江苏省南京市2022-2023学年高一上学期期末学情调研测试 英语含答案

2023-2024学年江苏省南京市江宁区四年级（上）期末英语试卷一、听录音，选出你所听到的内容。

（听两遍）（10分）1．（10分）（1）A.socks B.short C.skirts（2）A.door B.talk C.tall（3）A.sofa B.chairs C.chickens（4）A.small lion B.schoolbag C.snowman（5）A.small B.well C.ball（6）A.fifteen B.forty C.fourteen（7）e B.can C.some（8）A.tiger B.panda C.lion（9）A.hungry B.hurry C.hamburger（10）A.food B.fruit C.grapes二、听录音，给下列图片标序号。

（听两遍）（12分）2．（12分）三、听录音，判断所听内容是否与图意相符，正确的填“A”，错误的填“B”。

（听两遍）（6分）3．（6分）四、听录音，根据所听问题选择正确的应答。

（听两遍）（6分）4．（6分）（1）A.I'm four. B.It's four yuan. C.They're four yuan.（2）A.Ok.Thank you. B.They're nice. C.Here you are.（3）A.Yes，she does. B.Yes，she is. C.Yes，she can.（4）A.They are in the study. B.It's in the kitchen. C.He's in the living room.（5）A.I like lions. B.No，I don't. C.Yes.It's so cute.（6）A.Yes，please. B.They're twenty yuan. C.I have six oranges.五、听录音，根据短文内容判断下列句子正误，正确的填“A”，错误的填“B”。

2022-2023学年江苏省南京市第五高级中学高一上学期期末学情自测数学试题一、单选题1．已知集合A ={1，2，3}，B ={x ∈N |x ≤2}，则A ∪B =（ ） A ．{2，3} B ．{0，1，2，3} C ．{1，2} D ．{1，2，3}【答案】B【分析】先求出集合B ，再求A ∪B.【详解】因为{1,2,3},{0,1,2}A B ==，所以0,1,3}2,{A B =. 故选：B2．命题“(0,)sin 2x x x π∀∈≤，”的否定是（ ） A ．(0,)sin 2x x x π∀∈≥，B ．(0,)sin 2x x x π∀∈>，C ．(0,)sin 2x x x π∃∈≤，D ．(0,)sin 2x x x π∃∈>，【答案】D【分析】直接利用全称命题的否定为特称命题进行求解.【详解】命题“(0,)sin 2x x x π∀∈≤，”为全称命题， 按照改量词否结论的法则，所以否定为：(0,)sin 2x x x π∃∈>，， 故选：D 3．已知弧长为3π的弧所对的圆心角为6π，则该弧所在的扇形面积为（ ） AB ．1π3C ．2π3D ．4π3【答案】B【分析】先求得扇形的半径，由此求得扇形面积.【详解】依题意，扇形的半径为π32π6=，所以扇形面积为1ππ2233⋅⋅=.故选：B4．,x R ∀∈不等式2410ax x +-<恒成立，则a 的取值范围为（ ）A ．4aB ．4a 或0a =C ．4a ≤-D ．40a【答案】A【分析】先讨论系数为0 的情况，再结合二次函数的图像特征列不等式即可. 【详解】,x R ∀∈不等式2410ax x +-<恒成立， 当0a =时，显然不恒成立，所以0Δ1640a a <⎧⎨=+<⎩，解得：4a.故选：A.5．已知0.50.5e ,ln 5,log e a b c -===，则（ ） A ．c<a<b B ．c b a << C ．b a c << D ．a b c <<【答案】A【分析】借助指对函数的单调性，利用中间量0或1比较即可. 【详解】因为0.500=e <e 1a -<=，ln5lne=1b =>，0.50.5=log e<log 1=0c ， 所以c<a<b ， 故选：A.6．已知函数()f x 是定义在R 上的奇函数，()(4)f x f x =+，且(1)1f -=-，则(2020)(2021)f f +=（ ） A ．1- B ．0 C ．1 D ．2【答案】C【分析】由()(4)f x f x =+得函数的周期性，由周期性变形自变量的值，最后由奇函数性质求得值． 【详解】∵()f x 是奇函数，∴(0)0,(1)(1)1f f f ==--=， 又()(4)f x f x =+，∴()f x 是周期函数，周期为4． ∴(2020)(2021)(0)(1)011f f f f +=+=+=． 故选：C ．7．已知函数()e xf x x =+，()lng x x x =+，()sinh x x x =+的零点分别为,,,a b c 则,,a b c 的大小顺序为（ ） A ．c b a << B ．b a c << C ．a c b << D ．c<a<b【答案】C【分析】利用数形结合，画出函数的图象，判断函数的零点的大小即可．【详解】函数()e xf x x =+，()lng x x x =+，()sinh x x x =+的零点转化为e x y =，ln y x =，sin y x=与y x =-的图象的交点的横坐标，因为零点分别为,,,a b c在坐标系中画出e x y =，ln y x =，sin y x =与y x =-的图象如图： 可知a<0，0b >，0c ， 满足a c b <<． 故选：C ．8．已知函数()()sin f x A x ωϕ=+的图象的一部分如图1所示，则图2中的函数图象对应的函数解析式为（ ）A ．1(2)2y f x =+B ．(21)y f x =+C ．1()22x y f =+D ．(1)2xy f =+【答案】B【分析】利用三角函数的图象变换规律可求得结果.【详解】观察图象可知，右方图象是由左方图象向左移动一个长度单位后得到()1y f x =+的图象，再把(1)=-y f x 的图象上所有点的横坐标缩小为原来的12（纵坐标不变）得到的， 所以右图的图象所对应的解析式为(21)y f x =+. 故选：B二、多选题9．下列函数中，既是偶函数又在区间()0,∞+上是增函数的是（ ） A ．21y x =+ B ．3y x = C ．23y x = D ．3x y -=【答案】AC【分析】利用函数的奇偶性和单调性的概念进行判断. 【详解】对于A ： ()2211=-+=+y x x∴ 函数21y x =+ 是偶函数，在()0,∞+ 上是增函数，故A 正确；对于B ： ()33=-=-y x x∴ 函数3y x = 是奇函数，故B 错误；对于C ：()2233=-=y x x23y x ∴=是偶函数，在()0,∞+ 上是增函数，故C 正确； 对于D ： 33---==xxy3-∴=xy 是偶函数，在()0,∞+ 上是减函数，故D 错误.故选：AC 10．若110a b<<，则下列不等式正确的是（ ） A ．a b < B ．a b <C ．a b ab +<D ．2b aa b+>【答案】BCD【分析】利用不等式的基本性质求解即可【详解】由于110a b<<，则0b a <<，故a b <错误； 0a b ab +<<正确；||||a b <正确；2222a b a b abb a ab ab++=>=，∴2b a a b +>，正确故选：BCD ．11．若函数()tan(2)3f x x π=+，则下列选项正确的是（ ）A ．最小正周期是 πB ．图象关于点(,0)3π对称C ．在区间7(,)1212ππ上单调递增D ．图象关于直线12x π=对称【答案】BC【分析】利用正切函数的周期，对称中心，函数的单调性，判断选项即可． 【详解】函数tan(2)3y x π=+，函数的最小正周期为：2π，A 错误； 令23246k k x x k Z ππππ+=⇒=-∈，， 当2k =时，3x π=，所以图象关于点(,0)3π对称，B 正确；因为2232k x k πππππ-<+<+，Z k ∈，解得5(212k x ππ∈-，)212k ππ+，当1k =时，7(,)1212x ππ∈,所以在区间7(,)1212ππ上单调递增，C 正确；又正切函数不具有对称轴，所以D 错误故选：B C ．12．设x R ∈，用[]x 表示不超过x 的最大整数，则[]y x =称为高斯函数，也叫取整函数.令[]()22f x x x =-，以下结论正确的是（ ）A ．()1.10.8f -=B ．()f x 为偶函数C ．()f x 最小正周期为12D ．()f x 的值域为[]0,1【答案】AC【分析】根据高斯函数的定义逐项检验即可，对于A ，直接求解即可，对于B ，取 1.1x =-，检验可得反例，对于C ，直接求解()12f x f x ⎛⎫+= ⎪⎝⎭即可；对于D ，要求()f x 的值域，只需求102x ≤<时()f x 的值域即可.【详解】对于A ，()[]2.21 2.230..8.122f --=-+=-=-，故A 正确. 对于B ，取 1.1x =-，则()1.10.8f -=，而()[]2.2 2.2 2.221021..f --===， 故()()1.1 1.1f f -≠-，所以函数()f x 不为偶函数，故B 错误.对于C ，则[][]()1212121212f x x x x x f x ⎛⎫+=+-+=+--= ⎪⎝⎭，故C 正确.对于D ，由C 的判断可知，()f x 为周期函数，且周期为12， 要求()f x 的值域，只需求102x ≤<时()f x 的值域即可. 当0x =时，则()[]0000f =-=， 当102x <<时，()[]22202(0,1)f x x x x x =-=-=∈， 故当102x ≤<时，则有()01f x ≤<，故函数()f x 的值域为[)0,1，故D 错误. 故选：AC ．三、填空题135log 25=_____． 【答案】6【分析】利用根式性质与对数运算进行化简.5log 25426=+=， 故答案为：614．请写出一个同时满足下列两个条件的函数：____________.（1）12x x R ∀∈， ，若12x x >则12()()f x f x >（2）121212,,()()()x x R f x x f x f x ∀∈+= 【答案】()2x f x =，答案不唯一【分析】由条件（1）12x x R ∀∈， ，若12x x >则12()()f x f x >.可知函数()f x 为R 上增函数； 由条件（2）121212,,()()()x x R f x x f x f x ∀∈+=.可知函数()f x 可能为指数型函数. 【详解】令()2x f x =，则()2x f x =为R 上增函数，满足条件（1）.又1212()2x x f x x ++=，121212()()222x x x xf x f x +=⨯=故12()f x x +=12()()f x f x即121212,,()()()x x R f x x f x f x ∀∈+=成立.故答案为：()2x f x =，（()3x f x =，()4x f x =等均满足题意）15．在平面直角坐标系xOy 中，以Ox 轴为始边作两个锐角α，β，它们的终边分别与单位圆相交于P ，Q 两点，P ，Q 的纵坐标分别为35，45．则αβ+的终边与单位圆交点的纵坐标为_____________. 【答案】1【分析】根据任意角三角函数的定义可得3sin 5α=，4cos 5α=，4sin 5β=，3cos 5β=，再由()sin αβ+展开求解即可.【详解】以Ox 轴为始边作两个锐角α，β，它们的终边分别与单位圆相交于P ，Q 两点，P ，Q 的纵坐标分别为35，45所以3sin 5α=，α是锐角，可得4cos 5α=，因为锐角β的终边与单位圆相交于Q 点，且纵坐标为45，所以4sin 5β=，β是锐角，可得3cos 5β=， 所以()3344sin sin cos cos sin 15555αβαβαβ+=+=⨯+⨯=，所以αβ+的终边与单位圆交点的纵坐标为1. 故答案为：1.四、双空题16．已知函数()2log ,042,482x x f x cos x x π⎧<<⎪=⎨≤≤⎪⎩，,t R ∃∈使方程()f x t =有4个不同的解：1234,,,x x x x ，则1234x x x x 的取值范围是_________; 1234x x x x +++的取值范围是________.【答案】 (32,35) 65(14,)4【分析】先画出分段函数()f x 的图像，依据图像得到12,x x 之间的关系式以及34,x x 之间的关系式，分别把1234x x x x +++和1234x x x x 转化成只有一个自变量的代数式，再去求取值范围即可. 【详解】做出函数()2log ,042,482x x f x cos x x π⎧<<⎪=⎨≤≤⎪⎩的图像如下：()f x 在](0,1单调递减：最小值0；()f x 在[]1,4单调递增：最小值0，最大值2； ()f x 在[]4,8上是部分余弦型曲线：最小值2-，最大值2.若方程()f x t =有4个不同的解：1234,,,x x x x ，则02t << 不妨设四个解依次增大，则12341145,784x x x x <<<<<<<< 12,x x 是方程2log x t =(04)x <<的解，则2122log log x x =-，即121=x x ；34,x x 是方程2cos2x t π=(48)x ≤≤的解，则由余弦型函数的对称性可知3412x x +=.故2123434333(12)(6)36x x x x x x x x x ==-=--+，由345x <<得2332(6)3635x <--+<即12343235x x x x << 1234121111212x x x x x x x x +++=++=++当1114x <<时，1()12m x x x=++单调递减，则1116514124x x <++<故答案为：①(32,35)；②65(14,)4五、解答题 17．求值：(1)22log 33582lg 2lg 22+--(2)25π10π13πsincos tan 634⎛⎫-+- ⎪⎝⎭【答案】(1)6 (2)0【分析】(1)根据指数运算公式和对数运算公式求解即可； (2)根据诱导公式化简求值即可. 【详解】（1）22log 33582lg 2lg 22+--()()2lo 23g 3322lg5lg 22lg 2=+---223lg 5lg 22lg 2=+-+-7(lg5lg 2)=-+71=-6=；（2）25π10π13πsincos tan 634⎛⎫-+- ⎪⎝⎭πππsin 4πcos 3πtan 3π634⎛⎫⎛⎫⎛⎫=+-+-+ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭πππsincos tan 634=+- 11122=+- 0=．18．已知全集U =R ，集合{}2120A x x x =--≤，集合{}11B x m x m =-≤≤+.(1)当4m =时，求()U A B ⋃； (2)若()U B A ⊆，求实数m 的取值范围. 【答案】(1){4x x ≤或5}x >； (2)4m <-或5m >.【分析】（1）确定集合A ，B ，求出集合B 的补集，根据集合的并集运算，即可求得答案. （2）求出集合A 的补集，根据()U B A ⊆，列出相应不等式，求得答案.【详解】（1）集合{}{}212034A x x x x x =--≤=-≤≤，当4m =时，{}35B x x =≤≤，则{3UB x x =<或5}x >，故()UAB ={4x x ≤或5}x >；（2）由题意可知{3UA x x =<-或4}x > ，{}11B x m x m =-≤≤+≠∅,由UB A ⊆，则13m +<-或14m ->，解得4m <-或5m >.19．已知函数()()sin (0,0,)2f x A x A πωϕωϕ=+>><的部分图象如图.(1)求函数()f x 的解析式;(2)将函数()f x 的图象上所有点的横坐标变为原来的2倍，纵坐标不变，再将所得图象向左平移6π个单位，得到函数()g x 的图象，当,6x ππ⎡⎤∈-⎢⎥⎣⎦时，求()g x 值域.【答案】(1)()2sin 23f x x π⎛⎫=- ⎪⎝⎭；(2)[3,2]-.【分析】（1）根据图象由函数最值求得A ，由函数周期求得ω，由特殊点求得ϕ，即可求得解析式； （2）根据三角函数图象的变换求得()g x 的解析式，再利用整体法求函数值域即可.【详解】（1）由图象可知，()f x 的最大值为2，最小值为2-，又0A >，故2A =，周期453123T πππ⎡⎤⎛⎫=--= ⎪⎢⎥⎝⎭⎣⎦，2||ππω∴=，0ω>，则2ω=， 从而()2sin(2)f x x ϕ=+，代入点5,212π⎛⎫⎪⎝⎭，得5sin 16⎛⎫+=⎪⎝⎭πϕ， 则5262k ππϕπ+=+，Z k ∈，即23k πϕπ=-+，Z k ∈， 又||2ϕπ<，则3πϕ=-.()2sin 23f x x π⎛⎫∴=- ⎪⎝⎭.（2）将函数()f x 的图象上所有点的横坐标变为原来的2倍，纵坐标不变，故可得2sin 3y x π⎛⎫=- ⎪⎝⎭；再将所得图象向左平移6π个单位，得到函数()g x 的图象 故可得()2sin()6g x x π=-；[,]6x ππ∈-5[,]636x πππ∴-∈-，3sin ,162x π⎡⎤⎛⎫-∈-⎢⎥ ⎪⎝⎭⎣⎦， 2sin 3,26x π⎛⎫⎡⎤-∈- ⎪⎣⎦⎝⎭，()[3,2]g x ∴-的值域为. 20．已知函数sin()cos sin()cos(2)()cos tan()sin 2f πααπαπααπααα-+-=+-⎛⎫- ⎪⎝⎭（1）化简()f α；（2）若1(),052f παα=-<<，求sin cos ,s cos in αααα⋅-的值. 【答案】（1）()sin cos f ααα=+；（2）75- 【分析】（1）利用诱导公式进行化简即可；（2）由（1）结果两边平方，再利用同角三角函数的基本关系联立解方程组即可得出结果.【详解】解：（1）sin()cos sin()cos(2)sin cos (sin )cos ()cos tan()cos cos tan sin 2f πααπαπααααααπαααααα-+--=+=+--⎛⎫- ⎪⎝⎭所以()sin cos f ααα=+.（2）由1()sin cos 5f ααα=+=，平方可得221sin 2sin cos cos 25αααα++=， 即242sin cos 25αα⋅=-. 所以12sin cos 25αα⋅=-， 因为249 (sin cos )12sin cos 25αααα-=-=， 又02πα-<<，所以sin 0α<，cos 0α>，所以sin cos 0αα-<,所以7sin cos 5αα-=-. 【点睛】本题考查了诱导公式、同角三角函数的化简与求值，属于基础题.21．某市城郊有一块大约500m 500m ⨯的接近正方形的荒地，地方政府准备在此建一个综合性休闲广场，首先要建设如图所示的一个矩形体育活动场地，其中总面积为3000平方米，其中阴影部分为通道，通道宽度为2米，中间的三个矩形区域将铺设塑胶地面作为运动场地（其中两个小场地形状相同），塑胶运动场地占地面积为S 平方米.（1）分别用x 表示y 及S 的函数关系式，并给出定义域；（2）请你设计规划该体育活动场地，使得该塑胶运动场地占地面积S 最大，并求出最大值【答案】（1）1500030306S x x=--，定义域是(6，500)；（2）设计50m 60m x y ，==时，运动场地面积最大，最大值为2430平方米.【分析】（1）总面积为3000xy =，且26a y +=，可得3000y x =，15003a x =-（其中6500)x <<，从而运动场占地面积为(4)(6)S x a x a =-+-，代入整理即得；（2）由（1）知，占地面积150003030(6)S x x =-+，由基本不等式可得函数的最大值，以及对应的x 的值．【详解】解：（1）由已知30003000,,xy y x=∴=其定义域是(6,500). (4)(6)(210),S x a x a x a =-+-=-26a y +=，∴1500332y a x =-=-， 150015000(210)(3)30306S x x x x∴=--=--，其定义域是(6,500).（2）150003030(6)3030303023002430,S x x =-+≤--⨯= 当且仅当150006x x=，即50(6,500)x =∈时，上述不等式等号成立， 此时，50x =，60y =，2430max S =．答：设计50m 60m x y ，== 时，运动场地面积最大，最大值为2430平方米.22．已知函数1()ln 1x f x x -=+. (1)求证：()f x 是奇函数；(2)若对于任意[]3,5x ∈都有()3f x t >-成立，求t 的取值范围；(3)若存在,(1,)αβ∈+∞，且αβ<，使得函数()f x 在区间[],αβ上的值域为ln ,ln 22m m m m αβ⎡⎤⎛⎫⎛⎫-- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦，求实数m 的取值范围. 【答案】(1)证明见解析(2)(,3ln 2)-∞- (3)209m <<【分析】（1）利用奇偶性的定义求解即可；（2）对于任意[]3,5x ∈都有()3f x t >-成立，仅需min ()3f x t >-即可；（3）由()f x 单调性可得,αβ是方程112x m mx x -=-+，即22(2)20mx m x m +-+-=的两实数根，根据一元二次函数的图像和性质求解即可.【详解】（1）又101x x ->+即(1)(1)0x x -+>解得(,1)(1,)x ∈-∞-⋃+∞， 所以()f x 的定义域为(,1)(1,)x ∈-∞-⋃+∞，关于原点对称， 又因为1111()ln ln ln ()111x x x f x f x x x x -+--⎛⎫-===-=- ⎪-++⎝⎭， 所以()f x 是奇函数.（2）由题意2()ln 11f x x ⎛⎫=- ⎪+⎝⎭，令2()11u x x =-+(()0)u x >， 因为()u x 在(1,)+∞上为增函数，ln u 在(0,)+∞上为增函数，所以()f x 在(1,)+∞上为增函数，当[]3,5x ∈时，ln 2()ln 2ln3f x -≤≤-，对于任意[]3,5x ∈都有()3f x t >-成立，仅需min ()3f x t >-即可，所以ln 23t ->-，解得3ln 2t <-.（3）由（2）可知()f x 在(1,)+∞上为增函数，又因为()f x 在区间[],αβ上的值域为ln ,ln 22m m m m αβ⎡⎤⎛⎫⎛⎫-- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦， 所以0m >且1ln ln 121ln ln 12m m m m αααβββ⎧-⎛⎫=- ⎪⎪+⎝⎭⎪⎨-⎛⎫⎪=- ⎪⎪+⎝⎭⎩，所以112112m m m m αααβββ-⎧=-⎪+⎪⎨-⎪=-+⎪⎩， 则,αβ是方程112x m mx x -=-+，即22(2)20mx m x m +-+-=的两实数根， 令2()2(2)2h x mx m x m =+-+-， 则由题意对称轴214m x m-=>，2(2)42(2)0m m m ∆=--⨯⨯->，(1)0h m =>， 解得209m <<.。

全卷满分150分考试时间120分钟2022.12第一部分：听力（共两节，满分30分）（略）第二部分：阅读（共两节，满分50分）第一节（共15小题；每小题2.5分，满分37.5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。

ASouthwest China’s Guizhou Province made some projects in promoting (推动) high-quality development of both its culture and tourism industries.Building unique culture and tourism brandsGuizhou held the 2021 International Conference of Mountain Tourism and Outdoor Sports and the 16th Guizhou Tourism Industry Development Conference to improve cooperation in culture and tourism with southwest China’s Sichuan Province and Chongqing Municipality.Guizhou also organized a series of activities to promote its rich culture and tourism resources both online and offline.Reviving intangible cultural heritage (ICH) (非物质文化遗产)Eighteen items in Guizhou were listed in the fifth batch of national ICHs, bringing the province’s total number to 99. Song Shuixian, an ICH inheritor in the province, was named as one of China’s top 10 ICH inheritors of the Year 2020.Guizhou also held different online and offline ICH-related activities to increase the sales of ICH-related products.Increasing satisfaction in tourismGuizhou has improved the quality of tourism services recently. It put down illegal practices in tourism, ordered the closure of 50 tourism-related shopping places, and investigated nine travel agencies and four tour guides. It accepted and addressed 716 tourism-related complaints.Founding the Guizhou Vocational College of Culture and TourismOn Feb. 11, 2021, the people’s government of Guizhou planned to set up Guizhou V ocational College of Culture and Tourism, making it a new training base for professional tourism talents in the province.21.How many items were listed before the 5th batch of national ICHs?A.18.B. 99.C. 81.D. 10.22.Which projects turn to the Internet for the development?A. Building unique brands and reviving ICH.B. Building unique brands and increasing satisfaction in tourism.C. Reviving ICH and increasing satisfaction in tourism.D. Reviving ICH and founding the Guizhou V ocational College of Culture and Tourism.23. Who is likely to be punished in the promotion projects?A. A leader who attended the conference.B. A guide who charges extra fee.C. A student who receives tourism training.D. A shop which sells ICH-related products.BBefore I can tell you what happened to me, you first must learn about my job: performunderwater repairs. As you know, my office lies at the bottom of the sea. I wear a suit to office. It’s a wet suit.This time of year the water is quite cool. So what we do to keep warm is a water heater. This $ 20,000 piece of equipment that sucks the water out of the sea. It heats it to a delightful temperature. It then pumps it down to the diver (潜水员) through a garden pipe.Now, this all sounds like a wonderful plan, and I’ve used it several times.What I do when I get to the bottom is take the pipe and stuff it down the back of my wet suit. This floods my whole suit with warm water. It’s fantastic.Everything was going well until all of a sudden, my butt (屁股) started to itch (痒). So, of course, I scratched it. This, of course, only made things worse. Within a few seconds my butt started to feel hot. I pulled the pipe out from my back, thinking that maybe the water was too hot, but the damage was done.In pain, I realized what had happened. The hot water machine had sucked up a jellyfish (水母) and pumped it directly into my suit. Now, since I don’t have any hair on my back, the jellyfish couldn’t stick to it.I immediately informed the dive director of my situation over the communicator. His reply was unclear due to the fact that he, along with five other divers, were all laughing hysterically.Needless to say, I gave up the dive. As I climbed out of the water, the doctor, with tears of laughter running down his face, handed me a tube of cream and told me to rub it on my butt. The cream put the fire out, but I couldn’t poop (排泄) for two days because my bottom was swollen.So, next time you’re having a bad day at work, think about how much worse it would be if you had a jellyfish shoved up your bottom.Now repeat to yourself, “I love my job, I love my job, I love my job.” May you NEVER have a jellyfish bad day.24. What is the writer?A. A surfer.B. A swimmer.C. An officer.D. A repairman.25. What happened to the writer under the sea?A. He was bitten by a strange fish.B. The hot water burned his butt.C. His skin was cut by something.D. A jellyfish was attached to his back.26. Why didn’t the director make his reply clear?A. Because the doctor put in a word.B. Because he didn’t intend to help him.C. Because his voice mixed with laughter.D. Because voice ran slower under the sea.27. Which of the following can best describe the writer?A. Positive and humorous.B. Kind and friendly.C. Helpful and honest.D. Grateful and selfless.CMany Chinese home appliance enterprises (家电企业) saw demand for their heating products rise from the European market during the ongoing 132nd session of the China Importand Export Fair, or Canton Fair, the Shanghai Securities News reported on Friday.“This time we are focusing on heating products,” said Bu Zhiming, general manager Gd Shine Electric Appliances, who has had his hands full since the opening of this year’s on Oct 15. The major products on display are heaters and electric ovens. The orders be flooding in more than one month earlier than previous years, according to Bu. The company sold more than 500,000 heating equipment to the European market in the first nine months, up over 30 percent year-on-year.We’ve received many telephones for information on heating equipment from the customers,” said Zhang Wei, board director of the exhibitor, Honghuo Holdings, a heating equipment provider based in Guangdong Province. “Small-sized air blowers with lower price and electric blankets with lower energy consumption are popular among European customers.” To meet the demand from Europe, the company has set up a subsidiary in France to produce heating products under own brand and seen the export of own-brand products rising by four to five fold so far this year.“The European market is our focus during the Canton Fair this time. Since autumn, we’ve seen order rise,” said Peng Hongfei, head of the marketing department of the Vanward, a heating equipment provider. Peng said the sales of their heating equipment went beyond over 20 million yuan ($2.76 million) in the third quarter this year, up 102 percent on a yearly basis. “At present, we are also stepping up the preparation of new production lines to ensure order production.”28. How did the report prove the greater demand for Chinese home appliances?A. By advertising.B. By comparing.C. By interviewing.D. By analyzing.29. What does the underlined word in paragraph 3 mean?A. Lab.B. Branch.C. System.D. Standard.30. How will the Vanward deal with the rising order?A. Employ more workers.B. Add production lines.C. Learn new skills.D. Encourage extra work.31. What can be a suitable title for the text?A. Chinese heating equipment providers will make a big fortune.B. The European market greatly welcomes Chinese home appliances.C. Chinese heating products have changed fast over the past twenty years.D. Chinese heating products see orders from Europe rise during Canton Fair.DThere have been few positives during the Covid pandemic but British academics may have spotted one: People look more attractive in protective masks.Researchers at Cardiff University were surprised to find that both men and women were judged to look better with a face mask covering the lower half of their faces. They also discovered that a face covered with a disposable-type surgical mask (一次性外科罩) was likely to be considered the most attractive, which may be a blow for producers of fashionable coverings and the environment.Dr Michael Lewis, a reader from Cardiff University’s school of psychology, said research before the pandemic had found that medical face masks reduced attractiveness because theywere associated with illness.“We wanted to test whether this had changed since face coverings became popular,” he said.“Our study suggests faces are considered most attractive when covered by medical face masks. This may be because we’re used to healthcare workers wearing blue masks and now we associate these with people in caring or medical professions. At a time when we feel weak, we may find the wearing of medical masks secure and so feel more positive towards the wearer.”The first part of the research was carried out in February 2021 by which time the British population had become used to wearing masks in some places. Forty-three women were asked to rate on a scale of 1 to 10 ----the attractiveness of images of male faces without a mask, wearing a plain cloth mask, a blue medical face mask, and holding a plain black book covering the area a face mask would hide.The participants said those wearing a cloth mask were much more attractive than the ones with no masks or whose faces were partly covered by the book. But the surgical mask- which was just a normal, disposable kind--made the wearer look even better. Lewis said it was also possible that masks made people more attractive because they directed attention to the eyes. He said other studies had found that covering the left or right half of a face also made people look more attractive, partly because the brain fills in the missing gaps and beautify the overall effect.The results of the first study has been published in the journal Cognitive Research: Principles and Implications. A second study has been carried out, in which a group of men look at women in masks; it has yet to be published but Lewis said the results were probably the same.32. Why were protective masks not welcome before the pandemic?A. Because it reminded others of diseases.B. Because the masks wore uncomfortably.C. Because only doctors had the right to wear them.D. Because some wearers couldn’t take a smooth breath.33. Who looked more attractive according to the participants?A. Those with cloth masks.B. Those with surgical masks.C. Those without any covering.D. Those with “book” masks.34. Which of the following can’t explain the attraction?A. Those wearing masks may look more like doctors.B.Viewers focus more on eyes when judging a person.C.People like to imagine what a covered face look like.D.It’s recognized that masks can hide facial imperfections.35.Where is the text probably from?A.A book on biology.B. A medicine magazine.C. A newspaper.D. A guidebook.第二节（共2小题；每小题2.5分，满分12.5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。

2022-2023学年江苏省南京市金陵中学高一上学期12月学情调研测试数学试题一、单选题1．已知集合，则（ ）()(){}{}2210,log 1M x x x N x x =+-≤=≤∣∣M N ⋃=A ．B ．C ．D ．(]0,1[]22-,[]2,1-(],2-∞【答案】B【分析】解一元二次不等式，根据对数函数单调性解对数不等式即可.【详解】由题知，，解得，故，()()210x x +-≤21x -≤≤{}21M xx =-≤≤∣又因为，{}2log 1{02}N x x x x =≤=<≤∣∣所以，{}{}21{02}22M N x x x x x x ⋃=-≤≤⋃<≤=-≤≤∣∣∣即.[]2,2M N ⋃=-故选：B 2．“”是“函数的图像关于中心对称”的（ ）0sin 0x =tan y x =()0,0x A ．充分不必要条件B ．必要不充分条件C ．充分必要条件D ．既不充分又不必要条件【答案】A【分析】由充分条件必要条件的定义，结合三角函数的性质，作出判断.【详解】当时，，此时的图像关于中心对称，0sin 0x =0π,Z x k k =∈0tan 0,tan x y x ==()0,0x 当函数的图像关于中心对称时，，此时不一定为0.tan y x =()0,0x 0π,Z 2kx k =∈0sin x 所以“”是“函数的图像关于中心对称”的充分不必要条件.0sin 0x =tan y x =()0,0x 故选：A.3．平面直角坐标系中，角的终边经过点，则（ ）α(P cos 2πα⎛⎫+= ⎪⎝⎭A ．B ．CD ．12-12【答案】A【分析】根据给定条件，利用三角函数定义结合诱导公式计算作答.【详解】依题意，点到原点距离，(P 2r ==所以.cos sin 2παα⎛⎫+=-= ⎪⎝⎭故选：A 4．设，若，，，则（ ）()0,1m ∈lg a m =2lg b m =()2lg c m =A ．B ．C ．D ．a b c >>b c a >>c a b >>c b a>>【答案】C【分析】利用对数函数的性质即得.【详解】∵，()0,1m ∈∴，，，lg 0a m =<2lg 2lg lg b m m m a ==<=()2lg 0c m =>∴.c a b >>故选：C.5．已知函数满足，则解析式是（ ）()f x 2(1)43f x x x +=++()f x A ．B ．2()2f x x x =+2()2f x x =+C ．D ．2()2f x x x=-2()2f x x =-【答案】A【分析】利用换元法，求函数的解析式.【详解】设，故，则，1x t +=1x t =-()()()2214132f t t t t t =-+-+=+所以.()22f x x x=+故选：A6．王之涣《登鹳雀楼》：白日依山尽，黄河入海流.欲穷千里目，更上一层楼､诗句不仅刻画了祖国的壮丽河山，而且揭示了“只有站得高，才能看得远”的哲理，因此成为千古名句，我们从数学角度来思考：欲穷千里目，需上几层楼？把地球看作球体，地球半径，如图，设O 为地球6371km R =球心，人的初始位置为点M ，点N 是人登高后的位置（人的高度忽略不计），按每层楼高计算，3.3m “欲穷千里目”即弧的长度为，则需要登上楼的层数约为（ ）AM 500km （参考数据：，，）5000.07856371≈cos 0.07850.9969≈63716390.80.9969≈A ．1B ．20C ．600D ．6000【答案】D【分析】根据弧长公式可求得即的大小.在中，即可求得AOM ∠AON ∠5000.0785R θ=≈Rt OAN 的大小.ON 【详解】O 为地球球心，人的初始位置为点M ，点N 是人登高后的位置，的长度为.AM 500km 令，则.AON θ∠=5000.0785R θ=≈∵，，.OA AN ⊥OA R =63716390.8(km)cos cos 0.9969OA R ON θθ==≈≈∴，19.8(km)MN ON OM =-=又.19.8100060003.3⨯=所以按每层楼高计算，需要登上6000层楼.3.3m 故选：D.7．已知函数则函数的大致图象为（ ）())lnsin f x x x =⋅()f xA ．B ．C ．D ．【答案】A【分析】先利用函数的奇偶性排除部分选项，再根据时，函数值的正负判断.()0,x π∈【详解】易知函数为奇函数，也是奇函数，)lny x =sin y x =则函数为偶函数，故排除选项B ，C ；())lnsin f x x x=⋅因为，)ln ln y x ==当恒成立，所以恒成立，0x >1x >ln 0<且当时，，()0,x π∈sin 0x >所以当时，，故选项A 正确，选项D 错误，()0,x π∈()0f x <故选：A ．8．已知函数,且,则实数的取值范围是（ ）()3233e 1x f x x =-++()()2564f a f a +->a A ．B ．()6,1-()(),61,∞∞--⋃+C ．D ．()(),16,-∞-⋃+∞()1,6-【答案】B【分析】将化简为,令新函数,判断的奇偶()f x ()3e 132e 1x xf x x -=+++()()3e 13R e 1x x g x x x -=+∈+()g x 性和单调性,将不等式转化为关于的不等式,根据的函数性质转化为()()2564f a f a +->()g x ()g x 关于的不等式,解出即可.a 【详解】解:由题意得,函数,()332e 131232e 1e 1x x xf x x x -=-++=++++设,()()3e 13R e 1x x g x x x -=+∈+则,()()33e 1e 13()3e 1e 1x x x x g x x x g x --⎛⎫---=-+=-+=- ⎪++⎝⎭所以是上的奇函数,()g x R 因为,()()2f xg x =+由()()2564f a f a +->则()()2560,g a g a +->即,()()256g a g a >--因为,()()5665g a g a --=-所以,()()265g a g a >-又有,()33e 12331e 1e 1x x x g x x x -=+=-+++因为是上的增函数,33y x =R 是上的增函数,2e 1x y =-+R 所以是上的增函数;()g x R 则有,整理得:,265a a >-2560a a +->解得:或,1a >6a <-所以的取值范围为.a ()(),61,∞∞--⋃+故选:B二、多选题9．下列结论中，正确的是（ ）A ．函数的单调增区间是2213x xy -+⎛⎫= ⎪⎝⎭()1,+∞B ．命题“所有的素数都是奇数”的否定是假命题C ．是奇函数()πsin 2f x x ⎛⎫=- ⎪⎝⎭D ．函数的图像必过定点()23(0,1)x f x a a a -=->≠()2,2-【答案】AD【分析】根据复合函数的单调性可判断A 项；判断原命题真假即可判断B 项；化简为()cos f x x=偶函数，可说明C 项错误；令得，代入可判断D 项.20x -=2x =【详解】对于A 项，函数中，，在上递增，在2213x xy -+⎛⎫= ⎪⎝⎭222(1)1u x x x =-+=--+(),1-∞上递减；函数在R 上单调递减.()1,+∞13uy ⎛⎫= ⎪⎝⎭根据复合函数的单调性可知，函数的单调增区间是，故A 项正确：2213x xy -+⎛⎫= ⎪⎝⎭()1,+∞对于B 项，2是素数，但2不是奇数，所以“所有的素数都是奇数”命题错误，否定为真，故B 项错误；对于C 项，因为，是偶函数，故C 项错误；()πsin cos 2f x x x⎛⎫=-= ⎪⎝⎭对于D 项，函数中，由得，，即函数图象()23(0,1)x f x a a a -=->≠20x -=2x =()22f =-()f x 过点，故D 项正确.()2,2-故选：AD.10．下列不等式成立的是（ ）A ．B ．ππsin sin 108⎛⎫⎛⎫-<- ⎪ ⎪⎝⎭⎝⎭()cos400cos 50>-C ．D ．7π8πsin sin 87⎛⎫⎛⎫< ⎪ ⎪⎝⎭⎝⎭sin3sin2<【答案】BD【分析】利用正余弦函数的单调性可得出每个选项中两个三角函数值的大小，即可选出答案.【详解】因为，且函数在上单调递增，则，πππ02810-<-<-<sin y x =π,02⎛⎫- ⎪⎝⎭ππsin sin 810⎛⎫⎛⎫-<- ⎪ ⎪⎝⎭⎝⎭故选项A 错误；因为，且函数在上单调递减，则，()cos400cos40,cos 50cos50=-=cos y x =π0,2⎛⎫⎪⎝⎭cos40cos50> 即，故选项B 正确；()cos400cos 50>-因为，且函数在上单调递减，则，故选项C 错π7π8π3π2872<<<sin y x =π3π,22⎛⎫⎪⎝⎭7π8πsin sin 87⎛⎫⎛⎫> ⎪ ⎪⎝⎭⎝⎭误；因为，且函数在上单调递减，则，故选项D 正确.π3π2322<<<sin y x =π3π,22⎛⎫ ⎪⎝⎭sin3sin2<故选：BD 11．已知是定义在上的奇函数，且图象关于直线对称，当时，，()f x R 2x =[]0,2x ∈()2xf x a=+则不等式成立的一个充分条件是（ ）()()2f x f x ≤+A ．B ．15x ≤≤913x ≤≤C ．D ．1317x ≤≤2125x ≤≤【答案】CD【分析】先根据奇函数的性质求得，再根据函数图象的解析式与性质画出的图象，1a =-()y f x =结合函数图象平移的性质得出的图象，再根据函数的周期，数形结合分析即可得出结()2y f x =+果.【详解】由题意，因为的图象关于直线对称，故，()f x 2x =()()4f x f x -=+又为奇函数，所以有， 所以，()f x ()()f x f x -=-()()4f x f x =-+故，所以，故的周期为8.()()48f x f x +=-+()()8f x f x =+()f x 因为是定义在上的奇函数，故，解得，根据当时，()f x R ()0020f a =+=1a =-[]0,2x ∈，结合奇函数的性质与直线对称以及函数的周期性作图，且是()21x f x =-2x =()2y f x =+的图象往左平移2个单位所得，作图如下.()y f x =又不等式成立，即在的函数图象下方，()()2f x f x ≤+()y f x =()2y f x =+由对称性得，当时，与的交点的横坐标为，结合图象可得[]0,2x ∈()y f x =()2y f x =+1x =与的交点的横坐标满足，所以在这个周期内，满足题()y f x =()2y f x =+()14,x k k =+∈Z []4,4-意的解为，则所有符合题意的解为.[]3,1-[]83,81,k k k -+∈Z 当时，解为；当时，解为；当时，解为；当时解为，0k =[]3,1-1k =[]5,92k =[]13,173k =[]2125，故选：CD.【点睛】本题主要考查了根据函数的奇偶性、对称性与周期性，结合函数图象平移的方法数形结合解决函数不等式的问题，难度较大.12．已知函数，则下列结论正确的是（ ）()()()sin cos cos sin f x x x =+A ．函数的一个周期为B ．函数在上单调递减()f x 2π()f x π0,2⎛⎫ ⎪⎝⎭C ．函数D ．函数图象关于直线对称()f x ()f x πx =【答案】ABD【分析】根据三角函数的周期性、单调性、最值、对称性等知识对选项进行分析，从而确定正确答案.【详解】由，()()()()()()()()2πsin cos 2πcos sin 2πsin cos cos sin f x x x x x f x +=+++=+=所以是周期为的周期函数，A 正确：()f x 2π由在上单调递减及复合函数的单调性知，在上单调递减，cos y x =π0,2⎛⎫ ⎪⎝⎭()sin cos y x =π0,2⎛⎫⎪⎝⎭由在上单调递增，可知在上单调递减，sin y x =π0,2⎛⎫ ⎪⎝⎭()cos sin y x =π0,2⎛⎫⎪⎝⎭所以函数在上单调递减，故B 正确；()f x π0,2⎛⎫⎪⎝⎭当时，0x =()π0sin1cos01sin11sin1.56f =+=+>+=>故函数，故C 错误；()f x ，()()()()()()()()2πsin cos 2πcos sin 2πsin cos cos sin f x x x x x f x -=-+-=+= 关于直线对称，故D 正确.()f x \πx =故选：ABD三、填空题13．函数的定义域为______.()f x =【答案】()0,1【分析】利用具体函数定义域的求法与对数函数的性质求解即可.【详解】因为，()f x =所以，解得，0lg 010x x x >⎧⎪≠⎨⎪-≥⎩01x <<所以的定义域为.()f x ()0,1故答案为：.()0,114．函数的最大值为______.()22sin 3cos 1f x x x =--【答案】178【分析】利用三角函数的基本关系式将化为关于的二次函数，从而利用配方法即可得解.()f x cos x 【详解】因为()()2222sin 3cos 121cos 3cos 12cos 3cos 1f x x x x x x x =--=---=--+，23172cos 48x ⎛⎫=-++⎪⎝⎭又，所以当时，.1cos 1x -≤≤3cos 4x =-max 17()8f x =故答案为：.17815．设和是方程的两根，则______.log a m log b m 2430x x -+=log mba =【答案】23±【分析】利用对数的运算法则直接计算即可.【详解】因为和是方程的两根，log a m log b m 2430x x -+=所以或者，log 1,log 3a b m m ==log 3log 1a b m m ==112log log log .log log 3mm m b a b b a a m m =-=-=±所以故答案为：23±16．若函数为上的单调函数，则实数的取值范围是______.22, ()2, x x a f x x x x a ≥⎧=⎨-+<⎩R a 【答案】(]{},02-∞⋃【分析】分段函数是增函数，等价于每一段都是增函数，且“后一段”解析式在分段处的函数值不小于“前一段”解析式在分段处的函数值，据此列不等式求解即可【详解】因为当时，均为增函数，故只能为上的单调递增函数，1x <22,4y x y x x ==-+()f x R 在上为增函数，在上为减函数，故，根据分段函数是增函数可知，24y x x =-+(],2-∞[)2,+∞2a ≤，解得，结合可知，.224a a a ≥-+(][),02,a ∈-∞⋃+∞2a ≤(]{},02a ∞∈-⋃故答案为：(]{},02a ∞∈-⋃四、解答题17．已知角满足αsin cos αα-=(1)求的值；tan α(2)若角是第三象限角，，求的值.α()()()()()sin πtan 5πcos π3πtan 2πcos 2f αααααα-++=⎛⎫--- ⎪⎝⎭()f α【答案】(1)答案见解析【分析】（1）利用同角三角函数基本关系式列方程组求解即可；（2）利用诱导公式求解即可.【详解】（1）由题意和同角三角函数基本关系式，有，22sin cos sin cos 1αααα⎧-=⎪⎨⎪+=⎩消去得，解得sin α25cos 20αα-=cos α=cos α=当角是第一象限角时，，α1cos tan 2ααα==因为角是第三象限角，.αcos tan 2ααα===（2）由题意可得，()()sin tan cos cos tan sin f ααααααα--==--因为角是第三象限角，α所以cos α=()f α=18．已知函数的最小正周期为.()π2sin 1(0)3f x x ωω⎛⎫=++> ⎪⎝⎭π(1)求的值；π6f ⎛⎫⎪⎝⎭(2)求函数的单调递减区间：()f x (3)若，求的最值.π0,2x ⎡⎤∈⎢⎥⎣⎦()f x【答案】(1)π16f ⎛⎫=+ ⎪⎝⎭(2)π7ππ,π,Z 1212k k k ⎡⎤++∈⎢⎥⎣⎦(3)最大值为3，最小值为1+【分析】（1）由最小正周期，求得，得到，再求；ω()f x 6f π⎛⎫⎪⎝⎭（2）整体代入法求函数的单调递减区间；（3）由的取值范围，得到的取值范围，可确定最值点，算出最值.x π23x +【详解】（1）由最小正周期公式得：，故，2ππω=2ω=所以，所以.()π2sin 213f x x ⎛⎫=++ ⎪⎝⎭πππ2sin 211663f ⎛⎫⎛⎫=⨯++= ⎪ ⎪⎝⎭⎝⎭（2）令，解得，ππ3π2π22π,Z 232k x k k +≤+≤+∈π7πππ,Z 1212k x k k +≤≤+∈故函数的单调递减区间是.()f x π7ππ,π,Z 1212k k k ⎡⎤++∈⎢⎥⎣⎦（3）因为，所以，π0,2x ⎡⎤∈⎢⎣⎦ππ4π2,333x ⎡⎤+∈⎢⎥⎣⎦当，即时，的最大值为3，ππ232x +=π12x =()f x当，即时，的最小值为.π4π233x +=π2x =()f x 119．已知函数.()2121x f x =+-(1)判断的奇偶性并证明；()f x (2)判断在区间上的单调性，并利用函数单调性的定义证明.()f x ()0,∞+【答案】(1)奇函数，证明见解析(2)在区间上单调递减，证明见解析()f x ()0,∞+【分析】（1）根据奇函数的定义进行判断证明即可；（2）根据函数单调性的定义，结合指数函数的单调性进行判断证明即可.【详解】（1）函数为奇函数，理由如下：()f x 函数的定义域为，()2121x f x =+-{}0x x ≠∣对任意的，{}()()()22121120,1,21212112x x xx x x xx x x f x f x f x --+++∈≠=+=-===-----∣所以是奇函数；()f x （2）在区间上的单调递减，理由如下：()f x ()0,∞+对任意，且，()12,0,x x ∈+∞12x x <，()()())()2112121212222222211212121212121x x x x x x x x f x f x -⎛⎫-=+-+=-= ⎪------⎝⎭因为在单调递增，且，所以，2xy =()0,∞+120x x <<21121120,20,220x x x x --->>>所以，()()120f x f x ->所以在区间上的单调递减.()f x ()0,∞+20．已知函数，且不等式的解集为.()53f x ax x =+-()3xf x <{1}xx b <<∣(1)求实数的值；,a b (2)解关于的不等式（其中为常数）；x ()20ax ac b x bc -++<,c c ∈R (3)已知，若存在，使得成立，求实数的取()73g x mx m=+-[][]122,3,1,4x x ∈∈()()112x f x g x =m 值范围.【答案】(1)2b =(2)答案见解析(3)][(),21,∞∞--⋃+【分析】（1）由题意判断出是方程的两根，即可求解；1,b 2320ax x -+=（2）对a 分类讨论，分别写出不等式的解集；（3）设的值域为的值域为，判断出，()[],2,3y xf x x =∈()[],73,1,4A g x mx m x =+-∈B A B ⋂≠∅列不等式组，求出m 的范围.【详解】（1）因为，所以可化为，即，()53f x ax =+-()3xf x <2533ax x +-<2320ax x -+<因为不等式的解集为，即是方程的两根，()4xf x <{1}xx b <<∣1,b 2320ax x -+=将代入，得，故，1x =2320ax x -+=320a -+=1a =再由韦达定理得，故.122b ⨯==2b =（2）可化为，即，()20ax ac b x bc -++<()2220x c x c -++<()()20x c x --<当时，解得，2>c 2x c <<当时，不等式为，无解；2c =2(2)0x -<当时，解得；2c <2c x <<综上所述，当时，不等式解集为；2>c {2}xx c <<∣当时，不等式解集为；2c =∅当时，不等式解集为.2c <{2}xc x <<∣（3）因为存在，存在，使得成立，[]12,3x ∈[]21,4x ∈()()112x f x g x =设的值域为的值域为，则，()[],2,3y xf x x =∈()[],73,1,4A g x mx m x =+-∈B A B ⋂≠∅由（1）得，对称轴为，()235xf x x x =-+32x =故在上单调递增，所以，()y xf x =[]2,3[]3,5A =①当时，，不满足题意；0m =(){}7,7g x B ==②当时，在上单调递增，所以，所以，解得：；0m >()g x []1,4[]72,7B m m =-+72573m m -⎧⎨+⎩ 1m ③当时，在上单调递减，所以，所以，解得：0m <()g x []1,4[]7,72B m m =+-75723m m +⎧⎨-⎩ ；2m - 综上所述，.][(),21,m ∞∞∈--⋃+【点睛】方法点睛：常见解不等式的类型：(1)解一元二次不等式用图像法或因式分解法；(2)分式不等式化为标准型后利用商的符号法则；(3)高次不等式用穿针引线法；(4)含参数的不等式需要分类讨论．21．已知函数.()12f x x x =++-(1)求的最小值；()f x (2)设的最小值为，若正数满足，求的最小值；()f x m ,a b a b m +=13a a b +(3)设，若，求所有满足条件的的取值集合.()()()g x f x f x =+-()()2431g a a g a -+=-a 【答案】(1)3(2)73(3)或22a ≤≤4a =【分析】（1）将化简为，求出的单调性，即可求出的最小值；()f x ()21,23,1212, 1.x x f x x x x -≥⎧⎪=-<<⎨⎪-≤-⎩()f x ()f x （2）由（1）可知，即，由均值不等式即可的最小值.3m =3a b +=13a a b +（3）由题可知，为偶函数，分类讨论或，，解方程()g x 2431a a a -+=-1a -21431111a a a ⎧-≤-+≤⎨-≤-≤⎩和不等式即可得出答案.【详解】（1）因为，所以()12f x x x =++-()21,23,1212, 1.x x f x x x x -≥⎧⎪=-<<⎨⎪-≤-⎩当时，单调递增，最小值为3，2x ≥()f x 当时，单调递减，最小值为3，1x ≤-()f x 当时，，12x -<<()3f x =综上所述，的最小值为3.()f x （2）由（1）可知，即，3m =3a b +=因为均为正数，所以，,a b 1331317233333a a b a b a a ba b a b ++=+=++≥+=当且仅当即时取等号，3a b =39,44a b ==即的最小值为.13a a b +73（3）由题可知，的定义域为，关于原点对称，()g x R ()()()()g x f x f x g x -=-+=所以为偶函数，()g x 当时，，2x ≤-()12214g x x x x=---=-当时，，21x -<≤-()12324g x x x =-+=-+当时，，11x -<≤()6g x =当时，，12x <≤()31242g x x x =++=+当时，.2x >()21124g x x x x=-++=所以，()4,242,126,1124,214,2x x x x g x x x x x x >⎧⎪+<≤⎪⎪=-<≤⎨⎪-+-<≤-⎪-≤-⎪⎩则在上单调递增，在上单调递减，()g x ()1,+∞(),1-∞-①，解得或，2431a a a -+=-1a =4a =②，解得或，()2431-+=--a a a 1a =2a =③，解得，21431111a a a ⎧-<-+≤⎨-<-≤⎩22a ≤≤综上所述，或.22a ≤≤4a =22．已知函数在时有最大值和最小值，设.()221g x ax ax b =-++(),0a b ≥[]1,2x ∈10()()g x f x x =(1)求实数的值；,a b (2)若不等式在上恒成立，求实数的取值范围；()22log 2log 0f x k x -≤11,84x ⎡⎤∈⎢⎥⎣⎦k (3)若关于的方程有三个不同的实数解，求实数的取值范围.x ()22131021x x mf m -+--=-m【答案】(1)1,0a b ==(2)8,9∞⎛⎤-⎥⎝⎦(3)12m >-【分析】（1）根据已知条件列方程组，由此求得的值.,a b （2）结合换元法、分离常数法化简不等式，结合二次函数的性质求得的()22log 2log 0f x k x -≤k 取值范围.（3）利用换元法化简方程为一元二次方程的形式，结合指数型函数的()22131021x xmf m -+--=-图象、一元二次方程根的分布的知识求得的取值范围.m 【详解】（1）函数时不合题意，()()222111,0g x ax ax b a x b a a =-++=-++-=所以为，所以在区间上是增函数，0a >()g x []1,2故，解得.()()211110g b g b a ⎧=+=⎪⎨=+-=⎪⎩10a b =⎧⎨=⎩（2）由已知可得，则，()221g x x x =-+()()12g x f x x x x ==+-所以不等式，()22log 2log 0f x k x -≤转化为在上恒成立，2221log 22log 0log x k x x +--≤11,84x ⎡⎤∈⎢⎥⎣⎦设，则，即，在上恒成立，2log t x =[]3,2t ∈--1220t kt t +--≤[]3,2t ∈--即，[]22121111211,3,2,,23k t t t t t ⎛⎫⎡⎤≤+-=-∈--∴∈-- ⎪⎢⎥⎝⎭⎣⎦ 当时，取得最小值，最小值为，则，即.∴113t=-211t ⎛⎫- ⎪⎝⎭2116139⎛⎫--= ⎪⎝⎭1629k ≤89k ≤所以的取值范围是.k 8,9∞⎛⎤- ⎥⎝⎦（3）方程可化为：，，()22131021x x mf m -+--=-()()2213321120x x m m --+-++=210x -≠令，则方程化为，，21x t-=()()233120t m t m -+++=()0t ≠∵方程有三个不同的实数解，()22131021x x mf m -+--=-∴画出的图象如下图所示，21x t =-所以，，有两个根､，且或，.()()233120t m t m -+++=()0t ≠1t 2t 1201t t <<<101t <<21t =记，()()()23312h t t m t m =-+++则，即，此时，()()0120110h m h m ⎧=+>⎪⎨=--<⎪⎩121m m ⎧>-⎪⎨⎪>-⎩12m >-或得，此时无解，()()()012011033012h m h m m ⎧⎪=+>⎪⎪=--=⎨⎪-+⎪<-<⎪⎩121113m m m ⎧>-⎪⎪=-⎨⎪⎪-<<-⎩m 综上.12m >-【点睛】研究复杂的方程的根、函数的零点问题，主要考虑化归与转化的数学思想方法，将不熟悉、陌生的问题，转化为熟悉的问题来进行求解.如本题中，将方程有三个解的问，转化为指数型函数、二次型函数的知识来进行求解.。

马鞍山市2022~2023学年第一学期期末教学质量监测高一英语试题第一部分听力（共两节，满分30分）第一节（共5小题；每小题1.5分，满分7.5分）请听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. Who plays the best in the football team?A. Karen.B. John.C. Mike.【答案】B【解析】【原文】M: Hi, Karen. Who plays the best in your school football team?W: As far as I know, John plays the best, but Mike is the most popular one.2. How many people took part in the race?A. 60.B. 40.C. 20.【答案】C【解析】【原文】M: Linda, how many people signed up for the race last week?W: About 60, but only twenty of them took part in it.3. What is the man speaker probably?A. A teacher.B. A student.C. A headmaster.【答案】B【解析】【原文】M: Annie, I’m going to miss Mr Wang. Why isn’t he going to teach us this term?W: He just became our headmaster.4. Where are the speakers probably?A. In the supermarket.B. At home.C. In a park.【答案】A【解析】【原文】M: Hi, Mary. Long time no see. Do you often shop here?W: Yes, I live near this supermarket. My apartment is across from Yushanhu Park.5. What are the speakers talking about?A. Hong Kong Disneyland.B. Their schoolwork.C. Their vacation plan.【答案】C【解析】【原文】W: Daisy, what are you going to do for your winter vacation?M: I planned to go to Hong Kong and visit Hong Kong Disneyland. But I’ll have to stay at home and study because my schoolwork is not satisfactory.第二节（共15小题；每小题1.5分，满分22.5分）请听下面5段对话或独白。

南京市2021-2022学年度第一学期期末学情调研测试卷高一语文2022.01 注意事项：1．本试卷共150分，考试用时150分钟。

2．答题前考生务必将学校、班级、姓名、学号填写在答题卡上。

每题答案写在答题卡上对应题目的空格里或横线上。

考试结束，请将答题卡交回。

一、现代文阅读（35分）（一）现代文阅读Ⅰ（本题共5小题，19分）阅读下面的文字，完成1~5题。

材料一：中国民族音乐，从古到今，都是声乐占主导地位。

所谓“丝不如竹，竹不如肉，渐近自然也”（《世说新语》）。

中国古代所谓“乐”，并非纯粹的音乐，而是舞蹈、歌唱、表演的一种综合。

《乐记》上有一段记载：“故歌之为言也，长言之也。

悦之故言之，言之不足故长言之，长言之不足故嗟叹之，嗟叹之不足，故不知手之舞之，足之蹈之也。

”“歌”是“言”，但不是普通的“言”，而是一种“长言”，“长言”即入腔，成了一个腔调，从逻辑语言、科学语言走入音乐语言、艺术语言。

为什么要“长言”呢？就是因为这是一个情感的语言。

“悦之故言之”，因为快乐，情不自禁，就要说出，普通的语言不够表达，就要“长言之”和“嗟叹之”（入腔和行腔）。

这就到了歌唱的境界。

更进一步，心情的激动要以动作来表现，就走到了舞蹈的境界，所谓“嗟叹之不足，故不知手之舞之，足之蹈之也”。

这种思想在当时较为普遍。

《毛诗序》也说了相类似的话：“情动于中而形于言，言之不足故嗟叹之，嗟叹之不足故永歌之，永歌之不足，不知手之舞之，足之蹈之也。

”这也是说，逻辑语言，由于情感之推动，产生飞跃，成为音乐的语言，成为舞蹈。

那么，这推动逻辑语言使成为音乐语言的情感又是怎么产生的呢？古代思想家认为，情感产生于社会的劳动生活和阶级的压迫，所谓“男女有所怨恨，相从为歌。

饥者歌其食，劳者歌其事”。

这显然是一种进步的美学思想。

（节选自宗白华《美学散步》）材料二：盛唐本来就是一个音乐高潮。

当时传入的各种异国曲调和乐器，如龟兹乐、天竺乐、西凉乐、高昌乐等等，融合传统的“雅乐”“古乐”，出现了许多新创造，从宫廷到市井，从中原到边疆，从太宗的“秦王破阵”到玄宗的“霓裳羽衣”，从急骤强烈的跳动到徐歌曼舞的轻盈，正是那个时代的社会氛围和文化心理的写照。

绝密★启用前张家口市2022－2023学年度高一年级第一学期期末考试英语试卷班级姓名注意事项：1．答卷前，考生务必将自己的姓名、班级和考号填写在答题卡上。

2．回答选择题时，选出每小题答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号。

回答非选择题时，将答案写在答题卡上。

写在本试卷上无效。

3．考试结束后，将本试卷和答题卡一并交回。

第一部分听力（共两节，满分30分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例：How much is the shirt?A．￡19.15. B．￡9.18. C．￡9.15.答案是C。

1．What does Steven usually do on Saturday?A．Go swimming.B．Take a walk.C．Ride a bike.2．Where are the speakers going first?A．To the library.B．To the swimming pool.C．To the gym.3．How much is the coat?A．$200.B．$120.C．$80.4．What is the weather like now?A．Sunny.B．Windy.C．Rainy.5．What is the woman probably doing now?A．Hosting a program.B．Delivering a speech.C．Seeing a doctor.第二节（共15小题；每小题1.5分，满分22.5分）听下面5 段对话或独白。

江苏省南京市六校联合体2023-2024学年高二上学期期初调研英语试题（扫描版含答案无听力音频含听力原文）2023-2024学年第一学期六校联合体期初调研测试高二英语第一部分听力（共两节，满分30分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C 三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.What kind of TV programs does the man likeA.Quiz shows.B.Documentaries.C.Situationcomedies.2.Where does the conversation probably take placeA.In an office.B.In a restaurant.C.In a convenience store.3.What are the speakers talking aboutA.The weather.B.Writing skills.C.The plan for tomorrow.4.When will the speakers go to watch the matchA.At7:00.B.At7:30.C.At8:00.5.What is the probable relationship between the speakersA.Boss and secretary.B.Taxi driver and passenger.C.Shop assistant and customer.第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

西宁市2022—2023学年第一学期末调研测试卷高一英语注意事项：1.本试卷满分150分，考试时间120分钟。

2.本试卷为试题卷，不允许作为答题卷使用，答题部分请在答题卡上作答，否则无效。

3.答卷前，考生务必将自己的姓名、准考证号、考场、座位号填写在答题卡上，同时将学校、姓名、准考证号、考场填写在本试卷上。

4.选择题用2B铅笔把答题卡上对应题目的答案标号涂黑(如需改动，用橡皮擦干净后，再选涂其他答案标号)。

非选择题用0.5毫米的黑色签字笔答在答题卡相应的位置，书写工整，字迹清晰。

第一部分听力(共两节，满分30分)做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共5小题；每小题1.5分，满分7.5分)听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. When will the taxi arrive?A. In ten minutes.B. Right now.C. In half an hour.2. What is the relationship between the speakers?A. Nurse and patient.B. Boss and worker.C. Teacher and student.3. Where will the speakers have lunch?A. In the park.B. In the office.C. In the dining hall.4. What are the speakers most probably talking about?A. A book.B. An actor.C. A movie.5. What does the man need now?A. Ice cream.B. Milk.C. Water.第二节(共15小题；每小题1.5分，共22.5分)听下面5段对话或独白。