2018 级高一年级阶段性测试数学试题

2018年高一段考（3）数学试题参考答案一、选择题：ACBBD CDDDA CD 二、填空题：13、1 14、（1，3） 15、3 16、R ；（2，+∞） 三、解答题：17、解：由2)3(log 21-≥-x 得 2)21(30-≤-<x 解得 31<≤-x∴}31|{<≤-=x x A A C R }31|{≥-<=x x x 或 由125≥+x 移项得 0125≥-+x 023≤+-x x 即 ⎩⎨⎧≠-≤-+020)3)(2(x x x 解得 32≤<-x ∴}32|{≤<-=x x B∴}312|{=-<<-=x x x B A C R 或18、解：：∵10<<a ∴当21y y >即)52(log )132(log 22-+>+-x x x x a a 时，有323212152132132052132022222<<⇒⎪⎩⎪⎨⎧<<><⇒⎪⎩⎪⎨⎧-+<+-+-<⇒-+<+-<x x x x x x x x x x x x x x 或 ∴当32<<x 时，有21y y >19、解：设2121)0[,x x x x <∞+∈且，则 ]333)33[(21)]3131()33[(212332332112212121221121x x x x x x x x x x x x x x y y +---+-=-+-=+-+=-)311)(33(212121x x x x +--=∵210x x <≤ ∴2133x x < 即 03321<-x x 又0321>+x x ∴13121<+x x 即031121>-+x x∴021<-y y 即 21y y < ∴233xx y -+=在[0，+∞）上是增函数当0=x 时，y 有最小值1.20、解：① 由0322>-+x x 解得 31<<-x ∴ 函数的定义域是)3,1(- ② 设4)1(3222+--=-+=x x x u 当31<<-x 时 40≤<u ∴14log 4=≤y∴函数)32(log 24x x y -+=的值域是]1(，-∞ ③ 设4)1(3222+--=-+=x x x u当11≤<-x 时，u 单调递增；当31<≤x 时，u 单调递减∴)32(log 24x x y -+=的单调增区间为]1,1(-； 单调减区间为)3,1[ 21、解：当2014≤≤P 时，直线过 (14,22)，(20,10) 两点 ∴502+-=P Q 当2620≤<P 时，直线过 (20，10)，(26，1) 两点 ∴4023+-=P Q ∴⎪⎩⎪⎨⎧≤<+-≤≤+-=)2620( 4023)2014( 502P P P P Q(1)设利润为W 1，则Q PQ W 141-=，则该店至少能维持职工生活必须满足W 1≥56即 ⎪⎩⎪⎨⎧≤<≥+--≤≤≥+--)2602( 56)4023)(14()2014( 56)502)(14(P P P P P P 解得 2218≤≤P (2)设扣除职工生活费后余额为W 2，则56)14(2--=Q P W∴⎪⎩⎪⎨⎧≤+--≤≤+--=⎪⎩⎪⎨⎧-+--+-=⎪⎩⎪⎨⎧-+---+--=）＜（）（）　（26P 20625361P 2320P 145.4)5.19(26166123756782 56)4023)(14( 56)502)(14(22222P P P P P P P P P W ∴当P ＝19.5时，W 2最大，最大余额为4.5百元即450元. 22、解：（1）k x x x f +-=2)(由 2)(l o g 2=a f 得=)(a f 22＝4 ∴ 42=+-k a a ………… ①又k a f =)(log 2 ∴k k a a =+-222log )(log 即 0)1(log log 22=-a a ∵1≠a ∴0log 2≠a ∴1log 2=a ∴2=a把2=a 代入① 得 4222=+-k 解得 2=k ∴2)(2+-=x x x f（2）47)21(log 2log )(log )(log 222222+-=+-=x x x x f 当21log 2=x 即2=x 时，)(log 2x f 有最小值47.。

佛山一中2017-2018学年度高一第二次段考数学答案二、填空题. 13.49 14.53- 15.R x x x f ∈-=),62sin(2)(π 16.]21,(--∞ 三、解答题 17.解：（1）1sin cos 5θθ+=① ()21sin cos =1+2sin cos =25θθθθ∴+-----------------------------------------------------------------2分 242sin cos =-025θθ<------------------------------------------------------------------------------------------3分 ()0sin 0,cos 0sin cos 0θπθθθθ∈∴><∴->，-------------------------------------------4分()2497sin -cos =1-2sin cos =sin -cos =255θθθθθθ∴∴②--------------------------------------5分 由①②得：43sin ,cos 55θθ==-----------6分sin 4tan cos 3θθθ∴==-------------------------7分 （2）方法一：由（1）43sin ,cos 55θθ==-22224sin 165==.cos 2sin cos 33343--2-555θθθθ⎛⎫⎪⎝⎭∴-⎛⎫⎛⎫⨯⨯ ⎪⎪⎝⎭⎝⎭-----------------------------------------------10分 方法二：由（1）2244tan 163tan =.4312tan 33123θθθ⎛⎫- ⎪⎝⎭=-∴==-⎛⎫-- ⎪⎝⎭原式------------------------10分18.解：（1）()4f x ≤∴⎩⎨⎧<≤+--04142x x x 或⎩⎨⎧≥≤044x x ------------------------------2分解得：⎩⎨⎧<-≥-≤013x x x 或或⎩⎨⎧≥≤01x x 即：013<≤--≤x x 或或10≤≤x ----------------3分{}|3-11M x x x =≤-≤≤或---------------------------------------------------------------------------------5分21{|2530}{|3};2N x x x x x =-++≥=-≤≤ ---------------------------------------------7分 (2)11,2M N x x ⎧⎫⋂-≤≤⎨⎬⎩⎭----------------------------------------------------------------------------------9分 132R C N x x x ⎧⎫=<->⎨⎬⎩⎭或-------------------------------------------------------------------------------10分{}=13R M C N x x x ∴⋃≤>或---------------12分19.图象如下图--------------------------------------------------------------------------------------------------------------------------5分 最小正周期为22T ππ== ------------------------------------------------------------------------------- 6 分 令1)52sin(1=--=πx y得Z k k x ∈+=,210ππ对称中心为Z k k ∈+)1,210(ππ --------------------------------------------------------------------------8分 令Z k k x x ∈+=±=-,2207,1)52sin(πππ 对称轴为直线Z k k x ∈+=,2207ππ ---------------------------------------------------------------10分 73171110π20π5π20π10πxy 12由3222,252k x k k Z πππππ+≤-≤+∈得：717++,2020k x k k Z ππππ≤≤∈单调增区间为：717[++],2020k k k Z ππππ∈， --------------------------------------------------12分20.解：（1）因为()f x 是定义在R 上的奇函数， 所以()00f =，即1033a-+=+，解得1a =. ------------------------------------------------------- 1分 从而有()13133x x f x +-+=+. ---------------------- ------------------------------------------------------------- 2分此时R x ∈∀，都有)(331333133313)(x f x x x x f -=++--=+-=++--=-，所以()13133x x f x +-+=+为奇函数，1a =符合题意.-------------------------------------------------- 4分(2)由（1）知()()13112333331x x x f x +-+==-+++， ------------------------------------------------ 5分对于任意的12,x R x R ∈∈且12x x <，12210,330.xxx x ∴->-< --------------------------7分∴()()21f x f x ∴-()()21121233331331x x ⎛⎫⎛⎫ ⎪ ⎪=-+--+ ⎪ ⎪++⎝⎭⎝⎭-------------------------------8分 ()()2122331331x x =-++()()()1221233033131x x x x -=<++ -------------------------------------------- 11分所以()f x 在全体实数上为单调减函数. ------------------------------------------------------------- 12分21. 解：（Ⅰ）由题意：设当20020≤≤x 时)0()(≠+=a b ax x v ------------------------1分⎩⎨⎧==60)20(0)200(v v 所以⎩⎨⎧=+==+=6020)20(0200)200(b a v b a v ---------------------------------------------------3分 解得 3200,31=-=b a ---------------------------------------------------------------------------4分 ∴ 当20020≤≤x 时 )200(31)(x x v -= ---------------------------------------------------5分（Ⅱ）由（Ⅰ）可得⎪⎩⎪⎨⎧≤≤-≤≤=)20020()200(31)200(60)(x x x x v -----------------------------------6分∴ ⎪⎩⎪⎨⎧≤≤-≤≤=)20020()200(31)200(60)(x x x x xx f ------------------------------------------------------8分当200≤≤x 时，)(x f 是增函数，当20=x 时候其最大值为12002060=⨯；---------9分20020≤≤x 时，310000)100(31)200(31)(2+--=-=x x x x f -----------------------10分 当100=x 时，其最大值为3333310000≈（辆/小时） ---------------------------------11分 综上所述，当车流密度100=x (辆/千米)时，车流密度最大值为3333（辆/小时）------12分22.解：（Ⅰ）证明：xa a ax a x a a x x a f x f +--+-++--+=-++21221)2(2)(01221121=--+--+-+=-+-++--+=x a x a x a a x a x x a x a a x∴结论成立 …………………………………………………………………………………2分 （Ⅲ）解：)(|1|)(2a x a x x x g ≠-++=（1）当a x a x x x g a x a x -++=-++=≠-≥43)21(1)(,122时且…………………3分 如果211-≥-a 即21≥a 时，则函数在),(),1[+∞-a a a 和上单调递增2min )1()1()(-=-=a a g x g …………………………………………………………………4分 如果a g x g a a a -=-=-≠<-<-43)21()(,2121211min 时且即当……………………………5分当21-=a 时，)(x g 最小值不存在…………………………………………………………6分 （2）当45)21(1)(122-+-=+--=-≤a x a x x x g a x 时 …………………………7分如果45)21()(23211min -==>>-a g x g a a 时即…………………………………………8分如果2min )1()1()()1,()(23211-=-=--∞≤≤-a a g x g a x g a a 上为减函数在时即…9分又当0)21()43()1(210)23()45()1(232222>-=---<>-=--->a a a a a a a a 时当时……………………………………………………………………………………………11分综合得：当2121-≠<a a 且时,)(x g 最小值是a -43 当2321≤≤a 时,)(x g 最小值是2)1(-a , 当23>a 时 g （x ）最小值为45-a当21-=a 时,)(x g 最小值不存在…………………………………………………………12分。

分宜中学2018-2019学年度下学期高一年级第一次段考数 学 试 卷一、单选题(每小题5分，共60分） 1．下列角终边位于第二象限的是（ ）A ．420B ．860C ．1060D ．12602．已知扇形的弧长为4 cm ，圆心角为2 弧度，则该扇形的面积为 （ ）A ．4 cm 2B ．6 cm 2C ．8 cm 2D ．16 cm 23．在平面直角坐标系中，已知角始边与x 轴非负半轴重合，顶点与原点重合，且终边上有一点P 坐标为，则A ．1313B ．1313-C ．13134 D ．13134-4．已知θ为锐角，则)2sin()sin(21θπθπ--+=（ ）A ．θθsin cos -B ．θθcos sin -C ．)cos (sin θθ-±D ．θθcos sin + 5．下列函数的最小正周期为π且图象关于直线3π=x 对称的是（ ）A ．)32sin(2π+=x y B ．)62sin(2π-=x y C ．)32sin(2π+=x y D ．)32sin(2π-=x y6．要得到函数)32sin(π+=x y 的图象，只需将函数x y sin =的图象（ ）A ．先向左平移3π个单位长度，再横坐标伸长为原来的2倍，纵坐标保持不变 B ．先向左平移6π个单位长度，再横坐标缩短为原来的21倍，纵坐标保持不变 C ．先横坐标伸长为原来的2倍，纵坐标保持不变，再向左平移3π个单位长度D ．先横坐标缩短为原来的21倍，纵坐标保持不变，再向左平移6π个单位长度 7．函数1cos 2+=x y 的定义域是（ ）A ．)(32,32Z k k k ∈⎥⎦⎤⎢⎣⎡+-ππππ B ．)(322,32Z k k k ∈⎥⎦⎤⎢⎣⎡++ππππ C ．)(62,62Z k k k ∈⎥⎦⎤⎢⎣⎡+-ππππ D ．)(322,322Z k k k ∈⎥⎦⎤⎢⎣⎡+-ππππ8．已知43)2sin(=+βα，31cos =β，βα,为锐角，则)sin(βα+的值为（ ）A ．121423+ B ．121423- C ．122273+ D ．122273- 9．函数)3cos(2)(x x f -=π的单调递增区间是（ ）A ．)(342,32Z k k k ∈⎥⎦⎤⎢⎣⎡++ππππB ．)(322,32Z k k k ∈⎥⎦⎤⎢⎣⎡+-ππππ C ．)(32,322Z k k k ∈⎥⎦⎤⎢⎣⎡+-ππππ D ．)(342,322Z k k k ∈⎥⎦⎤⎢⎣⎡+-ππππ 10．在锐角ABC ∆中，已知C B A >>，则B cos 的取值范围为（ ）A. )22,0( B.)22,21( C.)21,0( D.)23,21(11．若33)6sin(=-απ，则=+)26sin(απ（ ） A ．31 B ．322 C ． 33 D ．3612．若在⎥⎦⎤⎢⎣⎡2,0π内有两个不同的实数x 满足m x x =+2sin 32cos ，则实数m 的取值范围是（ ）A ．21≤<mB ．21<≤mC ．21<≤-mD ．21≤<-m二、填空题(每小题5分，共20分）13．若31sin =α，则=+)2cos(απ______． 14．已知53)2sin(-=-απ，且πα<<0，则________．15．函数)(sin 2cos R y ∈+=θθθ的值域是_________. 16．设函数)42sin()(π-=x x f ，则下列结论正确的是______. (写出所有正确命题的序号) 函数)(x f y =的递减区间为)(87,83Z k k k ∈⎥⎦⎤⎢⎣⎡++ππππ； 函数)(x f y =的图象可由x y 2sin =的图象向右平移4π得到；函数)(x f y =的图象的一条对称轴方程为8π=x ；若⎥⎦⎤⎢⎣⎡∈2,247ππx ，则)(x f 的取值范围是⎥⎦⎤⎢⎣⎡1,22． 三、解答题（共70分）17．（本小题10分）已知扇形的周长为40，当它的半径和圆心角取何值时，才使扇形的面积最大？ 并求扇形面积的最大值。

太原市2018~2019学年第一学期高一年级阶段性测评数学试卷分析一、选择题（本大题共12小题，每小题3分，共36分） 1. 已知集合{}12A x x =−<<，{}01B x x =<<，则（ ）A. B A ⊆B.A B ⊆C.A B =D.A B =∅2. 函数()f x x =+的定义域为（ ） A. [)0,+∞ B. ()0,+∞ C. R D. {}0x x ≠3. 已知集合2=450M x x x −−=，21N x x ==，则MN =（ ）A. {}1,1,5−B. {}5,1−C. {}1− D. {}1,1−4. 已知函数()2log f x x =，且()2f a =，则a =（ ） A. 4 B. 2 C. 1D. 1 5. 已知集合{0,1}A =，若B A A =，则满足该条件的集合B 的个数是（ ）A. 1B. 2C. 3D. 4BA A =6. 下列函数中，既是偶函数又在(0,)+∞上是增函数的是（ ）A. ||1()x y =B. ||y x x =C. lg ||y x =D. 12y x =7. 已知30.4a =，0.43b =，4log 0.3c =，则（ ） A. a b c <<B. a c b <<C. c a b <<D. c b a <<{}1MN =−，所以选解析：()2log f x x =8. 已知全集U R =，{|09,}A x x x R =<<∈，{|44,}B x x x Z =−<<∈，则图中阴影部分所表示的集合中的元素个数是（ ） A. 3 B. 4C. 5D. 无穷多9. 已知集合2|320A x ax x =−+= 中有且只有一个元素，那么实数a 的取值集合是（ ）A. 98⎧⎫⎨⎬⎩⎭B. 90,8⎧⎫⎨⎬⎩⎭C. {}0D. 20,3⎧⎫⎨⎬⎩⎭10. 已知函数()2log 2f x x=+，则函数()f x 的图象（ ） A. 关于x 轴对称 B. 关于y 轴对称C. 关于直线y x =对称D. 关于原点对称11. 已知函数()31,12,1x x f x x x⎧+≤⎪=⎨>⎪⎩，若对任意的实数x ，都存在1x R ∈，使得()()1f x f x ≤成立，则1x =（ ）A. 1B. 2C. 3D. 412. 已知函数()()2ax bf x x c +=+的图象如图（1）所示，则函数()2g x ax bx c =+−的图象可能是（ ）二、填空题（本大题共4小题，每小题3分，共12分）13. 已知全集{}1,2,3,4,5U =,集合{}1,2,4A =,则U C A =_______.14. 函数21xy =−在[]1,3上的最大值为_______.15. 已知函数()f x 是定义在R 上的奇函数，当0x ≥时，()12xf x m ⎛⎫=+ ⎪⎝⎭，那么()1f −=_______.16. 已知R λ∈，函数()24,,43,,x x f x x x x λλ−≥⎧=⎨−+<⎩若函数()y f x =的图象与x 轴恰有两个交点，则实数λ的取值范围是_______.(]()1,34,+∞的取值范围为(]()1,34,+∞三、解答题（本大题共4小题，共48分）17. （本小题满分10分）已知集合{},,2A a b =，{}22,,2B b a =，若A B =，求实数,a b 的值18. （本小题满分10分） （1）已知log 86x =，求x 的值（2）已知()233log 101log x x −=+，求x 的值19. (本小题满分10分)已知幂函数()f x 的图像经过点13,3⎛⎫⎪⎝⎭（1）求函数()f x 的解析式；（2）设函数()()()2f x x g x =−⋅，求函数()g x 在区间1,12⎡⎤⎢⎥⎣⎦上的值域. 0x >20. (本小题满分10分) 说明：请同学们在（A ）、（B ）两个小题中任选一题作答. （A ）已知函数()2=2f x x ax a −+在区间(),2−∞上有最小值（1）求实数a 的取值范围； （2）当=1a 时，设函数()()=f x g x x，证明函数()g x 在区间()+∞1，上为增函数.（B ）已知函数()()2=log 4f x x ，()2=log g x x 的图像如图所示，点()11,A x y ，()22,B x y 在函数()y f x =的图像上，点()33,C x y 在函数()y g x =的图像上，且线段AC 平行于y 轴.（1）证明：13=2y y −；（2）ABC ∆为以角C 为直角的等腰直角三角形，求点B 的坐标.）()12+x x <∈∞1，21. （本小题满分12分）说明：请同学们在（A ）、（B ）两个小题中任选一题作答.（A ）已知函数()22x x f x k −=+⋅.（1）若函数()f x 为奇函数，求实数k 的值；（2）若对任意的[)0,x ∈+∞，都有()2xf x −>成立，求实数k 的取值范围. ）()11,A x y AC AC ABC ∆为以角（B ）已知函数()y f x =是定义在R 上的奇函数，且0x ≥时，()22f x x =−−.（1）求函数()y f x =的解析式，并在下图所示的坐标系中作出函数()y f x =的图像；（2）若对任意的x R ∈有()()()0f x a f x a −≤>恒成立，求实数a 的最小值.。

xOyxy Ox y Ox O y石门中学2018—2018学年度第一学期高一年级数学科中段考试题(全卷共3页,供高一级各班使用) 成绩______________ 说明:本卷必做题为第1题至第20题,满分为100分 . 一.选择题（每题4分共４0分，请把答案填在答题卷上）1. 已知集合{}|110,P x N x =∈≤≤集合{}2|60,Q x R x x =∈+-=则PQ 等于（ ）（A ）{}1,2,3 （B ）{}2,3 （C ）{}1,2 （D ）{}22.函数y ＝f (x )的图像与函数g (x )＝log 2x (x ＞0)的图像关于原点对称，则f (x )的表达式为 ( )(A )f (x )＝1log 2x(x ＞0) (B )f (x )＝log 2(－x )(x ＜0)(C )f (x )＝－log 2x (x ＞0) (D )f (x )＝－log 2(－x )(x ＜0)3. 已知函数x y e =的图象与函数()y f x =的图象关于直线y x =对称，则( )A ．()22()x f x e x R =∈B ．f(2x)=ln2·lnx (x>0)C ．()22()x f x e x R =∈D ．()2ln ln 2(0)f x x x =+> 4. 设()338x f x x =+-, 用二分法求方程3380(1,2)x x x +-=∈在内近似解的过程中, 计算得到(1)0,(1.5)0,(1.25)0,f f f <>< 则方程的根落在区间（ ）.A.（1,1.25）B.（1.25,1.5）C.（1.5,2）D.不能确定 5. 下列各式错误的是（ ）.A. 0.80.733>B. 0..50..5log 0.4log 0.6>C. 0.10.10.750.75-<D. lg1.6lg1.4>6. 已知定义在R 上的奇函数f (x )满足f (x+2)=－f (x ),则,f (6)的值为 ( )(A)－1 (B) 0 (C) 1 (D)2 7.函数()f x =13log 2,(0,3]x x +∈的值域为( )(A )[－1，1] (B )（－∞，1） （C ）[1，＋∞）(D )[3，＋∞） 8. 某文具用品店出售羽毛球拍和羽毛球，球拍每副定价20元，羽毛球每只定价5元，该店制定了两种优惠方法：①买一副球拍赠送一只羽毛球；②按总价的92%付款划购买4副球拍，羽毛球x 只(x 不于小于4)，总付款额y 元，若购买30只羽毛球，两种优惠方法中，哪一种更省钱？（ ） A B C D 9. 函数lg ||x y x=的图象大致是 （ ）A ．B ．C ．D ．10、设偶函数f(x)=log a |x-b|在()0，∞-上递增，则f(a+1)与f(b+2)的大小关系是（ ） A 、f(a+1)=f(b+2) B 、f(a+1)>f(b+2) C 、f(a+1)<f(b+2) D 、不能确定二.填空题 (每题4分共16分,请把答案填在答题卷上)11.设U={0，1，2，3，4}，A={0，1，2，3}，B={2，3，4}，则（C U A ）⋃（C U B ）=____12. 24,02(),(2)2,2x x f x f x x ⎧-≤≤==⎨>⎩已知函数则 ；若00()8,f x x ==则 .13．函数)23(log 221+-=x x y 的递增区间是___________________14、下列四个命题：⑴log ()a y x =--与log a y x =的图象关于原点对称；⑵log (2)x a y a =+在R 上是减函数；⑶()f x =)32lg(2++x x 的最小值为lg2；⑷将函数()f x =1xx -的图象左平移1个单位，再下平移一个单位后与函数()f x =1x的图象重合。

2018级高一下学期阶段性考试一本试卷分第I卷（选择题）和第II卷（非选择题）两部分，共10页。

满分150分，考试时间120分钟。

考试结束后，将本试卷以及答题卡和答题纸一并交回。

答卷前，考生务必将自己的姓名、准考证号、考试科目填涂在试卷、答题卡和答题纸规定的地方。

第I卷(选择题，共100分)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）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 did the woman do today?A. She did nothing.B. She wrote letters.C. She practiced the piano.2. What does the man say about Delta Restaurant?A. The service was better.B. The food was delicious.C. The menu was attractive.3. What vehicle (交通工具) will the woman probably take tomorrow?A. A bus.B. A taxi.C. The subwa y.4. How much will the woman pay?A. $3.B. $6.C. $9.5. What are the speakers mainly talking about?A. A book.B. A teacher.C. An exam.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。

江苏省高邮中学高一年级十月份第一次阶段测试数学试卷（考试时间：120分钟 试卷满分：150分）一．填空题：本大题共14小题，每小题5分，共70分。

请把答案填写在答题纸相应的位置．．．．．．．．上．. 1．已知集合{1,2},{2,3}A B ==，则AB = ▲ ．2．已知集合{}1,3P = ,则集合P 的子集共有 ▲ 个．3．设函数()23f x x =+的值域为{1,7,13}，则该函数的定义域为 ▲ ． 4．已知函数2(1)2f x x x -=-，则()f x = ▲ ．5．著名的Dirichlet 函数1,()0,x D x x ⎧=⎨⎩为有理数为无理数，则(D D = ▲ ．6．函数()f x x=的定义域为 ▲ ． 7．若函数2()(1)3f x kx k x =+-+是偶函数，则()f x 的递增区间是 ▲ ． 8．函数2()f x x ax =-的单调增区间为[1,)+∞，则a = ▲ ． 9．函数2()22,[2,2]f x x x x =-++∈-的值域为 ▲ ． 10．已知集合{}21,0,,2A m m =-+，{}|01B x x =<<，若A B ≠∅，则实数m 的取值范围是 ▲ ．11．122[(1]= ▲ ．12．已知函数3()1,,f x ax bx a b R =-+∈，则函数()f x 在区间[2018,2018]-上的最大值与最小值之和为 ▲ ．13．设2()0,()2f x g x x x ==-，若(),()()()(),()()f x f x g x F x g x f x g x <⎧=⎨≥⎩，则()F x 的最大值为 ▲ ．14．几位同学在研究函数()1||xf x x =+()x R ∈时，给出了下面几个结论：①()f x 的单调减区间是(,0)-∞，单调增区间是(0,)+∞；②若12x x ≠，则一定有12()()f x f x ≠； ③函数()f x 的值域为R ；④若规定1()()f x f x =，1()[()]n n f x f f x +=，则()1||n xf x n x =+对任意*n N ∈恒成立．上述结论中正确的是____▲____．二．解答题（本大题共6小题，共计80分，请在答题纸指定区域内作答，解答时应写出文字说明、证明或演算步骤） 15．(本题满分12分) 求值：⑴已知42121=+-xx ，求1x x -+的值；16．(本题满分12分)已知集合{|11}A x a x a =-≤≤+，{|0B x x =≤或3}x >． ⑴若0=a ，求AB ；⑵若R A C B ⊆，求实数a 的取值范围．17．(本题满分12分)高一某班共有学生43人，据统计原来每人每年用于购买饮料的平均支出是120元。

2018级高一第一学期阶段性测试 数学试卷高 一 年级 数学 学科 试题卷一、 填空题：（本大题共14小题，每小题5分，计70分）1．设集合{}{}1,2,3,2,4,5A B ==,则AB = ▲ ．2．函数()f x =的定义域是 ▲ ． 3．设函数()221,12,1x x f x x x x ⎧+≤⎪=⎨+->⎪⎩则((1))f f -= ▲ ． 4．已知函数(23)21f x x -=+，则函数(1)f = ▲ ．5．若函数()(21)()x f x x x a =+-为奇函数，则a = ▲ ．6．函数y =的单调递减区间是 ▲ ．7．函数24y x x =-的定义域为[]4,a -，值域为[]4,32-，则实数a 的取值范围为 ▲ ．8．函数211x y x -=-，[)2,x ∈+∞的值域为 ▲ ． 9. 已知⎩⎨⎧<-≥=0,10,1)(x x x f ，则不等式(2)(2)5x x f x ++⋅+≤的解集是 ▲ ． 10．函数2(31)4,1(),1a x a x f x x x a x -+<⎧=⎨-++≥⎩在区间()+∞∞-,内是减函数，则a 的取值范围是 ▲． 11. 已知函数)(x f 满足),()(x f x f =-当)0,(,-∞∈b a 时总有)(0)()(b a ba b f a f ≠>--， 若)2()1(m f m f >+，则实数m 的取值范围是 ▲ ．12. 已知函数2(),2x f x x R x +=∈+，则不等式)43()2(2-<-x f x x f 的解集为 ▲ ． 13．设函数()||f x x x a =-，若对于任意的1212,[2,),x x x x ∈+∞≠，不等式1212()(()())0x x f x f x -->恒成立，则实数a 的取值范围是 ▲ ．14．设函数f (x )的定义域为D ，若存在非零实数m 满足对任意的x ∈M (M ⊆D )，均有x+m ∈D ，且f (x+m )≥f (x )，则称f (x )为M 上的m 高调函数．如果定义域为R 的函数f (x )是奇函数，当x ≥0时，f (x )=|x-a 2|-a 2，且f (x )为R 上的8高调函数，那么实数a 的取值范围是 ▲ ．二、 解答题：（本大题共6小题，计90分，解答应写出文字说明证明过程或演算步骤）15.（本题满分14分）已知集合{}{}15,,20,A x x x R B x x m x R =-<≤∈=-<∈，（1）当3m =时，求()R A B ð； （2）若{}14AB x x =-<<，求实数m 的值．16.（本题满分14分）已知函数()f x ,当,x y R ∈时,恒有()()()f x y f x f y +=+.（1）求证: ()()0f x f x +-=；（2）若(3)f a -=，试用a 表示(9)f ；（3）如果0x >时, ()0f x <且1(1)2f =-,试求()f x 在区间[2,6]-上的最大值和最小值.17.（本题满分15分）已知函数2()1x f x x =-+. （1）用定义证明函数)(x f 在()1,-+∞上为单调递减函数；（2）若()()g x a f x =-，且当[]1,2x ∈时)(x g 0≥恒成立，求实数a 的取值范围.18．（本题满分15分）销售甲、乙两种商品所得利润分别是12,y y 万元，它们与投入资金x万元的关系分别为1y a =，2=y bx ，（其中,,m a b 都为常数），函数12,y y 对应的曲线1C 、2C 如图所示．（1）求函数1y 与2y 的解析式；（2）若该商场一共投资4万元经销甲、乙两种商品，求该商场所获利润的最大值．19. （本题满分16分）已知函数()f x 是定义在R 上的偶函数，当0x ≥时，2()21f x x x =--． （1）求()f x 的函数解析式；（2）作出函数()f x 的简图，写出函数()f x 的单调区间及最值；（3）当x 的方程()f x m =有四个不同的解时，求实数m 的取值范围． 20．（本题满分16分）已知函数2()21f x ax x a =-+-(a 为实常数)．（1）若1a =，求()f x 的单调区间；（2）若0a >，设()f x 在区间[]1,2的最小值为()g a ，求()g a 的表达式；（3）设()()f x h x x=，若函数()h x 在区间[]1,2上是增函数，求实数a 的取值范围．梁丰高级中学2016～2017学年第一学期阶段性测试参考答案一填空1. {}1,2,3,4,5;2. (,3)(3,2]-∞-⋃-；3. 4；4. 5 ；5.12； 6.[]1,3 7.[]2,8； 8.(]2,3； 9.3(,]2-∞； 10.11[,)63； 11. 1m >或13m <-； 12.（1,2）13.2a ≤；14.⎡⎣. 二 解答题：（本大题共6小题，计90分，解答题应写出文字说明证明过程或演算步骤）15. 解：(1）当3m =时，{}32302B x x x x ⎧⎫=-<=<⎨⎬⎩⎭…………2分 32R C B x x ⎧⎫=≥⎨⎬⎩⎭ …………………5分 352R A C B x x ⎧⎫∴=≤≤⎨⎬⎩⎭………………8分 ⑵2m B x x ⎧⎫=<⎨⎬⎩⎭……………………10分 {}14A B x x =-<< ，4,82m m ∴==…………14分 16.证明：（1）令0,(0)2(0),x y f f ===即(0)0f =令,y x =-()()(0)0,f x f x f +-==…………………3分（2）令3,x y ==(6)2(3),f f =同理：(9)3(3)3(3)3f f f a ==-=-…………………7分（3）任取12x x >令1x y x +=，2x x =，则120,y x x =->1212()()(),0,()0f x f x f x x x f x ∴-=-><12()()0f x f x ∴-<即()f x 在R 上单调递减………………12分且(2)1,(6)3,f f -==-∴()f x 在区间[2,6]-上的最大值为1和最小值为-3………………14分17. 解：（1）12,x x ∀∈()1,-+∞，且12x x <21122122()()11x x f x f x x x -=-++ …………2分 =21122()(1)(1)x x x x -++ ……………………5分 1212211,10,10,0x x x x x x -<<∴+>+>->，……6分12()()0f x f x -> ，12()()f x f x > …………7分∴函数)(x f 在()1,-+∞上为单调递减函数；…………8分 ⑵[]1,2x ∈时，()()g x a f x =-0≥恒成立∴()a f x ≥对[]1,2x ∈恒成立，只要max ()a f x ≥……10分由（1）已证函数)(x f 在()1,-+∞上为单调递减函数，所以在[]1,2x ∈ 上也为单调递减函数 …………12分max ()(1)1f x f ==- ………………14分1a ∴≥-，即实数a 的取值范围[)1,-+∞…………15分18. 1.由题意解得44,55m a ==-；14(0)45y x =≥分 又81855b b =∴= 21(0)75y x x ∴=≥分（不写定义域扣1分）； 2.设销售甲商品投资x 万元，则乙投资4-x 万元.41(4)(04)1055y x x =+-≤≤分(1t t =≤≤ 则有2141555y t t =-++………13分 当t=2即x=3时y 取到最大值为1.答：该商场所获利润的最大值为1万元. ………15分19.解：（1）当0x <时，0x ->，则()()2()21f x x x -=----2=21x x +-， 又()f x 是偶函数，()()f x f x -=，则0x <时，2()21f x x x =+-，所以2221,0()21,0x x x f x x x x ⎧+-<=⎨--≥⎩ ………6分 （2）简图略， ……10分单调减区间是(][],1,0,1-∞-，单调增区间是[][)1,0,1,-+∞，…12分当1x =±时，函数()f x 有最小值2- ………14分（3）由图得，()2,1m ∈-- ………16分20.解：(1)1=a 时，⎪⎪⎩⎪⎪⎨⎧<++≥+-=⎪⎩⎪⎨⎧<++≥+-=+-=0,43)21(0,43)21(0,10,11||)(22222x x x x x x x x x x x x x f ………2分 ∴)(x f 的单调增区间为(+∞,21)，(-21,0) )(x f 的单调减区间为(-21,-∞)，(21,0) ……4分 (2)当0>a ， x ∈[1,2]时，1412)21(12)(22--+-=-+-=a a a x a a x ax x f ………5分 10 1210<<a即21>a 为增函数在]2,1[)(x f 23)1()(-==a f a g …6分 20 2211≤≤a 即,2141时≤≤a 1412)21()(--==a a a f a g …7分 30 221>a 即410<<a 时上是减函数在]2,1[)(x f 36)2()(-==a f a g …8分 综上可得 ⎪⎪⎪⎩⎪⎪⎪⎨⎧>-≤≤--<<-=21,232141,1412410,36)(a a a a a a a a g ………………9分(3)112)(--+=xa ax x h 在区间[1,2]上任取1x 、2x ，且21x x < 则)112()112()()(112221--+---+=-x a ax x a ax x h x h )]12([)12)((2121122112---=---=a x ax x x x x x x a a x x (*) …11分 ∵上是增函数在]2,1[)(x h ∴0)()(12>-x h x h∴(*)可转化为0)12(21>--a x ax 对任意1x 、都成立且212]2,1[x x x <∈ 即 1221->a x ax …………12分 10 当上式显然成立时,0=a ………………13分20 0>a a a x x 1221->由4121<<x x 得 112≤-aa 解得10≤<a …14分 30 0<a a a x x 1221-< 412≥-a a 021<≤-a ………15分 所以实数a 的取值范围是]1,21[- ……16分。

2018级高一年级阶段性测试本试题卷共10页, 全卷满分150, 考试用时120分钟。

注意事项：1. 答题前, 先将自己的姓名、准考证号填写在试题卷和答题卡上。

用2B铅笔将答题卡上试卷类型A后的方框涂黑。

2. 选择题的作答：每小题选出答案后, 用2B铅笔把答题卡上对应题目的答案标号涂黑。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

3. 非选择题的作答：用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

第Ⅰ卷选择题（共90分）听力（共两节, 满分 30 分）第一节听下面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。

What is the boy looking for？A.A schoolbagB.A table.C.A notebook.2. What has the man bought？k.B.Eggs.C.Bread.3. Where is the man's mother now？A.At home.B.At a hotel.C.In a hospital.4. What is the woman trying to do？A.Read a book .B.Watch a movie.C.Work on the computer.5. How much does one ticket cost？A.$2.B.S3.C.$4.第二节听下面5段对话。

每段对话后有几个小题, 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。

听每段对话前, 你将有时间阅读各个小题, 每小题5秒钟；听完后, 各小题将给出5秒钟的作答时间。

每段对话读两遍。

听第6段材料, 回答第6、7题。

Where does the conversation probably take place？A.In a bookstore.B.In a library.C.In a classroom.7. What will the man do？A.Find a class schedule.B.Reserve a book for the woman.C.Wait for a phone call.听第7段材料, 回答第8、9题8.What are the speakers mainly talking about?A. Cultures.B. Countries.C. Books.9.What interests the man about China?A. The food.B. The history.C. The architecture.听第8段材料, 回答第10至12题How is the man getting to the museum？On foot. B.By bus.C.By car.What can the man see after turning right at the third cross?A. A road sign.B. A museum.C. A policeman.12. What does the woman advise the man to do?A. Call the policeman.B. Move his car right now.C. Park his car in a car park.听第9段材料, 回答第13至16题13. When will the speakers go to the movies?A. On Thursday.B. On Friday.C. On Saturday.14. Where have the speakers decided to meet?A. At the station.B. In a café.C. At the entrance to the cinema.15.When will the speakers meet?A. At 7:30.B. At 6:25.C. At 6:00.16.Why can the speakers get cheaper tickets?A. They are students.B. They can get tickets online.C. There are few people to see The Ship听第10段材料, 月答第17至20题When do the photography classes begin?A. At 2:00.B. At 2:30.C. At 2:45.18. How much does Emma spend on the photography course ?£80 B. £65C. £45.19.What will Emma photograph next?A. People.B. Animals.C. Flowers.20.What does Emma want to do after the course?A. Buy a digital camera.B. Take photos for a competition.C. Show her photos to her friends.第二部分阅读理解 (共15小题；每小题2分, 满分30分)ACan you imagine a cloud floating in the middle of your room? Do you want to know what it feels like to fly like a bird? Find your answers in Time Magazine's 25 "Best Inventions". Here, we have picked out the most interesting 4 to share with you. Talking GlovesEver felt confused by the sign language used by disabled peo ple? Here is the "helping hand"you need. Four Ukrainian stud ents have created a pair of gloves that helps people with heari ng and speech problems communicate with others. The glove s are equipped with sensors that recognize sign language and translate it into text on a smartphone. Then the smartphone c hanges the text to spoken words.Google GlassesGoogle Glasses are like a computer built into the frame of a p air of glasses. With its 1.3-centimeter display, the glasses allow you to surf the Internet a nd make calls without even lifting a finger. The glasses also h ave a camera and GPS mapping system. Users can take and share photos, check maps and surf the Internet just by lookin g up, down, left and right.Indoor CloudsIt's not virtual.That's a real world.Dutch artist Smilde managed to create a small but perfect white cloud in the middle of a ro om using a fog machine.But it required careful planning-the temperature, humidity（湿度）and lighting all had to be just right.Once everything was ready , the cloud formed in the air with the machine.But it only laste d for a short while.WingsuitsThe suit fulfills your dreams of flying like a bird.Well, not exact ly flying, but gliding（滑翔）through the air.It increases the surface area of the human bod y, which makes it easier for people to float in the air. Fliers we aring wingsuits can glide one kilometer in about 30 seconds. Who will probably be the users of “Talking Gloves”? Disabled people B. Normal people.C. Trained people.D. Working people.22. How can users of Google Glasses surf the Internet?A. By moving the mouse.B. By tapping the keyboard.C. By pressing the button.D. By moving the eyeballs.23. Which of the following can help you fly?A. Talking Gloves.B. Google Glasses.C. Wingsuits.D. Indoor Clouds.BAlmost all cultures celebrate the end of one year and the begi nning of another in some way. Different cultures celebrate the beginning of a new year in different ways, and at different tim es on the calendar.In Western countries, people usually celebrate New Year at m idnight on December 31st or January 1st.People may go to pa rties, sometimes dressed in formal clothes, and they may drin k champagne（香槟）at midnight.During the first minutes of the new year, people ch eer and wish each other happiness for the year ahead.But so me cultures prefer to celebrate the new year by waking up ear ly to watch the sunrise.They welcome the new year with the fir st light of the sunrise.Many cultures also do special things to get rid of bad luck at t he beginning of a new year.For example, in Ecuador, families make a big doll from old clothes.The doll is filled with old new spapers and firecrackers.At midnight, these dolls are burned t o show the bad things from the past year are gone and the ne w year can start afresh（重新）.Other common traditions to keep away bad luck in a new year include throwing things into rivers or the ocean, or saying special things on the first day of the new year.Other New Year traditions are followed to bring good luck in t he new year.One widespread Spanish tradition for good luck i s to eat grapes on New Year's Day.The more grapes a person eats, the more good luck the person will have in the new year .In France, people eat pancakes for good luck at New Year.In the United States, some people eat black-eyed peas for good luck-but to get good luck for a whole year you have to eat 365 of th em！Which of the following can be the best title of the text?A. The meaning of "Happy New Year!"B. Several different New Year traditionsC. What to eat on New Year's DayD. Why people dress up nicely on New Year's Day25. What do you know from the first two paragraphs？A. Different cultures celebrate the beginning and ending of a y ear in the same way.B. The Western people celebrate the New Year only by watch ing the sunrise.C. People hold parties, wear new clothes and drink champagn e for a whole day.D. People around the world celebrate the New Year at differe nt times.26. In some cultures, why do people throw things into rivers or oceans?A. To bring good luck.B. To avoid bad luck.C. To forget everything.D. To plan for the next year.27. Which of following is CORRECT if people want to escape bad luck and wish for good luck?A. Friends tell something special to each otherB. Families make big dolls filled with old clothesC. Some people get up early to watch the sunriseD. Europeans eat 365 grapes on New Year's DayCDale Carnegie(戴尔·卡耐基) was an American writer and lecturer, and the developer of famous courses in self-improvement, salesmanship, corporate training, public speaki ng and interpersonal skills.Born in 1888 in Maryville, Missouri, Carnegie was a poor farm er's boy.His family moved to Belton, Missouri when he was a small child.In his teens, though still having to get up at 4 am e very day to milk his parents' cows, he managed to obtain an e ducation at the State Teacher's College in Warrensburg.His fir st job after college was selling correspondence courses.He m oved on to selling bacon, soap, and lard（猪油）for Armour&Company.After saving $500, Dale Carnegie quit sales in 1911 in order t o achieve a lifelong dream of becoming a lecturer.He ended u p instead attending the American Academy of Dramatic Arts i n New York, but found little success as an ter he got t he idea to teach public speaking.In his first session, he sugge sted that students speak about"something that made them an gry", and discovered the technique that made speakers unafr aid to address a public audience.From its beginning, the Dale Carnegie Course developed.Carnegie had made use of the a verage American's desire to have more self-confidence. Perhaps one of Carnegie's most successful marketing moveswas to change the spelling of his last name from"Carnagey"to Carnegie, at a time when Andrew Carnegie was a widely rec ognized name.Carnegie's works include Lincoln the Unknown (1932), Public Speaking and Influencing Men in Business (1937), and How t o Stop Worrying and Start Living (1948). His greatest achieve ment, however, was when Simon &. Schuster published How to Win Friends and Influence People. The book was a bestsell er from 1936. By the time of Carnegie's death, the book had s old five million copies in 31 languages, and there had been 45 0,000 graduates of his Dale Carnegie Institute.Carnegie died at his home in New York in 1955.What do you think of Dale Carnegie's childhood?A. Difficult.B. Joyful.C. Lonely.D. Boring.29.Which of the following is important for Dale Carnegie?A. Encouraging one to trust himself.B. Helping people to get wealthy.C. Teaching people speaking skills.D. Advising people to live happily.30. Why did Dale Carnegie change his last name?A. To get more help and support.B. To replace Andrew Carnegie.C. To become more famous.D. To avoid misunderstanding.31. Which of the following is Dale Carnegie's most successful work?A. Lincoln the Unknown.B. How to Stop Worrying and Start Living.C. How to Win Friends and Influence People.D. Public Speaking and Influencing Men in Business.DDeveloping healthy eating habits starts from childhood, theref ore it is important for parents to teach and provide children wit h a healthy diet. DPHSS administrator of the Bureau of Nutriti on Services, Charlie Morris told KUAM News, “A healthy diet f or a child consists of a lot of fruits and vegetables in the diet li miting the amount of simple sugars in the diet and high fat foo d and highly processed(加工的) food.”This means staying away as much as possible from food such as chips, cookies, candies and sugary drinks, as all children need to have meals which involve a well-balanced diet. Community nutritionist(营养学家) Thelma Romoso said, “The fruit, the vegetable, the grain, t he protein, and also the milk, the diary product, so for the fruit s it's easy for a mother to go into the two plus three concept o f fruits and vegetables or five a day.”This concept means that there are at least two servings of frui ts a day, three servings of vegetables a day which can be ser ved for lunch and dinner. As for protein parents can make a v ariety of dishes from chicken, beef, fish or even dried beans t o make sure the child gets three servings a day.Morris said that the child's hunger level controls how much is eaten and the parent controls what and when the meal is offer ed, saying, “Mom needs to ensure that the food offered is goo d food for the child to eat and throughout the day depending o n how active they are, snacks(零食) are not a bad thing, so the mom should offer good kinds of snacks.”When it comes to preparing your child's lunch and snacks for school, make sure to keep in mind that the food you provide s hould not only be a source of energy but also be nutritionally beneficial such as fresh fruits, and whole grain bread. But par ents must be sure to remember that a good nutritional diet mu st be balanced with physical activity, namely it is important that children get outside and run around and play and get activity in addition to eating a good diet.What does the author intend to tell us in the text?Choose healthy food and snacks for your child.Fruits and vegetables are good for your child.C. Take steps to provide a good lunch for your child.D. Make various and delicious dishes for your child.33. According to Charlie Morris we can know that_________.A. mothers should prepare delicious food by themselvesB. highly processed food should be reduced in daily lifeC. sugars and high fat exist in few kinds of foodD. active children prefer more and more snacks34. Which is NOT true when preparing your child’s food for sc hool?A. The food must provide energy as well as nutrition.B. Physical exercise is as important as a good dietC. Running and playing when eating is beneficial.D. Fruits and whole grain foods should be included.35. It can be inferred from the text that _________.A. Fresh fruits are often ignored by childrenB. A good nutritional diet may cost too muchC. Choosing healthy food is linked with incomeD. Parents play an important role in children's diet第三部分：完形填空（共20题, 每题1.5分, 共30分）On a trip to California, my family stopped for lunch. As we wal ked toward the entrance to the restaurant, a man, with a 36 beard and dirty hair, jumped up from a bench and opened the door for us．Regardless of his 37 , he greeted us in a frien dly way．Once inside, my daughters whispered, "Mom, he 38 " After we ordered our lunch, I explained, telling the kids to look 39 the dirt．We then watched other customers approach the re staurant but many 40 him. Seeing this rudeness truly upset me. The day I became a mother, I had resolved to set a good 41 for my children. Yet sometimes when things didn't go rig ht, being a good example was 42 . When our meal arrived, I realized I had left the car-sick pills in the truck. With the windiest trip ahead, the kids ne eded them, so I 43 myself from the meal and went to get th em.Just then, the "doorman" was opening the door for a couple. T hey rushed past him without even acknowledging his 44 . L etting them in first, I said a loud "thank you" to him as I 45 .When I returned, we talked a bit. He said he was not allowed i nside 46 he purchased food. I went back and told my family his 47 . Then I asked our waitress to add one soup and san dwich. The kids looked 48 as we had already eaten, but wh en I said the order was for the "doorman" , they smiled. When it was time to 49 our trip, I noticed the "doorman" enjoying his meal. Upon seeing me, he stood up and thanked me heart ily. He then 50 his hand for a handshake and I gratefully acc epted. I suddenly noticed the tears in his eyes—tears of 51 .What happened next drew great astonishment: I gave the " doorman" a 52 .He pulled away, with tears 53 down his face.Back in truck, I fell into deep thought. While we can't choose many things in life, we can choose when to show gratitude(感恩). I said thanks to a man who had 54 help open a door for me, and also said thanks for that 55 to teach my children b y example.36．A．heavy B．long C．messy D．grey 37．A．service B．appearance C．status D．attitude38．A．smokes B．smiles C．sleeps D．smells 39．A．beyond B．over C．around D．into 40．A．hated B．ignored C．missed D．refused41．A．target B．rule C．record D．example 42．A．stressful B．genuine C．embarrassing D．tough43．A．excused B．freed C．prevented D．withdrew44．A．company B．presence C．effort D．attempt45．A．quitted B．marched C．exited D．approached46．A．before B．unless C．though D．since 47．A．story B．deed C．experience D．demand48．A．concerned B．shocked C．puzzled D．bored49．A．make B．start C．take D．continue 50．A．extended B．washed C．raised D．waved51．A．agreement B．devotion C．sympathy D．gratitude52．A．hug B．nod C．lift D．clap 53．A．gathering B．rolling C．lying D．drowning54．A．firmly B．instantly C．simply D．politely 55．A．journey B．wisdom C．opportunity D．challenge第 II 卷非选择题（共60分）单词拼写：请根据汉语和首字母提示或用所给单词正确形式完成句子,并将单词的完整形式写到答题卡上。