西安高新一中 2020高一期中考试试题(含答案)

陕西省西安市高新一中2019-2020学年上学期期中考试高一数学试题一、选择题：（本大题共10小题，每小题4分，共40分，在每小题给出的四个选项中，只有一项是符合题目要求的）1．下列函数中与函数y x =是同一函数的是（ ）．A．2y = B．3y = C．y = D ．2x y x= 2．若一次函数y kx b =+在R 上是增函数，则k 的范围为（ ）．A ．0k >B ．0k ≥C ．0k <D ．0k ≤3．已知集合A 满足{}{}1,2,31,2,3,4A =，则集合A 的个数为（ ）． A ．2 B ．4 C ．8 D ．164．函数2()1f x x =-在[2,0]-上的最大值与最小值之差为（ ）． A ．83 B ．43 C ．23 D ．15．如图是①a y x =；②b y x =；③c y x =，在第一象限的图像，则a ，b ，c 的大小关系为（ ）．6．已知函数2()8f x x kx =--在[1,4]上单调，则实数k 的取值范围为（ ）．A ．[2,8]B ．[8,2]--C ．(][),82,-∞--+∞D ．(][),28,-∞+∞7．已知函数()f x 是奇函数，在(0,)+∞上是减函数，且在区间[,](0)a b a b <<上的值域为[3,4]-，则在区间[,]b a --上（ ）． A ．有最大值4 B ．有最小值4- C ．有最大值3- D ．有最小值3- 8．设0.60.6a =， 1.50.6b =，0.61.5c =，则a ，b ，c 的大小关系是（ ）．A ．a b c <<B ．a c b <<C ．b a c <<D ．b c a <<9．设x ∈R ，定义符号函数1,0sgn 0,01,0x x x x >⎧⎪==⎨⎪-<⎩，则（ ）．A ．|sgn |x x x =-B ．sgn ||x x x =-C ．||||sgn x x x =D ．||sgn x x x =10．若在定义域内存在．．实数0x ，满足00()()f x f x -=-，则称()f x 为“有点奇函数”，若12()423x x f x m m +=-+-为定义域R 上的“有点奇函数”，则实数m 的取值范围是（ ）．A ．11m ≤B ．1m ≤C ．m -≤D ．1m -≤ 二、填空题：（本大题共4小题，每小题4分，共16分）11．若函数2(4)()1(4)x x f x x x ⎧=⎨+<⎩≥，则[(3)]f f =__________．12．设函数y =A ，函数ln(1)y x =-的定义域为B ，则R A B =ð__________．13．方程23x x k +=的解都在[1,2]内，则k 的取值范围为__________．14．已知函数11()log x a x f x -+=（0a >且1a ≠）有下列四个结论．①恒过定点；②()f x 是奇函数；③当1a >时，()0f x <的解集为{}|0x x >；③当1a >时，()0f x <的解集为{}|0x x >；④若m ，(1,1)n ∈-，那么()()1m n f m f n f mn +⎛⎫+= ⎪+⎝⎭． 其中正确的结论是__________（请将所有正确结论的序号都填在横线上）．三、解答题：（本大题共5小题，共44分，解答应写出文字说明、证明过程或演算步骤）．15．（本小题满分8分）求下列各式的值：（1）122.5053[(0.064)]π-．（2）2lg5++已知函数1()2axf x ⎛⎫= ⎪⎝⎭，a 为常数，且函数的图象过点(1,2)-． （1）求a 的值．（2）若()42x g x -=-，且()()g x f x =，求满足条件的x 的值．17．（本小题满分8分）已知集合{}2(,)|y 1A x y x mx ==-+-，{}(,)|3,03B x y y x x ==-≤≤．（1）当4m =时，求A B ． （2）若A B 是只有一个元素的集合，其实数m 的取值范围．18．（本小题满分10分）定义：已知函数()f x 在[,]()m n m n <上的最小值为t ，若t m ≤恒成立，则称函数()f x 在[,]()m n m n <上具有“DK ”性质．（1）判断函数2()22f x x x =-+在[1,2]上是否具有“DK ”性质？说明理由．（2）若2()2f x x ax =-+在[,1]a a +上具有“DK ”性质，求a 的取值范围．已知函数2()32log f x x =-，2()log g x x =．（1）当[1,4]x ∈时，求函数()[()1]()h x f x g x =+⋅的值域．（2）如果对任意的[1,4]x ∈，不等式2()()()f x f x k g x ⋅>⋅恒成立，求实数k 的取值范围．附加题：1．（本小题满分8分）若定义在(,1)(1,)-∞+∞上的函数()f x 满足2017()220171x f x f x x +⎛⎫+=- ⎪-⎝⎭，则(2019)f =__________． 2．（本小题满分12分）设()|lg |f x x =，a ，b 为实数，且0a b <<，若a ，b 满足()()22a b f a f b f +⎛⎫== ⎪⎝⎭，试写出a 与b 的关系，并证明这一关系中存在b 满足34b <<．陕西省西安市高新一中2019-2020学年上学期期中考试高一数学试题参考答案一、选择题：（本大题共10小题，每小题4分，共40分，在每小题给出的四个选项中，只有一项是符合题目要求的）1．下列函数中与函数y x =是同一函数的是（ ）．A ．2y =B ．3y =C ．y =D ．2x y x= 【答案】B【解析】A ．此函数的定义域是[)0,+∞与函数y x =的定义域不同，所以这是两个不同的函数； B ．此函数的定义域是一切实数，对应法则是自变量的值不变，与函数y x =的定义域和对应法则都相同，所以这是同一个函数；C ．此函数的值域是[)0,+∞与函数y x =的值域不同，所以这是两个不同的函数；D ．此函数的定义域是(,0)(0,)-∞+∞与函数y x =的定义域不同，所以这是两个不同的函数； 所以B 与函数y x =是同一个函数．2．若一次函数y kx b =+在R 上是增函数，则k 的范围为（ ）．A ．0k >B ．0k ≥C ．0k <D ．0k ≤【答案】A【解析】A ．法一：由一次函数的图象可知选A ．法二：设1x ∀，2x ∈R 且12x x <，∵()f x kx b =+在R 上是增函数，∴1212()(()())0x x f x f x -->，即212()0k x x ->，∵212()0x x ->，∴0k >．故选A ．3．已知集合A 满足{}{}1,2,31,2,3,4A =，则集合A 的个数为（ ）． A ．2 B ．4 C ．8 D ．16【答案】C【解析】∵{}{}1,2,31,2,3,4A =，∴{}4A =；{}1,4；{}2,4；{}3,4；{}1,2,4；{}1,3,4；{}2,3,4；{}1,2,3,4，则集合A 的个数为8，故答案为：8．4．函数2()1f x x =-在[2,0]-上的最大值与最小值之差为（ ）． A ．83 B ．43 C ．23 D ．1【答案】B【解析】由题意可得：∵20x -≤≤，∴22()0(1)f x x '=-<-， ∴()f x 在[2,0]-上单调递减， ∴max 2()(2)3f x f =-=-． min ()(0)2f x f ==-， ∴最大值与最小值之差为24(2)33---=， 综上所述，答案：43．5．如图是①a y x =；②b y x =；③c y x =，在第一象限的图像，则a ，b ，c 的大小关系为（ ）．A ．a b c >>B ．a b c <<C ．b c a <<D ．a c b << 【答案】A【解析】由幂函数图象和单调性可知：1a >，01b <<，0c <．∴a b c >>．6．已知函数2()8f x x kx =--在[1,4]上单调，则实数k 的取值范围为（ ）．A ．[2,8]B ．[8,2]--C ．(][),82,-∞--+∞D ．(][),28,-∞+∞【答案】D 【解析】22b k a -=，12k ≤或42k ≥，2k ≤或8k ≥．7．已知函数()f x 是奇函数，在(0,)+∞上是减函数，且在区间[,](0)a b a b <<上的值域为[3,4]-，则在区间[,]b a --上（ ）． A ．有最大值4 B ．有最小值4- C ．有最大值3- D ．有最小值3-【答案】B【解析】∵0a b <<，∴0a b ->->，∵函数()f x 是奇函数，在(0,)+∞上是减函数，∴()f x 在(,0)-∞上是减函数，∵在区间[,](0)a b a b <<上的值域为[3,4]-，∴()f x 在区间[,]b a --上的值域为[4,3]-，∴()f x 在区间[,]b a --上有最大值为3，最小值为4-，综上所述．故选B ．8．设0.60.6a =， 1.50.6b =，0.61.5c =，则a ，b ，c 的大小关系是（ ）．A ．a b c <<B ．a c b <<C ．b a c <<D ．b c a <<【答案】C【解析】解：∵00.61<<，0.6 1.5<，∴0.6 1.510.60.6>>，即a b >，∵1.51>，0.60>，∴0.61.51c =>，∴c a b >>．9．设x ∈R ，定义符号函数1,0sgn 0,01,0x x x x >⎧⎪==⎨⎪-<⎩，则（ ）．A ．|sgn |x x x =-B ．sgn ||x x x =-C ．||||sgn x x x =D ．||sgn x x x =【答案】A【解析】对于选项A ．右边,0|sgn |0,0x x x x x ≠⎧==⎨=⎩，而左边,0||,0x x x x x ⎧==⎨-<⎩≥，显然不正确；对于选项B ．右边,0sgn ||0,0x x x x x ≠⎧==⎨=⎩，而左边,0||,0x x x x x ⎧==⎨-<⎩≥，显然不正确； 对于选项C ，右边,0||sgn 0,0x x x x x ≠⎧==⎨≠⎩，而左边,0||,0x x x x x ⎧==⎨-<⎩≥，显然不正确； 对于选项D ，右边,0sgn 0,0,0x x x x x x x >⎧⎪===⎨⎪-<⎩，而左边,0||,0x x x x x ⎧==⎨-<⎩≥，显然正确．10．若在定义域内存在．．实数0x ，满足00()()f x f x -=-，则称()f x 为“有点奇函数”，若12()423x x f x m m +=-+-为定义域R 上的“有点奇函数”，则实数m 的取值范围是（ ）．A．11m ≤B．1m ≤C．m -≤ D．1m -≤ 【答案】B【解析】根据“局部奇函数”的定义可知，函数()()f x f x -=-有解即可，即1212()423(423)x x x x f x m m m m --++-=-+-=--+-，∴2442(22)260x x x x m m --+-++-=，即22(22)2(22)280x x x x m m --+-⋅++-=有解即可，设22x x t -=+，则222x x t -=+≥，∴方程等价为222280t m t m -⋅+-=在2t ≥时有解，设22()228g t t m t m =-⋅+-， 对称轴22m x m -=-=， ①若2m ≥，则2244(28)0m m ∆=--≥，即28m ≤，∴m -≤2m ≤≤②若2m <，要使222280t m t m -⋅+-=在2t ≥时有解，则2(2)00m f <⎧⎪⎨⎪∆⎩≤≥，即211m m m <⎧⎪⎨⎪-⎩≤≤，解得12m <，综上：1m -≤二、填空题：（本大题共4小题，每小题4分，共16分）11．若函数2(4)()1(4)x x f x x x ⎧=⎨+<⎩≥，则[(3)]f f =__________． 【答案】16【解析】∵函数2(4)()1(4)x x f x x x ⎧=⎨+<⎩≥， ∴(3)314f =+=，4[(3)](4)216f f f ===．12．设函数y =A ，函数ln(1)y x =-的定义域为B ，则R A B =ð__________．【答案】[1,2]【解析】240x -≥，22x -≤≤，10x ->，1x <，{}|1R B x x =ð≥，∴[1,2]R A B =ð．13．方程23x x k +=的解都在[1,2]内，则k 的取值范围为__________．【答案】[)5,10【解析】23x k x =-， 1x =时，32k -≥，5k ≥，2x =时，64k -<，10k <，[)5,10k ∈．14．已知函数11()log x a x f x -+=（0a >且1a ≠）有下列四个结论．①恒过定点；②()f x 是奇函数；③当1a >时，()0f x <的解集为{}|0x x >；③当1a >时，()0f x <的解集为{}|0x x >；④若m ，(1,1)n ∈-，那么()()1m n f m f n f mn +⎛⎫+= ⎪+⎝⎭． 其中正确的结论是__________（请将所有正确结论的序号都填在横线上）．【答案】①，②，④【解析】（1）解：∵1()log 1ax f x x -=+， ∴10111x x x->⇒-<<+， 故函数()f x 的定义域是|11x x -<<．（2）证明：∵m ，(1,1)n ∈-， ∴1111()()log log log 1111a a a m n m n f m f n m n m n ----⎛⎫+=+=⋅ ⎪++++⎝⎭， 11111log log log 111111a a a mn m n m n m n mn m n mn mn f mn m n m n m n mn mn mn mn+--+---++⎛⎫++==== ⎪++++++++⎝⎭+++， 故()()1m n f m f n f mn +⎛⎫+= ⎪+⎝⎭． （3）解：∵1111()()log log log log 101111aa a a x x x x f x f x x x x x+-+--+=+=⋅==-+-+， ∴()()f x f x -=-， 即()f x 在其定义域(1,1)-上为奇函数．三、解答题：（本大题共5小题，共44分，解答应写出文字说明、证明过程或演算步骤）．15．（本小题满分8分）求下列各式的值：（1）122.5053[(0.064)]π-． （2）2lg5++【答案】见解析．【解析】（1）原式12232.55327[(0.8)]18-⎛⎫=-- ⎪⎝⎭， 11=-0=．（2）2lg5++112222(lg 2)lg 2lg5=+⋅+2112lg 2lg 2lg522⎛⎫=+⋅+ ⎪⎝⎭2112lg 2lg 2lg522⎛⎫=+⋅ ⎪⎝⎭11lg 2(lg 2lg5)lg 2122=++- 11lg2lg(25)1lg222=⋅⋅+- 11lg21lg2122=+-=．16．（本小题满分8分） 已知函数1()2axf x ⎛⎫= ⎪⎝⎭，a 为常数，且函数的图象过点(1,2)-． （1）求a 的值．（2）若()42x g x -=-，且()()g x f x =，求满足条件的x 的值．【答案】见解析．【解析】（1）由已知得122a -⎛⎫= ⎪⎝⎭，解得1a =．（2）由（1）知1()2x f x ⎛⎫= ⎪⎝⎭， 又()()g x f x =，则1422x x -⎛⎫-= ⎪⎝⎭， 即112042x x ⎛⎫⎛⎫--= ⎪ ⎪⎝⎭⎝⎭，即2112022x x⎡⎤⎛⎫⎛⎫--=⎢⎥ ⎪ ⎪⎝⎭⎝⎭⎢⎥⎣⎦， 令12x t ⎛⎫= ⎪⎝⎭，则220t t --=， 即(2)(1)0t t -+=，又0t >，故2t =， 即122x⎛⎫= ⎪⎝⎭，解得1x =-， 满足条件的x 的值为1-．17．（本小题满分8分）已知集合{}2(,)|y 1A x y x mx ==-+-，{}(,)|3,03B x y y x x ==-≤≤． （1）当4m =时，求A B ． （2）若A B 是只有一个元素的集合，其实数m 的取值范围．【答案】见解析．【解析】（1）当4m =时，集合{}2(,)|41A x y y x x ==-+-， {}(,)|3,03B x y y x x ==-≤≤，联立得：2341y x y x x =-⎧⎨=-+-⎩， 消去y 得：2341x x x -=-+-， 即(1)(4)0x x --=，解得：1x =或4x =（不合题意，舍去）， 将1x =代入3y x =-得2y =， 则{}(1,2)A B =；综上所述：答案为{}(1,2)AB =． （2）集合A 表示抛物线上的点，抛物线21y x mx =-+-，开口向下且过点(0,1)-， 集合B 表示线段上的点，要使A B 只有一个元素，则线段与抛物线的位置关系有以下两种，如图： （i ）由图知，在函数2()1f x x mx =-+-中，只要(3)0f ≥，即9310m -+-≥， 解得：103m ≥． （ii ）由图知，抛物线与直线在[0,3]x ∈上相切，联立得：213y x mx y x ⎧=-+-⎨=-⎩， 消去y 得：213x mx x -+-=-， 整理得：2(1)40x m x -++=， 当2(1)160m ∆=+-=，∴3m =或5m =-，当3m =时，切点(2,1)适合， 当5m =-时，切点(2,5)-舍去， 综上所述：答案为m 范围为3m =或103m ≥．18．（本小题满分10分）定义：已知函数()f x 在[,]()m n m n <上的最小值为t ，若t m ≤恒成立，则称函数()f x 在[,]()m n m n <上具有“DK ”性质．（1）判断函数2()22f x x x =-+在[1,2]上是否具有“DK ”性质？说明理由． （2）若2()2f x x ax =-+在[,1]a a +上具有“DK ”性质，求a 的取值范围．【答案】见解析．【解析】（1）∵2()22f x x x =-+，[1,2]x ∈， 对称轴1x =，开口向上，当1x =时，取得最小值为(1)1f =， ∴min ()(1)11f x f ==≤，∴函数()f x 在[1,2]上具有“DK ”性质． （2）2()2g x x ax =-+，[,1]x a a ∈+， 其图象的对称轴方程为2a x =． ①当02a ≥，即0a ≥时，22min ()()22g x g a a a ==-+=． 若函数()g x 具有“DK ”性质，则有2a ≤总成立，即2a ≥． ②当12a a a <<+，即20a -<<时， 2min ()224a a g x g ⎛⎫==-+ ⎪⎝⎭． 若函数()g x 具有“DK ”性质，则有224a a -+≤总成立，解得a 无解． ③当12a a +≥，即2a -≤时，min ()(1)3g x g a a =+=+， 若函数()g x 具有“DK ”性质， 则有3a a +≤，解得a 无解． 综上所述，若2()2g x x ax =-+在[,1]a a +上具有“DK ”性质，则2a ≥．19．（本小题满分10分）已知函数2()32log f x x =-，2()log g x x =． （1）当[1,4]x ∈时，求函数()[()1]()h x f x g x =+⋅的值域． （2）如果对任意的[1,4]x ∈，不等式2()()()f x f x k g x ⋅>⋅恒成立，求实数k 的取值范围．【答案】见解析．【解析】（1）2222()(42log )log 2(log 1)2h x x x x =-⋅=--+，因为[1,4]x ∈，所以2log [0,2]x ∈，故函数()h x 的值域为[0,2]．（2）由2()()f x f k g x ⋅>⋅得222(34log )(3log )log x x k x -->⋅， 令2log t x =，因为[1,4]x ∈，所以2log [0,2]t x =∈，所以(34)(3)t t k t -->⋅对一切的[0,2]t ∈恒成立．1．当0t =时，k ∈R ；2．当(]0,2t ∈时，(34)(3)t t k t --<恒成立，即9415k t t<+-． 因为9412t t +≥，当且仅当94t t =，即32t =时取等号． 所以9415t t+-的最小值为3-， 综上，(,3)k ∈-∞-．附加题：1．（本小题满分8分）若定义在(,1)(1,)-∞+∞上的函数()f x 满足2017()220171x f x f x x +⎛⎫+=- ⎪-⎝⎭，则(2019)f =__________． 【答案】1344． 【解析】2018()2120171f x f x x ⎛⎫++=- ⎪-⎝⎭， 2x =：(2)2(2019)2015f f +=，① 2019x =：(2019)2(2)2f f +=-，②， ①⨯2-②3(2019)4032f ==， (2019)1344f =．2．（本小题满分12分）设()|lg |f x x =，a ，b 为实数，且0a b <<，若a ，b 满足()()22a b f a f b f +⎛⎫== ⎪⎝⎭，试写出a 与b 的关系，并证明这一关系中存在b 满足34b <<．【答案】见解析．【解析】（1）由()1f x =得，lg 1x =±，所以10x =或110． （2）结合函数图象，由()()f a f b =，可判断(0,1)a ∈，(1,)b ∈+∞， 从而lg lg a b -=，从而1ab =， 又122b a b b ++=， 因为(1,)b ∈+∞，所以12a b +>， 从而由()22a b f b f +⎛⎫= ⎪⎝⎭， 可得2lg 2lg lg 22a b a b b ++⎛⎫== ⎪⎝⎭， 从而22a b b +⎛⎫= ⎪⎝⎭． （3）由22a b b +⎛⎫= ⎪⎝⎭， 得2242b a b ab =++，221240b b b ++-=， 令221()24g b b b b =++-， 因为(3)0g <，(4)0g >，根据零点存在性定理可知， 函数()g b 在(3,4)内一定存在零点， 即方程221240b b b++-=存在34b <<的根．。

2020-2021学年陕西省西安市高新一中高一（上）期中化学试卷1. 2019年10月9日，本年度诺贝尔化学奖授予美、日三位科学家，表彰其在锂离子电池的发展方面作出的贡献。

在充电时，、在电池反应中常作还原剂。

下列说法错误的是( )A. 、都是复杂的氧化物B. 锂离子电池广泛用于新能源汽车、手机等产业C. 中Fe显价D. 在充电时，电池反应中某种元素化合价会升高2. 下列物质的化学用语使用正确的是( )A.硫酸钠化学式：B. 气体摩尔体积的单位：C. 钠在氯气中燃烧的反应电子转移情况：D. 氯离子结构示意图：3. 下列示意图与相关内容不相符水合离子用相应离子符号表示的是( )A B C D腐蚀品蒸馏NaCl的电离氯化氢分子形成示意图A. AB. BC. CD. D4. 下列叙述中，正确的是( )A. 熔融的能导电，所以是电解质B. 铜丝、石墨均能导电，所以它们都是电解质C.固体不导电，所以不是电解质D. NaCl溶于水，在通电条件下才能发生电离5. 当光束通过下列物质时，不会出现丁达尔效应的是( )①胶体②水③蔗糖溶液④溶液⑤云、雾A. ①③④B. ②③④C. ②④⑤D. ③④⑤6. 下列操作中不正确的是( )A. 过滤时，玻璃棒与三层滤纸的一边接触B. 过滤时，漏斗下方紧贴烧杯内壁C. 加热试管内物质时，试管底部与酒精灯外焰接触D. 向试管中滴加液体时，胶头滴管紧贴试管内壁7. 下列说法不正确的组合为( )①只含有一种元素的物质一定是纯净物②生石灰做干燥剂包含化学变化③是酸性氧化物，硫酸是它与水反应形成的水化物④碱性氧化物一定是金属氧化物⑤用纳米铁粉除去废水中的、、、等金属离子发生了化学变化⑥能与酸反应生成盐和水的氧化物一定是碱性氧化物⑦用鸡蛋壳膜和蒸馏水除去淀粉胶体中的食盐不涉及化学变化A. ①③⑥B. ①②③④⑥C. ②④⑤⑦D. ②④⑥8. 下列各组离子，能在溶液中大量共存的是( )A.、、、 B. 、、、C. 、、、D. 、、、9. 赤铜矿的成分是，辉铜矿的成分是，将赤铜矿和辉铜矿混合加热有以下反应：，对于该反应，下列说法正确的是( )A.该反应氧化剂只有B. Cu既是氧化产物又是还原产物C. 在反应中既是氧化剂又是还原剂D. 氧化产物与还原产物的物质的量之比为6：110. 下列离子方程式正确的是( )A. 向溶液中滴加适量的溶液恰好使沉淀完全：B. 过量与NaOH溶液反应：C. 醋酸滴在石灰石上：D. 铁和稀硫酸反应：11. 已知甲醇沸点和水合肼沸点可以互溶，要分离二者形成的混合物，可用蒸馏的方法。

2020-2021西安高新第一中学初中校区东区初级中学高中必修一数学上期中试题(附答案)一、选择题1．设集合{}1,2,4A =，{}240B x x x m =-+=．若{}1A B ⋂=，则B =( ) A ．{}1,3-B ．{}1,0C ．{}1,3D ．{}1,52．设()f x 是定义在R 上的偶函数，且当0x ≥时，()21,0122,1xx x f x x ⎧-+≤<=⎨-≥⎩，若对任意的[],1x m m ∈+，不等式()()1f x f x m -≤+恒成立，则实数m 的最大值是（ ） A ．1-B ．13-C ．12-D ．133．函数2xy x =⋅的图象是（ ）A ．B ．C ．D ．4．设函数()2010x x f x x -⎧≤=⎨>⎩，，，则满足()()12f x f x +<的x 的取值范围是（ ）A ．(]1-∞-，B ．()0+∞，C ．()10-，D ．()0-∞，5．函数()f x 在(,)-∞+∞单调递增，且为奇函数，若(1)1f =，则满足1(2)1f x -≤-≤的x 的取值范围是（ ）． A ．[2,2]-B ．[1,1]-C ．[0,4]D ．[1,3]6．已知函数224()(log )log (4)1f x x x =++，则函数()f x 的最小值是A ．2B ．3116C ．158D ．17．设x ∈R ，若函数f （x ）为单调递增函数，且对任意实数x ，都有f （f （x ）-e x）=e +1（e 是自然对数的底数），则f （ln1.5）的值等于（ ） A ．5.5B ．4.5C ．3.5D ．2.58．已知全集U ＝R ，集合A ＝{x |x 2－x －6≤0}，B ＝{x |14x x +-＞0}，那么集合A ∩(∁U B )＝( )A ．{x |－2≤x ＜4}B ．{x |x ≤3或x ≥4}C ．{x |－2≤x ＜－1}D ．{x |－1≤x ≤3}9．已知函数21(1)()2(1)ax x f x x x x x ⎧++>⎪=⎨⎪-+≤⎩在R 上单调递增，则实数a 的取值范围是 A ．[]0,1B ．(]0,1C ．[]1,1-D ．(]1,1-10．已知函数()y f x =在区间(),0-∞内单调递增，且()()f x f x -=，若12log 3a f ⎛⎫= ⎪⎝⎭，()1.22b f -=，12c f ⎛⎫= ⎪⎝⎭，则a 、b 、c 的大小关系为（ ）A ．a c b >>B ．b c a >>C ．b a c >>D ．a b c >>11．若0.23log 2,lg0.2,2a b c ===,则,,a b c 的大小关系为A ．c b a <<B ． b a c <<C ． a b c <<D ．b c a <<12．已知集合{|20}A x x =-<，{|}B x x a =<，若A B A =I ，则实数a 的取值范围是( ) A ．(,2]-∞-B ．[2,)+∞C ．(,2]-∞D ．[2,)-+∞二、填空题13．给出下列四个命题：（1）函数()f x x x bx c =++为奇函数的充要条件是0c =； （2）函数()20xy x -=>的反函数是()2log 01y x x =-<<；（3）若函数()()2lg f x x ax a =+-的值域是R ，则4a ≤-或0a ≥；（4）若函数()1y f x =-是偶函数，则函数()y f x =的图像关于直线0x =对称. 其中所有正确命题的序号是______.14．已知函数()(0,1)x f x a b a a =+>≠的定义域和值域都是[]1,0-，则a b += . 15．设函数21()ln(1||)1f x x x=+-+，则使得()(21)f x f x >-成立的x 的取值范围是_____.16．若函数()f x 满足()3298f x x +=+，则()f x 的解析式是_________. 17．若1∈{}2,a a, 则a 的值是__________18．已知函数()266,34,x x f x x ⎧-+=⎨+⎩ 00x x ≥<，若互不相等的实数1x ，2x ，3x 满足()()()123f x f x f x ==，则123x x x ++的取值范围是__________．19．已知函数()()212log 22f x mx m x m ⎡⎤=+-+-⎣⎦，若()f x 有最大值或最小值，则m的取值范围为______． 20．已知函数在区间，上恒有则实数的取值范围是_____.三、解答题21．计算下列各式的值：（1）()1110232710223π20.25927--⎛⎫⎛⎫--+ ⎪ ⎪⎝⎭⎝⎭．（2）()221log 3lg5ln e 2lg2lg5lg2-++++⋅．22．已知集合A ＝{x|2a ＋1≤x≤3a －5}，B ＝{x|x ＜－1，或x ＞16}，分别根据下列条件求实数a 的取值范围．（1）A∩B ＝∅；（2）A ⊆（A∩B ）． 23．计算下列各式的值：（Ⅰ)22log 3lg25lg4log (log 16)+- （Ⅱ）2102329273()( 6.9)()()482-----+24．已知函数()xf x b a =⋅，（其中,a b 为常数且0,1a a >≠）的图象经过点(1,6),(3,24)A B（1）求()f x 的解析式（2）若不等式11120xxm a b ⎛⎫⎛⎫++-≥ ⎪ ⎪⎝⎭⎝⎭在(],1x ∈-∞上恒成立，求实数m 的取值范围.25．已知函数())2log f x x =是R 上的奇函数，()2g x t x a =--.（1）求a 的值；（2）记()f x 在3,24⎡⎤-⎢⎥⎣⎦上的最大值为M ，若对任意的3,24x ⎡⎤∈-⎢⎥⎣⎦，()M g x ≤恒成立，求t 的取值范围.26．已知全集U ＝{1,2,3,4,5,6,7,8}，A ＝{x |x 2－3x ＋2＝0}，B ＝{x |1≤x ≤5，x ∈Z}，C ＝{x |2<x <9，x ∈Z}．求 (1)A ∪(B ∩C )；(2)(∁U B )∪(∁U C )．【参考答案】***试卷处理标记，请不要删除一、选择题 1．C 解析：C 【解析】∵ 集合{}124A ，，=，{}2|40B x x x m =-+=，{}1A B ⋂= ∴1x =是方程240x x m -+=的解，即140m -+= ∴3m =∴{}{}{}22|40|43013B x x x m x x x =-+==-+==，，故选C2．B解析：B 【解析】 【分析】由题意，函数()f x 在[0,)+∞上单调递减，又由函数()f x 是定义上的偶函数，得到函数()f x 在(,0)-∞单调递增，把不等式(1)()f x f x m -≤+转化为1x x m -≤+，即可求解. 【详解】易知函数()f x 在[)0,+∞上单调递减， 又函数()f x 是定义在R 上的偶函数， 所以函数()f x 在(),0-∞上单调递增， 则由()()1f x f x m -≤+，得1x x m -≥+，即()()221x x m -≥+，即()()22210g x m x m =++-≤在[],1x m m ∈+上恒成立，则()()()()()()3110121310g m m m g m m m ⎧=-+≤⎪⎨+=++≤⎪⎩，解得113m -≤≤-， 即m 的最大值为13-. 【点睛】本题主要考查了函数的基本性质的应用，其中解答中利用函数的基本性质，把不等式转化为1x x m -≤+ 求解是解答的关键，着重考查了分析问题和解答问题的能力，以及推理与运算能力，属于中档试题.3．A解析：A 【解析】 【分析】先根据奇偶性舍去C,D,再根据函数值确定选A. 【详解】因为2xy x =⋅为奇函数，所以舍去C,D; 因为0x >时0y >,所以舍去B ，选A. 【点睛】有关函数图象识别问题的常见题型及解题思路(1)由解析式确定函数图象的判断技巧：（1）由函数的定义域，判断图象左右的位置，由函数的值域，判断图象的上下位置；②由函数的单调性，判断图象的变化趋势；③由函数的奇偶性，判断图象的对称性；④由函数的周期性，判断图象的循环往复．(2)由实际情景探究函数图象．关键是将问题转化为熟悉的数学问题求解，要注意实际问题中的定义域问题．4．D解析：D 【解析】分析：首先根据题中所给的函数解析式，将函数图像画出来，从图中可以发现若有()()12f x f x +<成立，一定会有2021x x x <⎧⎨<+⎩，从而求得结果.详解：将函数()f x 的图像画出来，观察图像可知会有2021x x x <⎧⎨<+⎩，解得0x <，所以满足()()12f x f x +<的x 的取值范围是()0-∞，，故选D .点睛：该题考查的是有关通过函数值的大小来推断自变量的大小关系，从而求得相关的参数的值的问题，在求解的过程中，需要利用函数解析式画出函数图像，从而得到要出现函数值的大小，绝对不是常函数，从而确定出自变量的所处的位置，结合函数值的大小，确定出自变量的大小，从而得到其等价的不等式组，从而求得结果.5．D解析：D 【解析】 【分析】 【详解】()f x 是奇函数，故()()111f f -=-=- ；又()f x 是增函数，()121f x -≤-≤，即()(1)2(1)f f x f -≤-≤ 则有121x -≤-≤ ，解得13x ≤≤ ，故选D.【点睛】解本题的关键是利用转化化归思想，结合奇函数的性质将问题转化为()(1)2f f x -≤-(1)f ≤，再利用单调性继续转化为121x -≤-≤，从而求得正解.6．B解析：B 【解析】 【分析】利用对数的运算法则将函数()()()224log log 41f x x x =++化为()2221log 1log 12x x +++，利用配方法可得结果.【详解】化简()()()224log log 41f x x x =++()2221log 1log 12x x =+++22211131log log 224161616x x ⎛⎫=++-≥-= ⎪⎝⎭，即()f x 的最小值为3116，故选B.【点睛】本题主要考查对数的运算法则以及二次函数配方法求最值，属于中档题. 求函数最值常见方法有，①配方法：若函数为一元二次函数，常采用配方法求函数求值域，其关键在于正确化成完全平方式，并且一定要先确定其定义域；②换元法；③不等式法；④单调性法；⑤图象法.7．D解析：D 【解析】 【分析】利用换元法 将函数转化为f （t ）=e+1，根据函数的对应关系求出t 的值，即可求出函数f （x ）的表达式，即可得到结论 【详解】 设t=f （x ）-e x ，则f （x ）=e x +t ，则条件等价为f （t ）=e+1， 令x=t ，则f （t ）=e t +t=e+1， ∵函数f （x ）为单调递增函数， ∴t=1， ∴f （x ）=e x +1，即f （ln5）=e ln1.5+1=1.5+1=2.5， 故选：D ． 【点睛】本题主要考查函数值的计算，利用换元法求出函数的解析式是解决本题的关键．8．D解析：D 【解析】依题意A ＝{x |－2≤x ≤3}，B ＝{x |x ＜－1或x ＞4}，故∁U B ＝{x |－1≤x ≤4}，故A ∩(∁U B )＝{x |－1≤x ≤3}，故选D.9．C解析：C 【解析】x ⩽1时,f (x )=−(x −1)2+1⩽1， x >1时,()()21,10a a f x x f x x x=++'=-…在(1,+∞)恒成立， 故a ⩽x 2在(1,+∞)恒成立， 故a ⩽1，而1+a +1⩾1，即a ⩾−1， 综上,a ∈[−1,1]，本题选择C 选项.点睛：利用单调性求参数的一般方法：一是求出函数的单调区间，然后使所给区间是这个单调区间的子区间，建立关于参数的不等式组即可求得参数范围；二是直接利用函数单调性的定义：作差、变形，由f (x 1)－f (x 2)的符号确定参数的范围，另外也可分离参数转化为不等式恒成立问题．10．B解析：B 【解析】 【分析】由偶函数的性质可得出函数()y f x =在区间()0,∞+上为减函数，由对数的性质可得出12log 30<，由偶函数的性质得出()2log 3a f =，比较出2log 3、 1.22-、12的大小关系，再利用函数()y f x =在区间()0,∞+上的单调性可得出a 、b 、c 的大小关系. 【详解】()()f x f x -=Q ，则函数()y f x =为偶函数，Q 函数()y f x =在区间(),0-∞内单调递增，在该函数在区间()0,∞+上为减函数，1122log 3log 10<=Q ，由换底公式得122log 3log 3=-，由函数的性质可得()2log 3a f =，对数函数2log y x =在()0,∞+上为增函数，则22log 3log 21>=， 指数函数2xy =为增函数，则 1.2100222--<<<，即 1.210212-<<<， 1.22102log 32-∴<<<，因此，b c a >>. 【点睛】本题考查利用函数的奇偶性与单调性比较函数值的大小关系，同时也考查了利用中间值法比较指数式和代数式的大小关系，涉及指数函数与对数函数的单调性，考查分析问题和解决问题的能力，属于中等题.11．B解析：B 【解析】 【分析】由对数函数的单调性以及指数函数的单调性，将数据与0或1作比较，即可容易判断. 【详解】由指数函数与对数函数的性质可知，a =()3log 20,1,b ∈=lg0.20,c <=0.221>，所以b a c <<，故选：B.【点睛】本题考查利用指数函数和对数函数的单调性比较大小，属基础题.12．B解析：B 【解析】由题意可得{}|2A x x =<，结合交集的定义可得实数a 的取值范围是[)2,+∞ 本题选择B 选项.二、填空题13．（1）（2）（3）【解析】【分析】根据奇函数的定义得到（1）正确根据反函数的求法以及定义域值域得到（2）正确由函数的值域是得出其真数可以取到所有的正数由二次函数判别式大于等于0求解可判断出（3）正确解析：（1）（2）（3） 【解析】 【分析】根据奇函数的定义得到（1）正确，根据反函数的求法以及定义域值域得到（2）正确， 由函数()()2lg f x x ax a =+-的值域是R ，得出其真数可以取到所有的正数，由二次函数判别式大于等于0求解，可判断出（3）正确，根据函数图像平移可判断（4）不正确. 【详解】解：（1）当0c =时，()=+f x x x bx ，()()()-=---=-+=-f x x x bx x x bx f x ，当函数为奇函数时()()f x f x -=-，即()++=----+=+-x x bx c x x bx c x x bx c ，解得0c =，所以0c =是函数()f x x x bx c =++为奇函数的充要条件，所以（1）正确；（2）由反函数的定义可知函数()20xy x -=>的反函数是()2log 01y x x =-<<，所以（2）正确；（3）因为函数()()2lg f x x ax a =+-的值域是R ，所以2y x ax a =+-能取遍(0,)+∞的所有实数，所以240a a =+≥△，解得0a ≥或4a ≤-，所以（3）正确； （4）函数()1y f x =-是偶函数，所以()1y f x =-图像关于y 轴对称，函数()y f x =的图像是由()1y f x =-向左平移一个单位得到的，所以函数()y f x =的图像关于直线1x =-对称，故（4）不正确. 故答案为：（1）（2）（3） 【点睛】本题主要考查对函数的理解，涉及到函数的奇偶性、值域、反函数等问题.14．【解析】若则在上为增函数所以此方程组无解；若则在上为减函数所以解得所以考点：指数函数的性质解析：32-【解析】若1a >，则()f x 在[]1,0-上为增函数,所以11{10a b b -+=-+=，此方程组无解；若01a <<，则()f x 在[]1,0-上为减函数,所以10{11a b b -+=+=-，解得1{22a b ==-，所以32a b +=-.考点：指数函数的性质.15．【解析】试题分析：由题意得函数的定义域为因为所以函数为偶函数当时为单调递增函数所以根据偶函数的性质可知：使得成立则解得考点：函数的图象与性质【方法点晴】本题主要考查了函数的图象与性质解答中涉及到函数解析：1(1)3， 【解析】试题分析：由题意得，函数21()ln(1)1f x x x =+-+的定义域为R ，因为()()f x f x -=，所以函数()f x 为偶函数，当0x >时，21()ln(1)1f x x x =+-+为单调递增函数，所以根据偶函数的性质可知：使得()(21)f x f x >-成立，则21x x >-，解得113x <<. 考点：函数的图象与性质.【方法点晴】本题主要考查了函数的图象与性质，解答中涉及到函数的单调性和函数的奇偶性及其简单的应用，解答中根据函数的单调性与奇偶性，结合函数的图象，把不等式()(21)f x f x >-成立，转化为21x x >-，即可求解，其中得出函数的单调性是解答问题的关键，着重考查了学生转化与化归思想和推理与运算能力，属于中档试题.16．【解析】【分析】设带入化简得到得到答案【详解】设代入得到故的解析式是故答案为：【点睛】本题考查了利用换元法求函数解析式属于常用方法需要学生熟练掌握解析：()32f x x =+ 【解析】 【分析】设32t x =+，带入化简得到()32f t t =+得到答案. 【详解】()3298f x x +=+，设32t x =+ 代入得到()32f t t =+故()f x 的解析式是() 32f x x =+ 故答案为：()32f x x =+ 【点睛】本题考查了利用换元法求函数解析式，属于常用方法，需要学生熟练掌握.17．-1【解析】因为所以或当时不符合集合中元素的互异性当时解得或时符合题意所以填解析：-1 【解析】 因为{}21,a a∈，所以1a =或21a=，当1a =时，2a a =，不符合集合中元素的互异性，当21a =时，解得1a =或1a =-，1a =-时2a a ≠，符合题意．所以填1a =-．18．【解析】【分析】画出分段函数的图像由图像结合对称性即可得出【详解】函数的图像如下图所示不妨设则关于直线对称所以且满足则故的取值范围是【点睛】解决本题的关键是要会画分段函数的图像由图像结合对称性经过计解析：11(,6)3【解析】 【分析】画出分段函数的图像，由图像结合对称性即可得出。

2020~2021学年陕西西安雁塔区西安高新第一中学高一下学期期中物理试卷一、选择题（本大题共12小题，每小题4分，共48分。

其中1-8题为单选，9-12题为多选）1.A.B.C.D.关于万有引力定律的建立，下列说法中正确的是（ ）开普勒在研究行星运动规律的基础之上提出了万有引力定律卡文迪许测出了万有引力常量，从而使牛顿被称为“第一位称量地球的人”“月-地检验”表明地面物体所受地球引力与月球所受地球引力遵从同样的规律“月-地检验”表明物体在地球上受到地球对它的引力是它在月球上受到月球对它的引力的倍2.A.B.C.D.关于功和能的关系，下列说法正确的是（ ）一个重的物体，在的水平拉力的作用下，分别在光滑水平面和粗糙水平面上 发生相同的位移，拉力做的功相等一辆汽车的速度从增加到，或从增加到，两种情况下 牵引力做的功一样多载人飞船的返回舱在大气层以外向着地球做无动力飞行的过程中，机械能增大物体受拉力作用竖直向上运动，拉力做的功是，则物体动能的增加量是3.A.B.C.D.如图所示，某人把质量为的石块从距地面高处以初速度抛出，方向与水平方向夹角为（），石块最终落在水平地面上．若空气阻力可忽略，则下列说法的是（ ）抛出后石块和地球构成的系统的机械能守恒人对石块做功大小为石块落地时的动能大小为抛出后石块在运动过程中的动量变化量为错.误.4.A.B.C.D.静止在湖面上的小船中有两人分别向相反方向以相对于河岸相等的速率水平抛出质量相同的小球，先将甲球向左抛，后将乙球向右抛．水对船的阻力忽略不计，则下列说法正确的是（ ）两球抛出后，船向左以一定速度运动两球抛出后，船向右以一定速度运动抛出的过程中，人对甲球做的功大于人对乙球做的功抛出的过程中，人给甲球的冲量等于人给乙球的冲量5.A.B.C.D.质量为的小球以的速度在光滑水平面上运动，与质量为的静止小球发生对心碰撞，则碰撞后小球的速度大小和小球的速度大小可能为（ ）6.A.B.C.D.太空探测器常装配离子发动机，其基本原理是将被电离的原子从发动机尾部高速喷出，为探测器提供推力．若某探测器的质量为，离子以一定的速率（远大于探测器的飞行速率）向后喷出流量为，探测器获得的平均推力大小为，则离子被喷出的速率为（ ）7.A. B. C. D.如图所示，在轻弹簧的下端悬挂一个质量为的小球，若将小球从弹簧原长位置由静止释放，小球能够下降的最大高度为．若将小球换为质量为的小球，仍从弹簧原长位置由静止释放，重力加速度为，不计空气阻力，则小球下降时的速度为（ ）8.如图所示，小车静止于水平面上，端固定一个轻质弹簧，端粘有橡皮泥，小车质量为．质量为的木块放在小车上，用细线将木块连接于小车的端并使弹簧压缩．开始时小车与木块都处于静止状态，、间距为．现烧断细线，弹簧被释放，使木块离开弹簧向端滑去，并跟端橡皮泥粘在一起．对整个过程，以下说法正确的是（ ）A.B.C.D.若木块滑动中没有摩擦，整个系统的机械能和动量都守恒木块的速度最大时小车的速度一定最小小车向左运动的最大位移等于无论木块滑动中有没有摩擦，整个系统损失的机械能均相同9.A.地球的质量为B.地球自转的角速度为C.同步卫星的加速度为D.地球的平均密度为年月日，我国第颗北斗导航卫星成功发射，标志着北斗三号全球系统星座的部署已经全面完成．该卫星为地球同步轨道卫星．已知同步卫星围绕地球做匀速圆周运动的周期为、轨道半径为，地球半径为，引力常量为，下列说法正确的是（ ）10.A.B.C.D.将质量为的物体从地面竖直向上抛出，一段时间后物体又落回地面．在此过程中物体所受空气阻力大小不变，其动能随距离地面高度的变化关系如图所示，取重力加速度，下列说法正确的有（ ）抛出瞬间克服重力做功的瞬时功率为下降过程中重力做功为全过程中克服空气阻力做功空气阻力的大小为11.如图所示，在竖直平面内有固定的光滑轨道，其中是半径为的圆弧轨道，竖直轨道与相切于点，水平轨道与相切于点．一根长为的轻杆两端分别固定着两个质量均为的相同小球，（视为质点），开始时球处在圆弧上端点，由静止释放小球和轻杆，使其沿光滑轨道下滑，重力加速度为，下列说法正确的是（ ）A.B.C.D.球下滑过程中机械能守恒、滑到水平轨道上时速度为从释放到、滑到水平轨道上，整个过程中轻杆对球做的功为 从释放到、滑到水平轨道上，整个过程中轻杆对球做的功为12.A.B.C.D.如图所示，在光滑水平面上放置一个质量为的滑块，滑块的一侧是一个弧形凹槽，凹槽半径为，点切线水平．另有一个质量为的小球以速度从点冲上凹槽，重力加速度大小为，不计摩擦．下列说法中正确的是（ ）当时，小球不可能到达点当时，小球在弧形凹槽上运动的过程中，滑块的动能一直增大如果小球的速度足够大，小球将从滑块的左侧离开滑块后落到水平面上当时，小球返回点后可能做自由落体运动二、填空与实验题（本大题共4小题，共16分）13.足够长的固定光滑斜面倾角为，质量为的物体在斜面上由静止开始下滑，则秒末重力做功的功率为．（）14.甲、乙两小球在光滑水平面上发生正碰，碰撞前后两球的（位移时间）图像如图所示，碰撞时间极短．已知被碰小球的质量，则另一小球的质量为，甲、乙两小球碰撞后动能 （填“改变”或“不变”）．15.A.B.C.D.（1）（2）如图1所示，在“探究功与速度变化的关系”的实验中，主要过程如下：①设法让橡皮筋对小车做的功分别为、、、；②分析纸带，求出橡皮筋做功使小车获得的速度、、、；③作出图像；④分析图像．如果图像是一条直线，表明；如果不是直线，可考虑是否存在、、等关系．关于该实验，下列说法正确的有 ．通过增加橡皮筋的条数可以使橡皮筋对小车做的功成整数倍增加通过改变小车质量可以改变橡皮筋对小车做的功每次实验中，橡皮筋拉伸的长度必需保持一致先释放小车，然后再接通电源实验中打点计时器的工作频率为，根据第次实验的纸带（如图2所示）求得小车获得的速度为．16.用半径相同的两小球、的碰撞验证动量守恒定律，实验装置示意如图，斜槽与水平槽圆滑连接．实验时先不放球，使球从斜槽上某一固定点处由静止滚下，落到位于水平地面的记录纸上留下痕迹．再把球静置于水平槽前端边缘处，让球仍从Р处由静止滚下，球和球碰撞后分别落在记录纸上留下各自的痕迹，记录纸上的点是重垂线所指的位置．A.B.C.D.E.F.（1）（2）（3）本实验必须测量的物理量有以下哪些 ．小球、的质量﹑小球、的半径斜槽轨道末端到水平地面的高度小球、离开斜槽轨道末端后平抛飞行的时间记录纸上点到、、各点的距离、、球的固定释放点到斜槽轨道末端水平部分间的高度差按照本实验方法，验证动量守恒的验证式是 ．若测得各落点痕迹到点的距离：，，，并知两球的质量比为，系统碰撞前总动量与碰撞后总动量的百分误差（结果保留一位有效数字）．三、计算题（本大题共4小题，共36分）17.（1）（2）在一段平直的公路上，质量为的汽车从静止开始做匀加速运动，经过，速度达到随后汽车以的额定功率沿平直公路继续前进，又经过达到最大速度．设汽车所受的阻力恒定，大小为．求：汽车行驶的最大速度的大小．汽车从静止到达最大速度所经过的路程．18.如图所示，一游戏装置由安装在水平面上的固定轻质弹簧、竖直圆轨道（在最低点分别与水平轨道和相连）、斜轨道组成，各部分平滑连接．某次游戏时，滑块从高为的斜轨道端点由静止释放，沿斜轨道下滑经过圆轨道后压缩弹簧，然后被弹出，再次经过圆轨道并滑上斜轨道，循环往复．已知圆轨道半径，滑块质量且可视为质点，长，长，滑块与、之间的动摩擦因数，滑块与其它轨道摩擦及空气阻力忽略不计，取．求：（1）（2）求滑块第一次过最高点时，轨道对滑块的支持力大小．求弹簧获得的最大弹性势能．19.（1）（2）如图所示，一轻质弹簧一端固定在倾角为的光滑固定斜面的底端，另一端连接质量的小物块，小物块静止在斜面上的点，距点的处有一质量的小物块，由静止开始下滑，与小物块发生弹性正碰，碰撞时间极短，碰后当小物块第一次上滑至最高点时，小物块恰好第一次回到点．小物块、都可视为质点，重力加速度，，，求：碰后小物块的速度大小．从碰后到小物块第一次回到点的过程中，弹簧对小物块的冲量大小．20.（1）如图所示，以、为端点的圆形光滑轨道和以﹑为端点的光滑半圆轨道都固定于竖直平面内，一滑板静止在光滑水平地面上，左端紧靠点，上表面所在平面与圆弧轨道分别相切于、，一物体从点正上方处由静止开始自由下落，然后经沿圆弧轨道滑下，再经滑上滑板，滑板运动到时被牢固粘连．物体质量为（可视为质点），滑板质量，以、为端点的圆形光滑轨道半径为，板长，板右端到的距离，物体与滑板间的动摩擦因数为，重力加速度取．求：滑板右端运动到时，物体速度大小．（2）物体在轨道上运动时不脱离轨道，求半圆轨道半径的取值范围．。

2020-2021学年第一学期期中考试2023励离一数学试図H如L20牙时剛I I勿分斡一、站升•（并3小・肾那JS4分'A 40#・農制4、■越出曲西中.nw-«* 的）I■世金■</■仏2JA"J・《b 4-（2,Mh »-UA^7}+剛心硼沪（\A. （5| 氐aX4.5.6.7JI C. （XII D・tl.3»7]2.從八玉2}・F图噩很不从鼻钿1削1奋&的曲牧关巖的4H ）3.巳I■■台4{«1"-4*一|处0}・ f-(xlhg.fr-Qcd). M^n^» CA. |<|x<^}B.何I CHV J) C・{*iY"«c2) D. (>|x<2|€足/d-2)■如f M/W的解析贰息(k B. /(4>-3» + 2 c. m i»-* a /(x>-3x*2</<i)--3i-4B. tA. f(x) * J? •■的叫JJF fi- /(*)" I ・ f<f) ■ XC・ /(x>-jr+l ・ E0・W"M 0. f(x)»Jt—1a. My■兰的債慶为(BHf/UGF C. H 0. (-«.^)U^.**>*v •ir^r7.若・2一1韵陀义撼和Q"】・flUt为17—J]・D实(A. (031B. M\C. UliD.代氏H*哭IH#用更孟取町直硝区阎为< >A. (-«^1>& 0,*«) C. (I*®) D. (11)鼻已知0・护’上・0』工・g.九V ( >k b<^<a ft. b<a«C・c<a<b D・C <6< tf10・若关于工齡!方理聖解・删炙殴“的取鱼雄稱是< )A. <-^->|U|»<«)B. |«却G If “》D”f y.-ll二、大■井•小■・常小關4务・共16分〉1L布然合―卩丄3}・a-(X5|・用用举址叢胪川居仏句•一・()ott Y x v-i 皿 _ . , 1“昇血・W/U<j)l・__________ ..is. sfiluwHFH•、在区阎MH內♦底有_____ 十.14.已如曲做/匕卜;•:；:；'^•时任金的片屯畫R・斗*号•有”(时"佃)](斗活)"・帼实叶的髦值雄K ______________ -三MIFA I (車大■洪5水41・共44分•解符磴耳出文字说期、旺第过用蕈：M算敗・) IB.(本小H矗井8分)己知■會/■倒切,止・{ji|lo»r*>iLCD(2) *(G0)U儿te. 8 计ih(I)加宀*32+唤-E・W讥WDM)■仁*W J F17. 分）fiW文繼为K的RfiU/U"訣二他童"B L2r *«CD :ft•丄的AL（2）蔚对任魅的2H 不■式■初■灯"忸成比.求宜IH的取也彙■Id. 1未44・*分巾£0求・・/（0・一佩街一冊“-4（八 "佩”1］的鼻小值.忸（事不硏駕分10分〉己知・<L> »w>i!>a»r. /（«>-/（■）.球丄■丄的值,fW ff（2>若酬"A0时・的定义域与值锻均为卜・炳|”求所京n. Mana c車兀■共2小共游分・■杏盅引出丈字傀明.（£«««««>!»■）20.（專血需介8分》已MAft/U）-l（r・叩妊匚1" — 1旷“0・求关尸嚣的不悻式/Or* !）*/（»>> 202L C4flW^I2^)已如只巧羅宦风任R上的erttt ?ix<0W, /⑴越憎，fl. /(・□■©・j7+(2*-J；>»-2m+5,覽合$■[和时忏・"|giLjrUXS・r•(■!«(£«<6[(uir./(r<r)K0|・ #snr.2020-2021学牟JB-学期期屮考试2023届高一数学试-・堆丼lb «韋兀■共L0小电・W 如^4升，共柏好，I. D 2. D 3. B 4. B bull b. K 人 L B. C «・ U 10.D二.MI7M.4 4-0 分.K 16 ») II, j-L-14rl} 12a ~ l9>> 1 14.C*^MIKSX44» «WA!3H；m 文T*暮 还瞩址帕©■■斗・•' is. CMMI U)dnsri27imsrs3. a^-uiK>s3):ill bg^x^llHoS. M0>l(x| J>2|» 彳FIJI ・{*|2v>： S3} ・•・・——“唱 0 ⑵ 曲心・(1|1 >2}^u®«kl 1^21. /.(tfl)lJ/<-U|iSJ).............2J -|o|,32<*°bI ；：)-,o °- .......... 4 沪17,【"析1%)因为/(■)▲* 上的研廉《!•・Q. UMfr-L O<I 用.)■吕三 兄由/册・e«r ・2・ --------------------------------------- 2 0绘耐当"■詔甘・HE ・祖“― ⑺ m <1> ■益二一»“ 由 1 AMta/WttR 上JWt 曲氐f • 2 2 1*1又W 为从■和价Ff (P - 21)</卜M ... ••・一・“_・f $ 命 fiU>v/(4MJt由 I 氏耀用4力》3?“・ IP 时一浙“科如 U,ffiA«.4>ll2Jk<0・卄⑵叫帕才卜Ifr. tMIhl Cl) 4«a Ha18.【・紀】比二扶罰戰!fill 开口Mt.对禅粘万用为*・3-伽•汉区岡糊时耳樺 他的位・遢打分英，CU ^a*l<J-2yWa<pf. Aft/Car) to.-Ml 上由胡阳I ■埸lt ・ 腔为/Wh $ .. .......... .. ... ...... .. 3介 (2> ^s3-^£j+iW^SffSlnh •僮杠理点处槪押，检■小備为 (3> -|J-2^<al®u>lH.曲故/")衣歌•调灣描.为o 3 *4, a< j ・■W ** I "・；壬 <r 壬 L.. a 3 - 6a ♦ 9.a > ]./(0| o u* .6u ・¥. .....aJJDr /也・is.【•析““闵和g)・mo 用叫：»«4-|J 「・[・断以丄"・|口・陌叫7古|畦小叫很为…"，砂:JC2) ]•当055引时.加)*•討：|"1 tqg 運比冈加iiBUB 的址文即加期为寫::用曲•坏件g.— ............. ....' • 上 9 r‘j 卜) ••• ••• ••<••■•••••*• • •• »»^ ••••*• ••• •««••«•■• •-•^■••a»■•■••••>••••••«■••• ••4^所IAIIL *働通紐 霆q 炳]上・局恥绻W^Att/<>)的建1诚£・"为片•"”-*・■・/中■}•"所途・・£ •"苏 --- -- ------------ 囚.miufl q 本人■共2仆4L 決20分•・??邑卑出広字tn Mb 通舅址杜"H 少m2O ・(uni 令“何・ /(I) ICMxR).附&( •小 y(7) 4・ 一 [/V)-IQ 卜 7口)，斫以通離枳卩)为审■為 冷舅如旳歡矶町血K 上期艮囑5| --------- ------ ---- --- 4令 借以军*氏 H3i +IH /(1> >20« /C3t • H ・IO % -|/( <)-!#! » fO< i l» -r(i| Q 0(3r + D» g(7)o 3x^3>-io 衡以网农枷灣氏MH 冲{巾八扌21. WW1由己talim 町先樽■合旷间亿即由/賦训成7铮 5dr - «m I f£l^«|O.i|.g(x)<-i>. ■不Jt(s)< -I K fEBfh c(0J)CA&. <« 的取 fll 笊 Bl •…" . * ssb1同为用”-2“-1JM♦ 5- <r -^；y Hr -J；W-2w4J ................................. 8介令'■才・£€他字1 ・Wg(<)® -<8MW-2JW*3・ Hfi*；^F-/4-R/-2»+J<-|.”*U> 肾ai«w2dW住劉“0詁］恒戍立，于BKM的*tofiNK^*«). ................. . ............ —-12 e。

2020届西安高新第一中学国际高中高三英语期中考试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AStepping Out Into NatureThe classic road trip is more popular than ever. Here are several places to hit the open road.Colo-road TripsThe Colorado Tourism Office has made it easy for road-trippers to explore the state’s 24 Scenic & Historic Byways.A new microsite includes-an interactive map that enables travelers to explore options by region, interest or season. Travelers seeking inspiration can also access insider tips and side-trip suggestions for historic attractions, active adventures and highlight cultural opportunities.TheBeartooth Highway.Visitors of this extraordinary byway experience the grand sights ofMontana,WyomingandYellowstonePark. The windy 68-mile stretch introduces road explorers to one of the most diverse ecosystems accessible by auto. Breathtakingly beautiful, this All-American Road showcases wide, high alpine plateaus（高原）, painted with ice blue lakes, forested valleys, waterfalls and wildlife.SewardHighway,AlaskaThe road that connectsAnchorageto Seward is 127-mile treasure of natural beauty, wildlife and stories of adventure. The drive begins at the base of theChugach Mountains, hugs the scenic shores of Turnagain Arm and winds through gold mining towns, national forests and fishing villages. Expect waterfalls, glaciers, eagles and some good bear stories.The Lighthouse Trail,MaineTravel the 375 miles betweenKitteryandCalais,Maine, visiting lighthouses along the way. Hear tales of shipwrecks（海难）and of the difficult and lonely life led by those who kept the lights burning brightly. If possible, visit theMaineLighthouseMuseum. where artifacts and hands-on exhibits for children provide an appealing break.1.What makes Colo-road Tips special?A.Good bear stories.B.A scenic beach.C.Hands-on exhibits.D.An interactive map.2.Where can you explore state of gold miners?A.Colorado.B.Montana.C.Alaska.D.Maine3.Which place is suitable for a family with children?A.Colo-road Trips.B.TheBeartooth Highway.C.Seward Highway.D.The Lighthouse Trail.BAbout a month after I joined Facebook, I got a call from Lori Goler, a highly regarded senior director of marketing at eBay. She made it clear this was a business call. “I want to apply to work with you at Facebook,” she said. “Instead of recommending myself, I want to ask you: What is your biggest problem, and how can I solve it?”My jaw hit the floor. I had hired thousands of people over the previous decade and no one had ever said anything remotely like that. People usually focus on finding the right role for themselves, with the implication that their skills will help the company. Lori put Facebook’s needs front and center. It was a killer approach. I responded, “Recruiting is my biggest problem. And, yes, you can solve it.”Lori never dreamed she would work in recruiting, but she jumped in. She even agreed to trade earnings for acquiring new skills in a new field. Lori did a great job running recruiting and within months was promoted to her current job, leading People@Facebook.The most common metaphor for careers is a ladder, but this concept no longer applies to most workers. As of 2010, the average American had eleven jobs from the ages of eighteen to forty-six alone. Lori often quotes Pattie Sellers, who came up with a much better metaphor: “Careers are a jungle gym, not a ladder.”As Lori describes it, there’s only one way to get to the top of a ladder, but there are many ways to get to the top of a jungle gym. The jungle gym model benefits everyone, but especially women who might be starting careers, switching careers, getting blocked by external barriers, or reentering the workforce after taking time off. The ability to create a unique path with occasional dips, detours (弯路), and even dead ends presents great views of many people, not just those at the top. On a ladder, most climbers are stuck staring at the butt of the person above.4. Why did Lori make the call?A. She helped Facebook to solve the biggest problem.B. She wanted to make a business deal with Facebook.C. She tried to ask for a pay rise in Facebook.D. She wanted to become an employee in Facebook.5. What impressed “I” by Lori?A. Lori was good at running recruiting.B. Lori attached great importance to Facebook’s needs.C. Lori jumped in Facebook with no adequate experience.D. Lori was skilled in marketing at eBay.6. What can we infer from the passage?A. Now all people don’t tend to climb the ladder.B. None on the ladder can enjoy the great views.C. Jungle gyms offer limited exploration for employees.D. A pregnant woman, jobless, benefits little from the jungle gyms.7. What is the best title of the passage?A. It’s a Jungle Gym, Not a Ladder.B. Facebook’s Biggest Problem.C. Applying for a Job in Facebook.D. A Jungle Gym is Better than a Ladder.CDid you know that horses talk? Well, they do, and you can lean to understand “horse talk” if you pay close attention to the horses you see.When horses live in the wild, other animals try to eat them, so a lot of horse talk is about staying alive. Even now, when most horses live on farms, they watch for danger. For this reason, never walk behind a horse. If you surprise it, the horse might mistake you for a mountain lion or wolf and give a dangerous kick.By watching the ears of a horse, you can get clues to what it's hearing. A horse can tum each ear in a different direction. For a wild horse, this trick is important for survival. The horse can hear something sneaking up behind it while also checking out a threatening noise in front. When a horse lets its ears down, it's feeling safe and relaxed. If horses becomeisolated, they neigh, or “whinny,” calling for company. They're saying, “Where are you? I'm over here!” If a horse snorts(哼) while holding its head high and staring at something, it's saying, “That looks dangerous. Get ready to run!” When two horses meet, they put their noses together and smell each other's breath. It's their way of asking, “Are you a friend?” Horses nicker,too. Nickering is a quiet sort of sound. This friendly noise means they're feeling secure and saying, “Clad to see you.”In the wild, horses live in herds, with all members watching for danger. In a herd, only one horse is the leader, the “boss hoss”. The “boss hoss” is usually an older female. She watches for threats and teaches younger horseshow to behave. However, others may want her job. When that happens, she pins her ears back against her head and may even bite or kick to get challengers to back off. She's using body language to say, “Hey, I'm in charge here!” All horses know that the one who makes others move is the leader. Horses relate to people that way, too.Horses have a language of their own. Now you know a bit of what they might be saying.8. What is the general idea of this passage?A. Horses can talk with their owners.B. You can know a bit of horses' language.C. Horses can “talk” in their own way.D. Other animals can also learn language.9. What does the underlined word “isolated” in Paragraph 3 mean?A. Lonely.B. Glad.C. Sad.D. Frightened.10. What will a horse do when he wants to know whether another horse is a friend?A. He will let his ears down.B. He and another horse will put their noses together and smell each other's breath.C They both will hold their heads high and stare at something.D. He will give another horse a dangerous kick.11. How will the “boss hoss” deal with her challengers?A. By warning or fighting.B. By watching for threats.C. By teaching younger horses how to behave.D. By relating to people.DGetting drunk on ice cream used to be the stuff of dreams, but thanks to Will Rogers, inventor and owner of WDS Dessert Stations in Hinkley, Illinois, it has become a delicious reality. The Below Zero icecream machine uses a unique technique to freeze alcohol, which allows you to turn beers, cocktails and even spirits (烈酒) into delicious soft —serve ice cream.Rogers was trying to create a highly — caffeinated espresso ice cream flavor when he realized hecould use the same technique with alcoholic beverages. He started experimenting with various gums and stabilizers commonly used in the ice cream industry and eventually patented something called the NEA gel. It’s this magicalconcoction (调制品) that allows the alcohol to freeze to a near solid inside the Below Zero ice cream machine.Even though Below Zero changes the texture (质地) of beer, cocktails and even spirits, essentially turning them into soft —serve ice cream, it does not affect the alcohol contentat all. The ABV (酒精度) remains exactly the same, which means you can get drunk on ice — cream just as you would on the same concoctions in liquid form.Will Rogers claims that it takes around 30 minutes for beer to go from liquid to ice cream form, but higher alcohol content drinks take longer. Essentially, the higher the alcohol level, the longer the wait.The American inventor plans to sell Below Zero ice cream machines to bars and breweries wanting to surprise their patrons. Metro reports that machines will sell for about 6,000.12. What’s the name of the machine which can change beer and spirits into ice cream?A. Will RogersB. WDS Dessert StationsC. HinkleyD. Below Zero13. What makes alcohol to freeze to a near solid inside the machine?A. gums.B. stabilizers.C. NEA gel.D. ABV.14. What can we know from the passage?A. The machine can change all liquids into ice cream.B. It takes 20 minutes for beer to change into ice cream.C. The higher the alcohol level, the shorter the wait will be.D. The machine changes the texture of beer, cocktails and even spirits.15. What can we infer from the passage?A. The machine affects the alcohol content.B. You can get drunk if you have ice—creams made from spirits.C. The American inventor doesn’t want to sell themagical machine.D. Bars and breweries will not become potential buyers of the machine.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。

2020年西安高新第一中学国际高中高三语文上学期期中试卷及答案解析一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。

中华优秀传统文化是中华民族的精神命脉，其中最重要的就是先人正面的思维方式、生活态度和价值追求，这是传统文化的核心精神。

浩如烟海的经、史、子、集各类书籍，便是传统文化精神的重要载体。

然而古籍汗牛充栋，阅读应该从何入手呢？古诗是古人心声的真实记录，是展现先民人生态度的可靠文本，正如清人叶燮所说：“诗是心声，不可违心而出……故每诗以人见，人又以诗见。

”读诗就是读人，阅读那些长篇短什，古人音容笑貌如在目前，这是我们了解前人心态的最佳途径。

清人沈德潜说：“有第一等襟抱，第一等学识，斯有第一等真诗。

”中国古人评价文学家时有一个优良传统，就是人品与文品并重。

经过历代读者集体选择，凡是公认的大诗人，往往都是具备“第一等襟抱” 的人物，其作品必然也是第一等真诗，从中可以感受真实心跳和脉搏，从而沦肌浃髓地领会传统文化精神。

从《诗经》《楚辞》到明清诗词，都具有很高的阅读价值，如果兼顾作品的经典意义、阅读难度等因素，唐诗宋词应是我们的首选阅读对象。

唐诗宋词对于现代读者到底有什么价值？中国古典诗歌有一个最古老的纲领，就是“诗言志”。

到了西晋，陆机在《文赋》中又提出“诗缘情”。

有人认为二者是对立关系，但是初唐孔颖达在《左传正义》中说得很清楚：“情志一也”。

情志就是指一个人的内心世界，包括对生活的感受和思考，也包括对万事万物的价值判断。

唐诗宋词的内容跟现代人没有距离，因为诗词中表达的那些内容都是普通人的基本情感、基本人生观和基本价值观。

比如喜怒哀乐，比如对真善美的追求，比如对祖国大好河山的热爱、对保家卫国英雄行为的赞美，唐宋人如此，现代人也如此。

所以唐诗宋词中典范作品所表达的内心情感、思考和价值判断可以传递到今天，启发我们更细致地品味人生意义和美感。

当然，唐诗宋词对于现代人的最大意义，是其中的典范作品可以提升我们的情操、气质和人格境界，有深远教育作用。

2020届西安高新第一中学国际高中高三英语上学期期中考试试题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AIn his 402nd anniversary year, Shakespeare is still rightly celebrated as a great language master and writer. But he was not the only great master of play writing to die in 1616, and he is certainly not the only writer to have left a lasting influence on theater.While less known worldwide, Tang Xianzu is considered one of Chinas greatest playwrights and is highly spoken of in that country of ancient literary and dramatic traditions.Tang was born in 1550 inLinchuan,Jiangxiprovince. Unlike Shakespeare's large body of plays,poems and sonnets (十四行诗), Tang wrote only four major plays: The Purple Hairpin, Peony Pavilion (《牡丹亭》), A Dream under the Southern bough, and Dream of Handan. The latter three were constructed around a dream narrative, a way through which Tang unlocked the emotional dimension of human desires and ambitions and explored human nature beyond the social and political limits of that time.Similar to Shakespeare, Tang's success rode the wave of a renaissance (复兴) in theater as an artistic practice. As in Shakespeare'sEngland, Tang's works became hugely popular inChinatoo. During Tang'sChina, his plays were enjoyed performed, and changed. Kunqu Opera, a form of musical drama, spread from southernChinato the whole nation and became a symbol of Chinese culture. Combining northern tune and southern music, kunqu Opera was known for its poetic language, music, dance movements and gestures. Tang's works benefited greatly from the popularity of kunqu Opera, and his plays are considered classics of kunqu Opera.While Tang and Shakespeare lived in a world away from each other, there are many things they share in common, such e humanity of their drama, their heroic figures, their love for poetic language, a lasting popularity and the anniversary during which we still celebrate them.1. Why is Shakespeare mentioned in the first paragraph?A. To describe Shakespeare's anniversary.B. To introduce the existence of Tang Xianzu.C. To explain the importance of Shakespeare.D. To suggest the less popularity of Tang Xianzu.2. What's possibly one of the main theme of Tang's works?A. Social reality.B. Female dreams.C. Human emotions.D. Political environment.3. What does the author mainly tell us in Paragraph 4?A. The influence of Kunqu Opera on Tang's works.B. Tang's success in copying Shakespeare's styles.C. The way Kunqu Opera became a symbol of Chinese culture.D. Tang's popularity for his poetic language and music.BTourism is often about seeking deeper emotional and personal connections with the world around us. Not all travel experiences, however, need to take place in the real world. With the evolution of virtual reality(VR) technology, tourism will increasingly become a combination of physical and virtual worlds. VR may even remove the need to travel entirely.But can a VR experience really equal a real world one? Many experts believe it can. Studies have shown that our brains have an inbuilt VR-like mechanism that enables us to live imagined experiences. Much of our waking life is spent thinking about either the past or the future. This is known as" mind wandering". During these events we' re not paying attention to the current world around us. Instead, we' re recalling memories, or creating and processing imagined futures.When engaged in mind wandering, our brains process these mental images using the same pathways used to receive inputs from the real world. So, the imagined past or future can create emotions and feelings similar to how we react to everyday life. VR can create these same feelings.While critics might argue that a virtual experience will never match reality, there are several ways VR tourism could make a positive contribution. Firstly it could help protect sensitive locations from over-tourism. In recent years famous sites such as Maya Bay in Thailand, and Cambodia's Angkor Wat Temples have had to limit the number of visitors because of their negative impact. These places are now producing their own VR experiences that will allow tourists to pass through virtual models of the sites.Virtual reality may also allow people back in time, to experience historical events, visit ancient cities, and even to walk among dinosaurs.Finally, in a world where many people suffer from stress and depression due to overwork, virtual tourism may provide a cheap and convenient way for people to take brief holidays to otherwise unreachable destinations andrecharge their batteries, without ever leaving their homes.It sounds like science fiction but it's already happening. As virtual technology improves and as people continue to demand new and interesting experiences, expect more virtual tourism, both in combination with the real world and instead of it.4. What is driving the development of virtual tourism?A. Companies seeking to make more money.B. Improvements in virtual reality technology.C. People's demand for more shared experiences.D. People's deeper understanding of the physical world.5. Which of the following best describes "mind wandering"?A. The brain processes which help people think VR is real.B. The way the brain processes inputs from the real world.C. Brain activities focusing on past or future events.D. Experiences coming from a person's imagination.6. What does the underlined word "it" in the last paragraph refer to?A. Science fiction.B. Virtual technology.C. Virtual tourism.D. The real world.7. What is the purpose of the passage?A. To describe the advantages of VR tourism.B. To give suggestions for reducing over-tourism.C. To encourage people to develop VR technology.D. To argue VR tourism will replace the real world travel.CSix Neanderthals who lived in what is now France were eaten by their fellow Neanderthals some 100,000 years ago, according to fearful evidence of the cannibalistic (食人的) event discovered by scientists in a cave in the 1990s. Now, researchersmay have figured out why the Neanderthals, including two children, became victims of cannibalism: Global warming.While previous studies have examined Neanderthal remains to find proof of cannibalistic behavior, this is thefirst study to offer clues as to what may have led Neanderthals to become cannibals. Scientists found that rapid changes in local ecosystems as the planet warmed may have wiped out the animal species that Neanderthals ate, forcing them to look elsewhere to fill their stomachs.The researchers examined a layer of sediment (沉积物) in a cave known as Baume Moula-Guercy, in southeastern France. In that layer, charcoal (碳) and animal bones were so well-preserved that scientists could reconstruct an environmental picture representing 120,000 to 130,000 years ago. They discovered that the climate in the area was likely even warmer than it is today, and that the change from a cold, dry climate to a warmer one happened quickly. “Maybe within a few generations”, study co-author Emmanuel said. As the animals that once populated the landscape disappeared, some Neanderthals ate what they could find — their neighbors.Cannibalism is by no means unique to Neanderthals, and has been practiced by humans and their s “from the early Palaeolithic to theBronze Age and beyond,” the study authors reported. The behavior adopted by the starving Neanderthals in the Baume Moula-Guercy should therefore not be viewed as “a mark of bestiality (兽性) or sub-humanity”, but as an emergency adaptation to a period of severe environmental stress, according to the study.8. What does the study mainly focus on?A. The social behavior of Neanderthals.B. The reason for cannibalism among Neanderthals.C. The climate change in southeasternFrance.D. The influence of global warming on ancient animals.9. What can possibly be used to describe the climate in southeasternFrance120,000 to 130,000 years ago?A. It was no warmer than it is today.B. It was first warm while later cold and dry.C. Its change was mild and went through quite a long process.D. Its change is a chief factor contributing to cannibalism.10. Which of the following might the study authors agree with?A. Neanderthals’ cannibalism showed their bestiality.B. Cannibalism was actually a measure the Neanderthals had to adopt to survive.C. Neanderthals’ cannibalism guaranteed their rule over other tribes.D. Only Neanderthals were found to have cannibalism in human history.11. Where can you most possibly find this passage?A. In a science journal.B. In a travel brochure.C. In a history book.D. In a geography book.DTaking an afternoon nap could keep your brain sharp, a new study has said. Adults ages 60 and older who took afternoon naps showed signs of better mental ability compared to those who didn’t nap, according to a study published in General Psychiatry earlier this week.Researchers analyzed napping habits in 2, 214 older Chinese people and measured their cognitive abilities using several cognitive tests. Participants took the Mini-Mental State Examination and theBeijingversion of the Montreal Cognitive Assessment, both of which test for memory, language and other cognitive abilities. In every category listed in the study, nappers tested statistically higher on average compared to their non-napping counterparts.Researchers did not gather data from people under 60, so a correlation cannot be drawn between napping and younger generations. The study also noted that there are conflicting studies about the benefits and risks of napping.Sleeping behaviors can be affected by a multitude of factors, said Dr. David Neubauer, associate professor atJohnsHopkinsUniversity. “Daily routines, medication use and sleep disorders can all play a role in how frequently someone takes a nap.” he said.Neubauer recommended taking a shorter “power nap” of up to 20 minutes to decrease the chances of transitioning into slow-wave sleep, which makes people feel groggy when they wake up. “Napping can be a healthy part of an older adult’s day”, Neubauer acknowledged, but make sure “sleepiness isn’t due to a treatable nighttime sleep disorder.” Older individuals who want to do all they can to preserve their cognitive functioning should put nighttime sleep first.12. What is the best way to keep a better cognitive ability for the old?A. Take a nap every day.B. Take sleep disorder seriously.C. Have a good sleep at night.D. Do exercise regularly.13. How did the researchers reach the conclusion?A. By taking the examinations and assessment.B. By analyzing napping habits and giving tests.C. By measuring nappers’ ability and analyzing them.D. By doing an experiment.14. Which has NO effect on one’s sleeping behaviors?A. Taking a nap.B. Medicine treatmentC. Sleep problemsD. Everyday activities.15. What can we know about “slow-wave sleep”?A. It appears in short nap up to 20 minutes.B. People should avoid it in their naps.C. It is a healthy part of an elder’s day.D. It was recommended by Neubauer.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。

2020年西安高新第一中学国际高中高三英语上学期期中考试试题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ASan Francisco Bay Area is a great place if you're a sports fan as you'll find several events all year round and plenty of team pride. If you are anywhere close to the area during a game,these fantastic sports events are here for you.San Francisco Giants BaseballThe San Francisco Giants baseball team plays in SF at Oracle Park. This is a fun ballpark because it's always packed with great energy and offers views of the bay. It's one of the most popular San Francisco sports events. The Giants are part of the National League West Division. Since their arrival here in 1958,they have been World Series Champions three times.Golden State Warriors BasketballThe fan base of the Golden State Warriors distributes the whole San Francisco Bay Area as this region's only NBA team.Their regular season runs from late October through mid-April, and all home games are played at the Chase Center in San Francisco.In total, the Warriors has won six NBA championships.San Francisco 49ers FootballThe 49ers are San Francisco's NFL team, though they have recently moved to Levi's Stadium in Santa Clara, about an hour south of SF. The football team was named for the prospectors (探矿者) who arrived in the area in 1849 for the Gold Rush. They’ve won 5 Super Bowl championships, all between 1981 and 1994.San Jose Sharks HockeyThe San Jose Sharks represent the Bay Area in hockey (冰球).They were founded in 1991 as the only Bay Area team to compete in the NHL. Sharks fans love going to these San Francisco sports events at the SAP Center,which they call the Shark Tank,located about an hour southeast of SF.1.Where can a sports fan have a good view of the area?A.The Oracle Park.B.The Chase Center.C.Levi's Stadium.D.The SAP Center2.Which team has claimed the most titles according to the text?A.The Giants.B.The Golden State Warriors.C.The 49ers.D.The San Jose Sharks.3.Where is the passage probably taken from?A.A book review.B.A news report.C.A science fiction.D.A tourist magazine.BPaper is an important part of modern life. People use it in school, at work, to make artwork and books, to wrap presents and much more. Trees are the most common material for paper these days.So how do people make paper out of trees today? People first cut trees, load them onto trucks and bring them to a factory. Machines cut open the outer coverings of the trees, and cut the trees into pieces. Those pieces are boiled into a soup. After that, it is hit flat, dried and cut up into sheets of paper.The entire process, from planting a small tree to buying your school notebook, takes a very long time. Just growing the trees takes 10 to 20 years.Making tons of paper from trees can harm the planet. Humans cut down 80, 000 to 160,000 trees around the world every day, and use many of them to make paper. Some of those trees come from tree farms. But people also cut down forests for paper, which means that animals and birds lose their homes.Cutting forests down also contributes to climate change, and paper factories pollute the air. After you throw paper, it often takes the paper six to nine years to break down. That's why recycling is important. It saves a lot of trees, slows climate change and helps protect endangered animals, birds and all creatures that rely on forests for their homes and food.So if paper isn't good for the environment, why don't people write on something else?The answer: They do. With computers, tablets and cellphones, people use much less paper than in the past. Maybe a day will come when we won't use paper at all — or will save it for very special books and artworks.4. What can we know about making paper out of trees?A. It costs much money.B. It takes a lot of time.C. It is very easy and fast.D. It is dangerous and difficult.5. What is the impact of paper production?A. It promotes the recycling.B. It does harm to the environment.C. It slows down the climate change.D. It protects the animals from losing homes.6. How will we use paper someday in the future according to the text?A. Use it for books only.B. Use the recycled paper.C. Treasure it occasionally.D. Use it for artworks.7. What idea does the author want to express from the text?A. The influence of making paper on environment.B. The wonderful experience of making paper.C. The necessary process of making paper.D. The good reasons for making paper.CThere is an old Chinese proverb that states “One generation plants the trees; another gets the shade,” and this is how it should be with mothers and daughters. The relationship between a mother and a daughter is sometimes confusing. The relationship can be similar to friendship. However, the mother and daughter relationship has unique characteristics that distinguish it from a friendship. These characteristics include responsibilities and unconditional love, whichprecludemothers and daughters from being best friends.Marina, 27 years old, said, “I love spending time with my mom, but I wouldn’t consider her my best friend. Best friends don’t pay for your wedding. Best friends don’t remind you how they carried you in their body and gave you life! Best friends don’t tell you how wise they are because they have been alive at least 20 years longer than you.” This doesn’t mean that the mother and daughter relationship can’t be very close and satisfying. This generation of mothers and adult daughters has a lot in common, which increases the likelihood of shared companionship. Mothers and daughters have always shared the common experience of being homemakers, responsible for maintaining(保持) and passing on family values and traditions. Today contemporary mothers and daughters also share the experience of work and technology, which may bring them even closer together.Best friends may ormay not continue to be best friends, but for better or worse; the mother and daughter relationship is permanent, even if for some unfortunate reason they aren’t speaking. Sometimes this is not an equal relationship. Daughters don’t always feel responsible for their mother’s emotional well-being. But mothers never stop being mothers, which includes frequently wanting to protect their daughters and often feeling responsible for their happiness. The mother and daughter relationship is a relationship that is not replaceable by any other. Mothers always “trump(胜过)” friends.8. What does the underlined word “preclude” in paragraph 1 probably mean?A. differ.B. benefit.C. prevent.D. change.9. What can we learn from what Marina said?A. Best friends will not spend money on her wedding.B. Best friends will not remind her of important issues in life.C. Her mother is wiser on account of her age.D. Her mother is definitely not her best friend.10. Why can a mother and a daughter build a even closer relationship today?A. Because they share advanced technology with each other.B. Because they work together to support the whole family.C. Because they experience the same values and traditions.D. Because they have common experience in life and work.11. What is the text mainly about?A. How to build a good mother and daughter relationship.B. A mother-daughter relationship is irreplaceable.C. Mothers want to be daughters’ friends.D. A daughter is a mother’s best friend.DAs a rider, Anna Kiesenhofe’s Olympics victory might be a surprise. The winner of the road race at the Tokyo Olympics left the sport at the end of 2017 when she found herself out of contract (合同). She came into Tokyo without a professional team and left as an Olympic champion.The 30-year old began her cycling career in 2014 after running injuries that prevented her from continuing her pursuits of triathlon (铁人三项). She later joined a Catalan team and won the Spanish National Cup in 2016.The then-26 year old signedher first professional contract with Lotto Soudal Ladies for the following season. However, she ended her 2017 campaign in April and did not sign a contract for 2018, eventually taking a year off the bike. In 2019, Kiesenhofer came back to the sport as a rider, winning the Austrian national road race. Despite her results, Kiesenhofer sill had no professional contract while going into the Tokyo Olympics.Kiesenhofer was the first rider to attack in the Olympic road race, eventually forming a breakaway along with Carl Oberholzer, Omer Shapira, Vera Looser and Anna Plichta, which went on to reach a gap of 11 minutes. After Looser and Oberholzer were dropped, Kiesnhofer ataced her two remaining breakaway companions.After Shapira and Plichta were caught by the peloton (主车群), the rest of the riders seemed to believe thatthey were racing among themselves for Gold, not knowing that Remehofere was still in front. While it might be a misjudgment from the rest of the peloton that allowed Kiesenhofer to keep her lead of more than two minutes, other riders’ mistakes should not detract from the Austrian’s efforts.Off the bike the new Olympic Champion has a PhD in mathematics after studying at the Technical University of Vienne as well as at Cambridge University. She currently works at the University of Lausanne.12. Why did Anna give up triathlon?A. She got injured.B. She lost interest in it.C. She had to attend university.D. She never won a medal.13. Which is the right order of the following events?①She ended her campaign.②She took a year off the bike.③She began her cycling career.④She won the Austrian national road race.⑤She won the Spanish National Cup.A. ③④①②⑤.B. ②③④①⑤.C. ③⑤①②④.D. ④②③①⑤.14. What were the riders of the peloton unaware of at the Tokyo Olympics?A. The road race was so difficult.B. Anna was a new rider.C. They had caught up with Anna.D. Anna took the lead of them.15. What is Anna’s present job?A. A cycling coach.B. A university teacher.C A professional rider. D. A college student.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。