填空题（本大题共4小题，共20.0分）13.方程x22−k +y22k−1=1表示焦点在x轴上椭圆，则实数k的取值范围是______．14.已知正实数a，b，c满足1a +1b=1，1a+b+1c=1，则c的取值范围是_____．15.已知数列{a n}中，a1=1，a na n+1−a n =n(n∈N∗)，则a2016=______ ．16.已知F1，F2是椭圆x2m2+y2m2−4=1(m>2)的左，右焦点，点P在椭圆上，若|PF1|⋅|PF2|=2√3m，则该椭圆离心率的取值范围为______ ．三、解答题（本大题共6小题，共70.0分）17.已知m∈R，命题p：对∀x∈[0,8]，不等式恒成立；命题q：对∀x∈(−∞,−1)，不等式2x2+x>2+mx恒成立．(1)若命题p为真命题，求实数m的取值范围；(2)若p∧q为假，p∨q为真，求实数m的取值范围．18.已知双曲线x24−y2b2=1的右焦点与抛物线y2=12x的焦点重合，求该双曲线的焦点到其渐近线的距离．19.已知数列{b n}前n项和为S n，且b1=1，b n+1=13S n．(1)求b2，b3，b4的值；(2)求{b n}的通项公式；(3)求b2+b4+b6+⋯+b2n的值．20.某房地产商建有三栋楼宇A,B,C，三楼宇间的距离都为2千米，拟准备在此三楼宇围成的区域ABC外建第四栋楼宇D，规划要求楼宇D对楼宇B，C的视角为120∘，如图所示，假设楼宇大小高度忽略不计．(1)求四栋楼宇围成的四边形区域ABDC面积的最大值；(2)当楼宇D与楼宇B，C间距离相等时，拟在楼宇A，B间建休息亭E，在休息亭E和楼宇A，D间分别铺设鹅卵石路EA和防腐木路ED，如图，已知铺设鹅卵石路、防腐木路的单价分别为a，2a(单位：元千米，a为常数).记∠BDE=θ，求铺设此鹅卵石路和防腐木路的总费用的解析式．若θ=60∘时，求出具体费用．21.如图，椭圆E：x2a2+y2b2=1(a>b>0)的左、右顶点分别为A、B，离心率e=√53，长轴与短轴的长度之和为10．(Ⅰ)求椭圆E的标准方程；(Ⅱ)在椭圆E 上任取点P(与A 、B 两点不重合)，直线PA 交y 轴于点C ，直线PB 交y 轴于点D ，证明：OC ⃗⃗⃗⃗⃗ ⋅OD⃗⃗⃗⃗⃗⃗ 为定值．22. 已知数列{b n }的前n 项和为S n ，且b 1=1，b n+1=13S n ．(1)求b 2，b 3，b 4的值; (2)求{b n }的通项公式．-------- 答案与解析 --------1.答案：C解析：解：命题p：∀x∈N∗，2x>x2，则￢p是∃x∈N∗，2x≤x2，故选：C．欲写出命题的否定，必须同时改变两个地方：①：“∀”；②：“>”即可，据此分析选项可得答案．这类问题的常见错误是没有把全称量词改为存在量词，或者对于“>”的否定用“<”了．这里就有注意量词的否定形式．如“都是”的否定是“不都是”，而不是“都不是”.特称命题的否定是全称命题，“存在”对应“任意”．2.答案：B解析：解：抛物线y2=2px的准线方程是x=−2，则p的值：4．故选：B．利用抛物线的准线方程求出p，即可．本题考查抛物线的简单性质的应用，是基础题．3.答案：C解析：【分析】本题主要考查等比数列的应用，属于简单题．【解答】解：由题意得，(2+√3)(2−√3)=1，2+√3和2−√3的等比中项是±1，故选C．4.答案：A解析：解：由双曲线x2a2−y2b2=1(a,b>0)的渐近线方程为：y=±bax，双曲线x24−y216=1的a=2，b=4，可得渐近线方程为y=±2x．由双曲线x2a2−y2b2=1(a,b>0)的渐近线方程为y=±bax，求得双曲线的a，b，即可得到所求渐近线方程．本题考查双曲线的渐近线方程的求法，注意运用双曲线方程和渐近线方程的关系，考查运算能力，属于基础题．5.答案：C解析：解：∵x、y均为正数，2x +8y=1，∴2x +8y=1≥2√2x⋅8y，化为xy≥64，当且仅当y=4x=16时取等号．∴xy有最小值64．故选：C．利用基本不等式的性质即可得出．本题考查了基本不等式的性质，属于基础题．6.答案：A解析：【分析】由于1a2>1b2⇔a2<b2⇔|a|<|b|，因此利用充分不必要条件的概念对A，B，C，D四个选项逐一判断即可．本题考查不等式的基本性质，考查必要条件、充分条件与充要条件的判断，正确理解充分不必要条件的概念是判断的关键，属于中档题．【解答】解：∵a，b∈R，1a2>1b2⇔0<a2<b2⇔0<|a|<|b|，∴对于A，若b>a>0，则1a2>1b2，即充分性成立；反之，当|a|<|b|时，不能⇒b>a>0，即必要性不成立．∴b>a>0是1a2>1b2成立的一个充分不必要的条件，即A满足题意；同理可分析B，C，D，均是1a >1b成立的既不充分也不必要的条件；故可排除B，C，D；故选A．7.答案：B解析：解：∵在等差数列{a n}中，a3+a8=8，∴S10=102(a1+a10)=102(a3+a8)=5×8=40．由已知利用等差数列的通项公式和前n项和公式求解．本题考查等差数列的前10项和的求法，是基础题，解题时要认真审题，注意等差数列的性质的合理运用．8.答案：C解析：【分析】本题考查指数函数以及对勾函数的图象和性质，考察导数的应用，函数的单调性问题，属于中档题．由∀x1∈[12,3]，都∃x2∈[2,3]，使得f(x1)≥g(x2)，可得f(x)在x1∈[12,3]的最小值不小于g(x)在x2∈[2,3]的最小值，构造关于a的不等式，可得结论．【解答】解：当x1∈[12,3]时，由对勾函数的性质可得：f(x)在[12,2]上单调递减，在(2,3]上单调递增，∴f(2)=4是函数的最小值；当x2∈[2,3]时，g(x)=2x+a为增函数，∴g(2)=a+4是函数的最小值，又∵∀x1∈[12,3]，都∃x2∈[2,3]，使得f(x1)≥g(x2)，可得f(x)在x1∈[12,3]的最小值不小于g(x)在x2∈[2,3]的最小值，即4≥a+4，解得：a≤0．故选C．9.答案：A解析：【分析】本题考查等差数列的性等基础知识，考查运算求解能力，考查函数与方程思想，是基础题．由等差数列{an}的前n项和公式、通项公式得到−11<=11a6≤22，由此能求出a6的取值范围．【解答】解：∵等差数列{a n}的前n项和为S n，且−11<S11≤22，∴−11<112(a1+a11)=11a6≤22，解得−1<a6≤2，∴a6的取值范围是(−1,2]．故选：A．解析：【分析】本题考查了双曲线的几何意义，根据到焦点的距离等于到准线的距离进行解答．【解答】解：抛物线y2=4x的准线l的方程为x=−1，焦点坐标为(1,0)，由抛物线几何性质知|MF|为点M 到直线l的距离，则有点P到直线l的距离d=4，如下图所示，则有|MP|+|MF|≥d=4，当点M 为点P到直线l的垂线与抛物线的交点时等号成立．故选B．．11.答案：D解析：【分析】本题考查了对数的运算和基本不等式，属基础题．根据log2x+log2y=2，求出xy的值，然后直接利用基本不等式求解x+2y．【解答】解：∵log2x+log2y=2，∴log2xy=2，∴xy=4，x>0，y>0，∴x+2y≥2√2xy=4√2，当且仅当x=2y=2√2，即x=2√2，y=√2时取等号．∴x+2y的最小值为4√2．故选D．12.答案：B解析：【分析】本题考查了数列递推关系、等差数列的通项公式与求和公式，考查了推理能力与计算能力，属于中档题．对于任意n>1，n∈N∗，满足S n+1+S n−1=2(S n+1)，可得S n+1−S n=S n−S n−1+2，可得a n+1−a n=2.利用等差数列的通项公式与求和公式即可得出．【解答】解：∵对于任意n>1，n∈N∗，满足S n+1+S n−1=2(S n+1)，∴S n+1−S n=S n−S n−1+2，∴a n+1−a n=2．∴数列{a n}在n≥2时是等差数列，公差为2．则S10=1+9×2+9×82×2=91．故选：B．13.答案：(12,1)解析：解：∵方程x22−k +y22k−1=1表示焦点在x轴上的椭圆，∴2−k>2k−1>0，解得12<k<1．∴实数k的取值范围是(12,1)．故答案为：(12,1)．直接由题意列关于k的不等式组得答案．本题考查椭圆的简单性质，考查了曲线方程表示椭圆的条件，是基础题．14.答案：(1,43]解析：【分析】本题主要考查利用基本不等式求最值，属于基础题．因为a，b是正实数，且1a +1b=1，则a+b=ab⩾2√ab，当且仅当a=b=2时等号成立，解得ab≥4.又由1a+b +1c=1得1ab+1c=1，可得c的取值范围．【解答】解：因为a，b是正实数，且1a +1b=1，则a+b=ab⩾2√ab，当且仅当a=b=2时等号成立，解得ab≥4．又由1a+b +1c =1得1ab +1c =1， c =abab−1=1+1ab−1∈(1,43]． ∴c 的取值范围为(1,43]. 故答案为(1,43].15.答案：2016解析： 【分析】本题考查了递推关系、数列的通项公式，考查了推理能力与计算能力，属于中档题．由an a n+1−a n =n(n ∈N ∗)，可得：na n+1=(n +1)a n ，又a 1=1，可得a n =n.即可得出．【解答】解：∵an a n+1−a n =n(n ∈N ∗)，∴na n+1=(n +1)a n ，化为：a n+1a n=n+1n，∴n ≥2时，a n =a na n−1·a n−1a n−2⋯a 2a 1·a 1=nn−1·n−1n−2⋯21·1，∴a n =n ． ∴a 2016=2016．故答案为：2016．16.答案：[√7−√32,√33]解析：解：由椭圆的定义可得|PF 1|+|PF 2|=2m ， ∴2m =|PF 1|+|PF 2|≥2√|PF 1||PF 2|=2√2√3m ， 化为m 2≥2√3m ，又m >2， 解得m ≥2√3．另一方面：设∠F 1PF 2=θ，由余弦定理可得：|PF 1|2+|PF 2|2−2|PF 1||PF 2|cosθ=(2c)2=16． |PF 1|2+|PF 2|2+2|PF 1||PF 2|=4m 2． 相减可得：1+cosθ=2√3m．∵θ∈[0,π)，∴0<2√3m≤2.m ≥2√3∴2≤m ≤√3+√7．∴e =c a =√1−m 2−4m 2=2m ∈[√7−√32,√33]， ∴该椭圆离心率的取值范围为[√7−√32,√33]， 故答案为：[√7−√32,√33]． 由椭圆的定义可得|PF 1|+|PF 2|=2m ，利用基本不等式的性质可得：|PF 1|+|PF 2|≥2√|PF 1||PF 2|，化简整理即可得出．另一方面：设∠F 1PF 2=θ，由余弦定理可得：|PF 1|2+|PF 2|2−2|PF 1||PF 2|cosθ=(2c)2=16．|PF 1|2+|PF 2|2+2|PF 1||PF 2|=4m 2.相减利用三角函数的单调性、不等式的解法即可得出．本题考查了椭圆的定义标准方程及其性质、基本不等式的性质，考查了推理能力与计算能力，属于中档题．17.答案：解：(1)令，则f (x )在上为减函数， 因为x ∈[0,8]，所以当x =8时，， 不等式恒成立，等价于−2≥m 2−3m ，解得1≤m ≤2，故命题p 为真，实数m 的取值范围为[1,2]；(2)若命题q 为真，则，对上恒成立， 令，因为g(x)在上为单调增函数， 则，故m ≥1，即命题q 为真，m ≥1，若p ∧q 为假，p ∨q 为真，则命题p ，q 中一真一假，①若p 为真，q 为假，那么{1<m <2m <1，则无解， ②若p 为假，q 为真，那么{m <1或m >2m ≥1，则m >2． 综上m 的取值范围为．解析：本题主要考查全称量词命题，存在量词命题的真假判定，复合命题真假的判定．(1)令，则f(x)在上为减函数，，结合已知条件，问题等价于−2≥m 2−3m ，即可解得实数m 的取值范围；(2)由已知可得命题p ，q 中一真一假，然后分类讨论即可解得实数m 的取值范围．18.答案：解：∵抛物线y 2=12x 的p =6，开口方向向右，∴焦点是(3,0)，∵双曲线x 24−y 2b 2=1的右焦点与抛物线y 2=12x 的焦点重合， ∴4+b 2=9，∴b 2=5∴双曲线的渐近线方程为y =±√52x ，即√5x ±2y =0∴双曲线的焦点到其渐近线的距离为|3√5−0|3=√5．解析：先求出抛物线y 2=12x 的焦点坐标，由此得到双曲线的右焦点，从而求出b 的值，进而得到该双曲线的离心率与渐近线方程，从而可求该双曲线的焦点到其渐近线的距离．．本题考查双曲线的性质和应用，考查了学生对基础知识的综合把握能力，属于中档题． 19.答案：解：(1)b 2=13S 1=13b 1=13，b 3=13S 2=13(b 1+b 2)=49， b 4=13S 3=13(b 1+b 2+b 3)=1627 ;(2)∵b n+1=13S n ． ∴b n =13S n−1(n ≥2)， 两式相减可得，b n+1−b n =13b n ，∴b n+1=43b n ，∵b 2=13， ∴b n =13⋅(43)n−2(n ≥2)， ∴b n ={1,n =113⋅(43)n−2,n ≥2 ; (3)b 2，b 4，b 6…b 2n 是首项为13，公比(43)2的等比数列，∴b 2+b 4+b 6+⋯+b 2n=13[1−(43)2n ]1−(43)2=37[(43)2n −1]．解析：本题主要考查了等比数列的通项公式及求和公式，等比数列的性质的应用，数列的递推公式的应用是解答本题的关键．(1)由b 1=1，b n+1=13S n .分别令n =1，2，3可求;(2)由题意可得b n+1=13S n .b n =13S n−1(n ≥2)，两式相减，结合等比数列的通项公式可求;(3)由(2)可得b 2，b 4，b 6…b 2n 是首项为13，公比(43)2的等比数列，结合等比数列的求和公式可求． 20.答案：解：(1)由题意，AB =AC =BC =2，∠BDC =120°，由余弦定理得，BC 2=BD 2+DC 2−2BD ·DC ·cos120°，即4=BD 2+DC 2+BD ·DC ，即4−BD ·DC =BD 2+DC 2≥2BD ·DC ，(当且仅当BD =DC =23√3时取等号)所以，BD·DC≤43，所以，≤√3+12×43×√32=43√3，即四边形区域ABCD面积的最大值为43√3平方千米．(2)当楼宇D与楼宇B、C间距离相等时，由(1)得：BD=DC=23√3，则∠DBC=∠DCB，又∠BDC=120°，所以∠DBC=30°，由等边三角形ABC得∠ABC=60°，所以∠ABD=∠ABC+∠DBC=90°，在RtΔEBD中，∠BDE=θ，所以，，则，所以铺设此鹅卵石路和防腐木路的总费用为：，当时，．答：铺设此鹅卵石路和防腐木路的总费用的解析式为；当时，铺设鹅卵石路和防腐木路的总费用为8√33a元.解析：本题考查解三角形的实际应用、余弦定理和利用基本不等式求最值，属较难题型．(1)根据题意，结合余弦定理求得4=BD2+DC2+BD·DC，结合基本不等式得BD·DC≤43，即可求得；(2)由题意可得∠DBC=30°，∠ABD=∠ABC+∠DBC=90°，从而在RtΔEBD中得铺设此鹅卵石路和防腐木路的总费用为，代入即可得出结果．21.答案：解：(Ⅰ)由题可知e =c a =√53，2a +2b =10，解得a =3，b =2． 故椭圆E 的标准方程为E ：x 29+y 24=1证明(Ⅱ)：设P(x 0,y 0)，直线PA 交y 轴于点C(0,y 1)，直线PB 交y 轴于点D(0,y 2).则x 029+y 024=1，即9y 029−x 02=4． 易知OC ⃗⃗⃗⃗⃗ 与OD ⃗⃗⃗⃗⃗⃗ 同向，故OC ⃗⃗⃗⃗⃗ ⋅OD⃗⃗⃗⃗⃗⃗ =y 1y 2． 因为A(−3,0)，B(3,0)，所以得直线PA 的方程为y−y 0−y 0=x−x 0−3−x 0，令x =0，则y 1=3y 03+x 0； 直线PB 的方程为为y−y 0−y 0=x−x 03−x 0，令x =0，则y 2=3y03−x 0 所以故OC ⃗⃗⃗⃗⃗ ⋅OD ⃗⃗⃗⃗⃗⃗ =y 1y 2=9y 029−x 02=4，为定值．解析：(Ⅰ)由e =c a =√53，2a +2b =10，解得a =3，b =2.，进而得到椭圆方程； (Ⅱ)设P(x 0,y 0)，直线PA 交y 轴于点C(0,y 1)，直线PB 交y 轴于点D(0,y 2)，求得直线PA ，PB 的方程，分别求出y 1，y 2，再根据向量的数量积即可证明本题考查椭圆的方程的求法，注意运用联立直线求交点，考查向量的数量积的坐标表示，考查化简整理的运算能力，属于中档题．22.答案：解：(1)∵b 1=1，b n+1=13S n ，∴b 2=13b 1=13，b 3=13(b 1+b 2)=49，b 4=13(b 1+b 2+b 3)=1627． (2)当n ≥2时，b n+1−b n =13S n −13S n−1=13b n ，可得b n+1=43b n ，∴b n =13×(43)n−2．∴b n ={1,n =113×(43)n−2,n ≥2．解析：本题主要考查数列通项公式b n 与前n 项和S n 的关系，以及数列通项公式的求法，属于基础题．(1)根据b 1=1，b n+1=13S n ，可求出b 2，b 3，b 4的值；(2)利用b n 与前n 项和S n 的关系b n ={S 1,n =1S n −S n−1,n ≥2可求出通项公式．。
江苏省盐城中学2019-2020学年高二上学期期中考试物理试题(必修)(解析版)
江苏省盐城中学2019—2020学年度第一学期期中考试高二年级物理（必修）试卷（2019.11）一、选择题：本大题共27小题，每小题3分，共计81分。
在每小题的四个选项中，只有一个选项符合题目要求。
1. 将近1000年前，宋代诗人陈与义乘着小船在风和日丽的春日出游时写下诗句“飞花两岸照船红，百里榆堤半日风，卧看满天云不动，不知云与我俱东．”，请问诗句中的“云与我俱东”所对应的参考系是A. 两岸 B. 船 C. 云 D. 诗人【答案】A【解析】云和我（诗人）都在向东运动，是假定两岸不动，以两岸为参考系。
所以答案选A。
2. 关于质点，下列说法正确的是（）A. 体积大、质量小的物体就是质点B. 质量大、体积大的物体不可视为质点C. 汽车可以视为质点，火车不可视为质点D. 研究月球绕地球运行，可将月球视为质点【答案】D【解析】试题分析：解决本题要正确理解质点的概念：质点是只计质量、不计大小、形状的一个几何点，是实际物体在一定条件的科学抽象，能否看作质点物体本身无关，要看所研究问题的性质，看物体的形状和大小在所研究的问题中是否可以忽略．解：A、质量很小的物体它的体积不一定能够忽略，不一定能看成质点，如原子的质量很小，在研究原子内部结构的时候是不能看成质点的，故A错误．B、质量大、体积大的物体也可能看成质点，例如研究地球的公转，故B错误C、物体能否看作质点，不是看本身体积和质量大小，而是看物体的形状和大小在所研究的问题中是否可以忽略．故C错误D、研究月球绕地球运行，此时能看成质点，故D正确故选D．【点评】质点是运动学中一个重要概念，要理解其实质，不能停在表面．3.下列各组物理量中，包含一个标量与一个矢量的是A. 位移、加速度B. 力、动能C. 功率、速率D. 功、势能【答案】B。
【解析】【分析】考查矢量和标量【详解】A．位移和加速度都是矢量，A错误；B．力是矢量，动能是标量，B正确；C．功率和速率都是标量，C错误；D．功和势能都是标量，D错误。
2019-2020学年江苏省盐城中学高三英语上学期期中试卷及答案
2019-2020学年江苏省盐城中学高三英语上学期期中试卷及答案第一部分阅读（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ANothing beats live music, but the venue makes a difference. When you're able to score tickets to an incredible concert in an incredible place, you won't forget the experience. Here are some of the coolest music venues from around the world. If you haven't been to any of these, you've got some traveling to do.Red Rocks, Morrison, the United StatesRed Rocks might be the most beautiful and famous venue in the United States. At 6,450 feet above sea level, Red Rocks is a geologically formed natural stage. Its massive sandstone provides a perfect stage for jam bands. If you're into the blues and jazz, you'll have no trouble finding something in line with your interests.Meet Factory, Prague, Czech RepublicSmallest venues on this list, Meet Factory is an art gallery, theater, and music venue. The venue only accommodates 1,000 people, so you won't see any huge names come through. Still, it's a great place to see up-and-coming local acts, and if you've got an eye for contemporary art, you'll love your time here.Arena of NÎmes, NÎmes, FranceOriginally built around A. D. 70, the Arena of Nimes presents concertgoers with an interesting question: Should they enjoy the music, or marvel at the architecture? The Arena is, after all, one of the world's best-preserved Roman theaters. Many major touring acts plan stops at the Arena of Nimes, especially during the venue's annual festival.Sydney Opera House, Sydney, AustraliaThe Sydney Opera House is one of the world's most famous performing venues. I's also one of the most distinctive buildings in Sydney, thanks to the breathtaking design by Danish architect Utzon. It hosts about 40 events per week, so whether you're into jazz, rock, classical music, or opera, you'll find something to watch.1.Where can you enjoy music in natural beauty?A.At Red Rocks.B.At Meet Factory.C.At Arena of Nimes.D.At Sydney Opera House.2.What is special about Meet Factory?A.It enjoys breathtaking scenery.B.It hosts both musical and artistic events.C.It is the largest venue of all.D.It is famous for contemporary music.3.What do the listed music venues have in common?A.They have a long history.B.They are built near the sea.C.They accommodate thousands of people.D.They are beautiful tourist attractions.BOne day, when I was working as a psychologist（心理学家）in England, an adolescent boy showed up in my office. It was David. He kept walking up and down restlessly, his face pale, and his hands shaking slightly. His head teacher had referred him to me.“This boy has lost his family,” he wrote. “He is understandably very sad and refuses to talk to others, and I’m very worried about him. Can you help?”I looked at David and showed him to a chair. How could I help him? There are problems psychology doesn’t have the answer to, and which no words can describe. Sometimes the best thing one can do is to listen openly and sympathetically（同情）The first two times we met, David didn’t say a word. He sat there, only looking up to look at the children’s drawings on the wall behind me. I suggested we play a game of chess. He nodded. After that he played chess with me every Wednesday afternoon---in complete silence and without looking at me. It’s not easy to cheat in chess, but I admit I made sure David won once or twice.Usually, he arrived earlier than agreed, took the chess board and pieces from the shelf and began setting them up before I even got a chance to sit down. It seemed as if he enjoyed my company（陪伴）. But why did he never look at me?“Perhaps he simply needs someone to share his pain with,” I thought. “Perhaps he senses that I respect his suffering.” Some months later, when we were playing chess, he looked up at me suddenly.“It’s your tum,” he said.After that day, David started talking. He got friends in school and joined a bicycle club. He wrote to me a few times, about his biking with some friends, and about his plan to get into university. Now he had really started to live his own life.Maybe I gave David something. But I also learned that one…without any words---can reach out to another person. All it takes is a hug, a shoulder to cry on, a friendly touch, and an ear that listens.4. When he first met the author, David .A. felt a little excitedB. looked a little nervousC. walked energeticallyD. showed up with his teacher5. David enjoyed being with the author because he .A. wanted to ask the author for adviceB. liked the children’s drawings in the officeC. beat the author many times in the chess gameD. needed to share sorrow with the author6. What can be inferred about David?A. He liked biking before he lost his family.B. He recovered after months of treatment.C. He went into university soon after starting to talk.D. He got friends in school before he met the author.7. What made David change?A. The author’s silent communication with him.B. His teacher’s help.C. The author’s friendship.D. His exchange of letters with the author.CAs I work in theLarkwhistle Garden in Dyer's Bay, Ontario, I take my time and the garden is teaching me about working with the earth. I recognize that there will be both successes and failures and there are many variables(变量)that affect them both.The quality of the seeds planted has a bearing on how the plants will grow. The weather can be too hot, too cold, or exactly right, and usually varies between all three. Weeds seem to grow in the garden and need to be taken care of, pulled, and removed to ensure they do not eat into the fruits, vegetables, and flowers we have so lovingly planted.I take time to stand back and rest, and to observe the plants and how they are growing. Each plant is unique and develops in the way that is best for them. Some have large broad leaves to shadow their fruit from the severe rays of the sun, while other plants are more open, their fruit needing the light to grow and ripen.Getting my hands dirty, feeling the sweat on my forehead, and the strength and flexibility of my body as I dig, bend and work under the warm summer sun, reminds me I am alive in ways I would not have remembered sitting on the couch.It is attractive to move things around, to transplant, and to disturb the natural order of how plants grow. The garden teaches me that it is important to know when to disturb things and when to let them be. The garden's life cycle follows a pattern that is repeated according to laws of nature, birth, growth, and then dies and it teaches us to accept this fact.8. How many variables may affect the growth of plants in the garden?A. Two.B. Three.C. Four.D. Five.9. What can we learn about the author?A. He feels exhausted while gardening.B. He enjoys life from working practice.C. He takes pleasure in harvesting fruits.D. He dreams away his time in the garden.10. How is the garden tended by the author?A. Its rank grass is got rid of.B. Its plants are left to grow freely.C. Its temperature is controlled properly.D. Its plants with large broad leaves are cut off.11. What fact does the author learn from gardening?A. Life takes its course.B. Hard work benefits health.C. Time is short and precious.D. Gardening brings good harvest.DEach year, the women of Olney and Liberal compete in an unusual footrace. Dressed in aprons (围裙) and headscarves, they wait at both towns’starting lines. Each woman holds a frying pan with one pancake inside. At the signal, the women flip (轻抛) pancakes and they’re off!This “pancake racing” tradition is said to have started on Shrove Tuesday, 1445, in Olney. Shrove Tuesday is the day beforethe Christian season of Lent (大斋戒) begins. During Lent, many people decide to give up sugary or fatty foods.Legend says that in 1445, an Olney woman was making pancakes to use up some of her sugar and cooking fats before Lent. She lost track of time and suddenly heard the church bells ring, signaling the beginning of the Shrove Tuesday service. Realizing that she was going to be late for church, she raced out the door still wearing her apron and headscarf and holding her frying pan with a pancake in it. In the following years, the woman’s neighbors imitated her dash to church, and pancake racing was born.The rules are simple. Racers must wear the traditional headscarf and apron. They must flip their pancakes twice - once before starting and once after crossing the finish line. After the race, there are Shrove Tuesday church services. Then Liberal and Olney connect through a video call to compare race times and declare a winner.In both towns, the races have grown into larger festivals. Olney’s festival is an all-day event starting with a big pancake breakfast. Liberal’s festival lasts four days and includes a parade, a talent show, and contests that feature eating and flipping pancakes. Although the women’s race is still the main event, both towns now hold additional races for boys and girls of all ages.12. How did pancake racing start?A. A woman in Olney created it.B. Women made pancakes before Lent.C. A woman dashed to church with a pancake.D. People followed the suit of an interesting incident.13. What should racers obey during the race?A. They can wear fashionable headscarves and aprons.B. They must flip their pancakes once in the race.C. They must flip their pancakes at the beginning of the race.D. They can flip their pancakes in the middle of the race.14. What can we learn about the race from the last paragraph?A. People can show their talent in Olney festival.B. People can enjoy a one-day holiday in Liberal.C. The race is not only intended for women now.D. People can have a big pancake breakfast in both towns.15. What is the text mainly about?A. The origin of pancake racing.B. The history of pancake racing.C. The development of pancake racing.D. The introduction to pancake racing.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。
江苏省盐城中学2019—2020学年度第一学期期中考试高二化学(必修)试卷(含答案)
江苏省盐城中学2019—2020学年度第一学期期中考试高二化学（必修）试卷一、选择题：本大题共26小题，每小题3分，共计78分。
在每小题的四个选项中，只有一个选项符合题目要求。
1.水(H2O)是生命之源。
下列物质的化学式可用“H2O”表示的是( )A. 水晶B. 可燃冰C. 干冰D. 冰【答案】D【解析】A. 水晶是二氧化硅，A错误；B. 可燃冰是甲烷水合物，B错误；C. 干冰是二氧化碳，C错误；D. 冰是水，化学式为H2O，D正确，答案选D。
2.摩尔是下列哪个物理量的基本单位A. 物质数量B. 物质质量C. 物质的量D. 微粒个数【答案】C【解析】【详解】物质的量是国际单位制中的基本物理量，摩尔是物质的量的单位。
答案选C。
3.下列过程中不属于．．．化学变化的是A. 煤的干馏B. 钢铁锈蚀C. 橡胶老化D. 碘的升华【答案】D【解析】【详解】A、煤干馏：煤通过干馏获得苯、甲苯等重要化工原料，有新物质生成，属于化学变化，选项A不选；B、在钢铁锈蚀过程中，有新物质铁锈生成，是化学变化，选项B不选；C、橡胶的老化是指橡胶或橡胶制品在加工、贮存或使用过程中，由于受到热、氧、光、机械力等因素的影响而发生化学变化，导致其性能下降，所以橡胶的老化属于化学变化，选项C不选；D、碘的升华是固体碘变为气态，只改变物质的状态，没有新的物质生成，属于物理变化，不属于化学变化，选项D选；答案选D。
【点睛】本题考查物理变化与化学变化的判断，要注意化学变化和物理变化的本质区别是否有新物质生成，化学变化是指有新物质生成的变化，物理变化是指没有新物质生成的变化，化学变化和物理变化的本质区别是否有新物质生成。
4.当光束通过下列分散系时，能产生丁达尔效应的是A. 淀粉溶液B. 稀硝酸C. 氨水D. 氮气和氧气的混合物【答案】A【解析】【详解】A、淀粉溶液属于胶体分散系，当光束通过时，可能产生丁达尔效应，选项A正确；B、稀硝酸属于溶液分散系，当光束通过时，不可能产生丁达尔效应，选项B错误；C、氨水属于溶液分散系，当光束通过时，不可能产生丁达尔效应，选项C错误；D、氮气和氧气的混合物不形成胶体，当光束通过时，不可能产生丁达尔效应，选项D错误；答案选A。
江苏省盐城市初级中学2019-2020学年度第一学期期中考试初二年级数学试题(答案)
盐城市初级中学2019/2020学年度第一学期期中考试初二年级数学试题（2019.11）（卷面总分：100分考试时间：100分钟）一、选择题：（本大题共8小题，每小题2分，共16分）1. 下列倡导节约的图案中，是轴对称图形的是（）A. B. C. D.2. 如图，在平面直角坐标系中，点P的坐标为（）A. （-1，2）B. （-1，-2）C. （2，-1）D. （1，2）3. 体积是2的立方体的边长是（）A. 2的平方根B. 2的立方根C. 2的算术平方根D. 2开平方的结果4. 如图，在△ABC中，AB=AC，BD=CD，下列结论不一定正确的是（）A. ∠B=∠CB. AD⊥BCC. AD平分∠BACD. AB=2BD5. 下列四组线段中，能组成直角三角形的是（）A. a=2，b=3，c=4B. a=3，b=4，c=5C. a=4，b=5，c=6D. a=7，b=8，c=96. 下列说法正确的是（）A. 实数与数轴上的点一一对应B. 无理数与数轴上的点一一对应C. 整数与数轴上的点一一对应D. 有理数与数轴上的点一一对应7. （）A. 1和2之间B. 2和3之间C. 3和4之间D. 5和6之间8. 若点P（a，b）在第四象限，则点Q（b，-a）在第几象限（）A. 第一象限B. 第二象限C. 第三象限D. 第四象限二、填空题：（本大题共10小题，每小题2分，共20分）9. 的相反数是.10. 点Q（1，4）到x轴的距离是.11. 在△ABC中，∠A=∠B=60°，AB=3，则BC等于.12. 小亮用天平称得一个罐头的质量为2.019 kg，请用四舍五入法将2.019 kg精确到0.01 kg的近似值为kg.13. 有意义，则m的值可以是.（填一个你喜欢的数）14. 在平面直角坐标系中，将点M（2，-1）向上平移2个单位长度得到点N的坐标是.15. 图书馆现有1500本图书供学生借阅，如果每个学生一次借3本，则剩下的数y （本）和借书学生人数x （人）之间的函数关系式是 .16. 如图，在△ABC 中，AB=BC ，AB 的垂直平分线交AB 于点D ，交BC 于点E ，AD=3，△ACE 的周长为11，则AC 的长为 .17. 已知等腰三角形的一个内角等于40°，则该三角形的顶角度数是 . 18. 如图，Rt △ABC 中，∠C=90°，AB=5，AC=4，分别以Rt △ABC 三边为直径作半圆，则阴影部分面积为 . 三、解答题：（本大题共有8小题，其中第19题4分，第20题~第24题每题8分，第25~26题每题10分，共64分. 解答需写出必要的文字说明、演算步骤.）19. （本题满分4分）计算：2-20. （本题满分8分）求下列各式中的x 的值： （1）290x -=（2）()311x +=21. （本题满分8分）已知：如图，∠C=90°，点A 、B 分别在∠C 的两直角边上，AC=1，BC=2.是 .（填“有理数”或“无理数”）画图：人类经历了漫长、曲折的历史过程，发现了无理数是客观存在的.（1（2）在射线CA 上画出长度为1+的线段.（注：保留画图痕迹，并把所画线段标注出来）22. （本题满分8分）如图，方格纸中每个小方格都是边长为1个单位的正方形，A（-3，1），B（3，2），解答以下问题：（1）在图中标出平面直角坐标系的原点O，并建立直角坐标系；（2）点A关于x轴的对称点A’坐标为，并在坐标系中画出点A’；（3）点P是x轴上一点，当PA+PB最小时，在图中画出点P的位置.23. （本题满分8分）盐城市初级中学为了缓解校门口的交通堵塞，倡导学生步行上学. 小丽步行从家去学校，图中的线段表示小丽步行的路程s（米）与所用时间t（分钟）之间的函数关系. 试根据函数图像回答下列问题：（1）小丽家离学校米；（2）小丽步行的速度是米/分钟；（3）求出m的值.24. 已知：如图，在△ABC中，∠ABC和∠ACB的角平分线相交于点P，且PE⊥AB，PF⊥AC，垂足分别为E、F （1）求证：PE=PF；（2）若∠BAC=60°，连接AP，求∠EAP的度数.25. （本题满分10分）已知：在平面直角坐标系xOy中，点A、B的坐标分别为（3，0），（0，4），点C（t，0）是x轴上一动点，点M是BC的中点.（1）当点C和点A重合时，求OM的长；（2）若S△ACB=10，则t的值为；（3）在（2）的条件下，直线AM交y轴于点N，求△ABN的面积.26. （本题满分10分）在“学本课堂”的实践中，王老师经常让学生以“问题”为中心进行自主、合作、探究学习. 【课堂提问】王老师在课堂中提出这样的问题：如图1，在Rt△ABC中，∠ACB=90°，∠BAC=30°，那么BC 和AB有怎样的数量关系？【互动生成】经小组合作交流后，各小组派代表发言.（1）小华代表第3小组发言：AB=2BC. 请你补全小华的证明过程.证明：把△ABC沿着AC翻折，得到△ADC.∴∠ACD=∠ACB=90°，∴∠BCD=∠ACD+∠ACB=90°+90°=180°，即：点B、C、D共线.（请在下面补全小华的证明过程）（2）受到第3小组“翻折”的启发，小明代表第2小组发言：如图2，在△ABC中，如果把条件“∠ACB=90°”改为“∠ACB=135°”，保持“∠BAC=30°”不变，若BC=1，求AB的长.【能力迁移】我们发现，翻折可以探索图形性质，请利用翻折解决下面问题.如图3，点D是△ABC内一点，AD=AC，∠BAD=∠CAD=20°，∠ADB+∠ACB=210°，则AD、DB、BC三者之间的数量关系是.【课后拓展】如图4，在四边形ABCD中，∠BCD=45°，∠BAD=90°，∠ADB=∠CDB=60°，且AC=1，则△ABD的周长为 .。