高新一中本部第一学期期中高一试题(北师大版)
陕西省西安市高新一中2019-2020学年高一上学期期中考试数学试题Word版含答案
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．。
北师大版英语高一上学期期中试卷word版2
一第一学期期中考试（英语）1025本试卷分第I卷（选择题）和第II卷（非选择题）两部分，共120分，考试时间120分钟。
第I卷(选择题三部分共85分)第一部分听力(共两节，满分20分)第一节(共5小题；每小题1分，满分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的阅读下一小题，每段对话仅读一遍。
1. What is the woman doing on the Internet?A. Chatting.B. Listening to music.C. Learning English.2. What subject is Jack good at?A. Maths.B. Chinese.C. English.3. What’s the weather like now?A. Cloudy.B. Windy.C. Rainy.4. Where did they go just now?A. To a café.B. To a restaurant.C. To the airport.5. What do they plan to buy for their mother?A. Clothes.B. Shoes.C. A cap.第二节(共15小题；每小题1分，满分15分)听下面5段对话或独白。
每段对话或独白后有几个小题，从每题所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听每段对话或独白前，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。
每段对话或独白读两遍。
听第6段材料，回答第6—7题。
6. Why was the woman late for work?A. She missed the bus.B. The weather was bad.C. Her car broke down.7. What day is it today?A. Wednesday.B. Tuesday.C. Friday.听第7段材料，回答第8至10题。
西安高新第一中学2023-2024学年高一上学期期中数学试题(教师版)
西安市高新第一中学2023-2024学年高一上学期期中数学试题一、单选题1.集合A=1,2,3，B=y y=2x-1,x∈A，则A∩B等于( )A.∅B.2C.1,3D.1,3,5【答案】C【详解】由题设B={1,3,5}，故A∩B={1,3}.故选：C2.命题“∃x≥3，x2-2x+3<0”的否定是( )A.∀x≥3，x2-2x+3<0B.∀x≥3，x2-2x+3≥0C.∀x<3，x2-2x+3≥0D.∃x<3，x2-2x+3≥0【答案】B【详解】解：因为命题“∃x≥3，x2-2x+3<0”为存在量词命题，所以其否定为“∀x≥3，x2-2x+3≥0”．故选：B．3.设α∈-1, 12, 1, 2, 3，则使函数y=xα的定义域为R且为奇函数的所有α值为( )A.-1，1B.1，3C.1，2，3D.12，1，3【答案】B【详解】因为y=x-1,y=x12的定义域都不是R，函数y=x2是定义域为R的偶函数，所以y=x-1,y=x12，y=x2均不满足题意，而y=x，y=x3均符合题意，所以满足题意的α的值为1,3.故选：B4.如今我国物流行业蓬勃发展，极大地促进了社会经济发展和资源整合．已知某类果蔬的保鲜时间y(单位：小时)与储藏温度x(单位：℃)满足函数关系y=e ax+b(a，b为常数)，若该果蔬在6℃的保鲜时间为216小时，在24℃的保鲜时间为8小时，那么在12℃时，该果蔬的保鲜时间为( )A.16小时B.24小时C.36小时D.72小时【答案】D【详解】由题设216=e6a+b8=e24a+b⇒e18a=127⇒a=-ln36，b=4ln3+3ln2，所以x=12时，ax+b=-2ln3+4ln3+3ln2=ln72，此时y=e ln72=72小时.故选：D5.我国著名数学家华罗庚先生曾说：“数缺形时少直观，形缺数时难入微，数形结合百般好，隔裂分家万事休.”在数学的学习和研究中，常用函数的图像来研究函数的性质，也常用函数的解析式来琢磨函数的图像的特征，如函数f(x)=3x1-x2的图像大致是( )A. B.C. D.【答案】C【详解】由f (x )=3x 1-x 2可知，当x ∈0,1 时，f x >0，故排除A ；当x >1时，f x <0，排除BD .故选：C 6.已知函数f x 是偶函数，当0≤x 1<x 2时，f x 2 -f x 1 x 2-x 1 >0恒成立，设a =f 55 ，b =f -2 ，c =f 33 ，则a ，b ，c 的大小关系为( )A.a <b <cB.c <b <aC.b <c <aD.b <a <c 【答案】A【详解】当0≤x 1<x 2时，f x 2 -f x 1 x 2-x 1 >0恒成立，可知函数f x 在0,+∞ 上单调递增，又因为函数f x 是偶函数，所以b =f -2 =f 2 ，设a 1=55,b 1=2,c 1=33，则a 1 10=55 10=25,b 1 10=2 10=32，所以a 1<b 1，又b 1 6=2 6=8,c 1 6=33 6=9，所以b 1<c 1，所以a 1<b 1<c 1，又因为函数f x 在0,+∞ 上单调递增，所以a <b <c .故选：A .7.已知二次函数y =ax -1 x -a ．甲同学：y >0的解集为-∞,a ∪1a ,+∞ ；乙同学：y <0的解集为-∞,a ∪1a ,+∞ ，丙同学：y =ax -1 x -a 的对称轴在y 轴右侧．在这三个同学的论述中，只有一个假命题，则实数a 的取值范围为( )A.a <-1B.-1<a <0C.0<a ≤1D.a >1【答案】C【详解】若y >0的解集为-∞,a ∪1a ,+∞ ，则a >01a≥a ⇒0<a ≤1；若y <0的解集为-∞,a ∪1a ,+∞ ，则a <01a≥a ⇒a ≤-1；若y =ax -1 x -a 的对称轴在y 轴右侧，则a +1a 2>0⇒a +1a =a 2+1a>0⇒a >0；又这三个同学的论述中，只有一个假命题，故乙同学为假，综上，0<a ≤1.故选：C8.定义在R 上的函数f (x )满足f (x +1)=13f (x )，且当x ∈[0,1)时，f (x )=1-|2x -1|.若对∀x ∈[m ,+∞)，都有f (x )≤281，则m 的取值范围是( )A.103,+∞ B.113,+∞C.133,+∞D.143+∞ 【答案】B【详解】因为当x ∈[0,1)时，f (x )=1-|2x -1|，所以f (x )=2x ,0≤x <122-2x ,12≤x <1，又因为函数f (x )满足f (x +1)=13f (x )，所以函数f (x )的部分图像如下，由图可知，若对∀x ∈[m ,+∞)，都有f (x )≤281，则m ≥113.故A ，C ，D 错误.故选：B .二、多选题9.已知a <b <c ，且ac <0，则下列不等式中一定成立的是( )A.ac <bcB.ab 2<cb 2C.a a -b >0D.ac a -b >0【答案】ACD【详解】因为a <b <c ，且ac <0，所以c >0，a <0，故ac <bc ，A 正确．当b =0时，ab 2=cb 2，B 错误．a -b <0，a a -b >0，C 正确．a -b <0，ac a -b >0，D 正确．故选：ACD .10.下列四个不等式中，解集为-∞,1 ∪3,+∞ 的是( )A.2x -4x -3≥1 B.4x -5⋅2x +1+17≥x -3 0C.x 2-4x +3≥0 D.x -3+1 x +3≥0【答案】AB【详解】A ：由2x -4x -3-1=x -1x -3≥0，则x -1 x -3 ≥0x -3≠0 ⇒x ≤1或x >3，符合；B ：由4x -5⋅2x +1+17≥x -3 0，则22x -10⋅2x +16≥0x -3≠0 ⇒(2x -2)(2x -8)≥0x ≠3 ，所以x ≤1或x >3，符合；C ：x 2-4x +3=(x -1)(x -3)≥0，可得x ≤1或x ≥3，不符合；D ：x -3+1 x +3=(x -1)(x -3)≥0，则x ≤1或x ≥3，且x ≥0，所以0≤x ≤1或x ≥3，不符合.故选：AB .11.函数f x 的定义域为D ，若存在闭区间a ,b ⊆D ，使得函数f x 同时满足①f x 在a ,b 上是单调函数；②f x 在a ,b 上的值域为ka ,kb k >0 ，则称区间a ,b 为f x 的“k 倍值区间”．下列函数存在“3倍值区间”的有( )A.f x =2x (x ≤0)B.f x =1x (x >0)C.f x =x 2(x ≥0)D.f x =x 1+x 2(0≤x ≤1)【答案】BC【详解】A ：f x =2x 在(-∞,0]上递增，令2a=3a 2b =3b ，由于y =2x ,y =3x 在(-∞,0]上无交点，所以不存在a ,b 上的值域为3a ,3b ，不符合；B ：f x =1x 在(0,+∞)上递减，令1a =3b 1b =3a且b >a >0，即ab =13，故a =13,b =1时，存在a ,b 上的值域为3a ,3b ，符合；C ：f x =x 2在[0,+∞)上递增，令a 2=3a b 2=3b 且b >a ≥0，可得a =0b =3 ，故a =0,b =3时，存在a ,b 上的值域为3a ,3b ，符合；D ：在0<x ≤1，f x =11x +x ，而y =1x +x 在(0,1]上递减，则f x 在(0,1]上递增，又f 0 =0，所以f x 在[0,1]上的值域为0,12 ，令a 1+a 2=3a b 1+b 2=3b 且0≤a <b ≤1，可得a =0b =0 ，不合题设；故选：BC12.已知函数f x 的定义域是0,+∞ ，且f xy =f x +f y ，当x >1时，f x <0，f 2 =-1，则下列说法正确的是( )A.f 1 =0B.函数f x 在0,+∞ 上是减函数C.f 12023 +f 12022 +⋯+f 13 +f 12 +f 2 +f 3 +⋯+f 2022 +f 2023 =2023D.不等式f 1x-f x -3 ≥2的解集为4,+∞ 【答案】ABD【详解】对于A ，令x =y =1，得f 1 =f 1 +f 1 =2f 1 ，所以f 1 =0，故A 正确；对于B ，令y =1x >0，得f 1 =f x +f 1x =0，所以f 1x =-f x ，任取x 1,x 2∈0,+∞ ，且x 1<x 2，则f x 2 -f x 1 =f x 2 +f 1x 1 =f x 2x 1 ，因为x 2x 1>1，所以f x 2x 1<0，所以f x 2 <f x 1 ，所以f x 在0,+∞ 上是减函数，故B 正确；对于C ，f 12023 +f 12022 +⋅⋅⋅+f 13 +f 12 +f 2 +f 3 +⋅⋅⋅+f 2022 +f 2023 =f 12023×2023 +f 12022×2022 +⋅⋅⋅+f 13×3 +f 12×2 =f 1 +f 1 +⋅⋅⋅+f 1 +f 1 =0，故C 错误；对于D ，因为f 2 =-1，且f 1x =-f x ，所以f 12 =-f 2 =1，所以f 14 =f 12 +f 12 =2，所以f 1x -f x -3 ≥2等价于f 1x +f 1x -3≥f 14 ，又f x 在0,+∞ 上是减函数，且f xy =f x +f y ，所以1x x -3 ≤141x >01x -3>0，解得x ≥4，即不等式f 1x-f x -3 ≥2的解集为4,+∞ ，故D 正确,故选：ABD ．三、填空题13.函数f (x )=x +1x -1的定义域为．【答案】0,1 ∪1,+∞【详解】由题意得x ≥0x -1≠0 ，解得x ≥0且x ≠1，故答案为：0,1 ∪1,+∞14.我校召开秋季运动会，高一某班有28名同学参加比赛，有15人参加集体项目，有8人参加田赛，有14人参加径赛，同时参加集体项目和田赛的有3人，同时参加集体项目和径赛的有3人，没有人同时参加三个项目的比赛，则只参加径赛的有人．【答案】8【详解】假设只参加径赛的有x 人，又没有人同时参加三个项目的比赛，所以同时参加田赛和径赛人数为14-3-x ，只参加田赛人数为8-3-(14-3-x )，综上，9+x +x -6+3+3+11-x =28，可得x =8.故答案为：815.已知f x =x 2-2x +3，g x =122x +1-m ，若对任意x 1∈0,3 ，都存在x 2∈-2,-1 ，使得f x 1 ≥g x 2 ，则实数m 的取值范围是.【答案】[0,+∞)【详解】f x =x 2-2x +3=x -1 2+2，f x 在-∞,1 上单调递减，在[1,+∞)上单调递增.所以当x ∈0,3 时，f x min =f (1)=2.g x =122x +1-m 在R 上单调递减，所以当x ∈-2,-1 时，g x min =g (-1)=2-m .因为对任意x 1∈0,3 ，都存在x 2∈-2,-1 ，使得f x 1 ≥g x 2 ，所以只需f x min ≥g x min 即可，即2≥2-m ，解得m ≥0，即m 的取值范围是[0,+∞).故答案为：[0,+∞)16.已知函数f x =x +1x +a ，若对任意实数a ，关于x 的不等式f x ≥m 在区间12,3 上总有解，则实数m 的最大值为．【答案】23【详解】函数y =x +1x 在区间12,3 上的图象如下图所示：根据题意，对任意实数a ，关于x 的不等式f x ≥m 在区间12,3上总有解，只要找到其中一个实数a ，使得f x =x +1x+a 的最大值最小即可，如图，函数y =x +1x向下平移到一定的程度时，函数f x 的最大值最小，此时只有当f 1 =f 3 时，才能保证函数f x 的最大值最小，设函数y =x +1x 的图象向下平移了t 个单位，其中t >0，则103-t =-2-t ，解得t =83，此时函数f x max =103-83=23，∴m ≤23.因此，实数m 的最大值为23.故答案为：23.四、解答题17.集合A =x x -1x +3<0 ，B =x x 2-4x -5<0 ，C =x x <2m -1,m ∈R ．(1)求A ∩B ；(2)若x ∈B 是x ∈C 的充分条件，且x ∈C 是x ∈A 的必要条件，求实数m 的取值范围．【答案】(1)(-1,1)(2)3,+∞【详解】(1)由x -1x +3<0⇔x +3 x -1 <0⇔-3<x <1，则A =(-3,1)，由x 2-4x -5<0⇔(x +1)(x -5)<0⇔-1<x <5，则B =(-1,5)，故A ∩B =(-1,1)；(2)x ∈B 是x ∈C 的充分条件，则B ⊆C ；x ∈C 是x ∈A 的必要条件，即x ∈A 是x ∈C 的充分条件，则A ⊆C ；故A ∪B ⊆C ,由A ∪B =(-3,5)，C =x x <2m -1,m ∈R ，则5≤2m -1，解得m ≥3,故实数m 的取值范围是3,+∞ .18.已知x >0，y >0，且满足4x +1y =2．(1)求x +y 的最小值；(2)求1x +4 y +1的最大值．【答案】(1)92；(2)116.【详解】(1)由题设x +y =12(x +y )4x +1y =125+4y x +x y ≥125+24y x ⋅x y =92，当且仅当4y x =x y ，即x =3,y =32时等号成立，所以x +y 的最小值为92.(2)由4x +1y =2⇒4y +x =2xy ，则1x +4 y +1 =1xy +4y +x +4=13xy +4，又4y +x =2xy ≥24xy =4xy ，故xy (xy -2)≥0，即xy ≥4，当且仅当4y =x ，即x =4,y =1时等号成立，所以3xy +4≥16，故1x +4 y +1≤116，仅当x =4,y =1时等号成立，所以1x +4 y +1的最大值116.19.已知函数f x =3x +m 3x +1为奇函数．(1)判断函数f x 的单调性，并加以证明．(2)若不等式f at 2+2t -2 +f 1-t ≥0对一切t ∈1,4 恒成立，求实数a 的取值范围．【答案】(1)f x 在R 上单调递增，证明见解析(2)[0,+∞)【详解】(1)函数f x 的定义域为R ，f x =3x +m 3x +1=3x +1+m -13x +1=1+m -13x +1，因为f x 为奇函数，所以∀x ∈R ,f -x =-f (x )，所以1+m -13-x +1=-1-m -13x +1，则2=-(m -1)13x +1+13-x +1=(1-m )13x +1+3x 1+3z =1-m 所以m =-1；函数f x =1-23x +1，在R 上单调递增.下面用单调性定义证明：任取x 1,x 2∈R ，且x 1<x 2，则f (x 1)-f (x 2)=1-23x 1+1-1-23x 2+1 =23x 2+1-23x 1+1=2(3x 1-3x 2)(3x 1+1)(3x 2+1)因为y =3x 在R 上单调递增，且x 1<x 2，所以3x 1-3x 2<0，又(3x 1+1)(3x 2+1)>0，所以f (x 1)<f (x 2)，所以函数f x 在R 上单调递增.(2)因为f x 为奇函数，所以f -x =-f (x )，由f at 2+2t -2 +f 1-t ≥0得f at 2+2t -2 ≥-f 1-t ，即f at 2+2t -2 ≥f t -1 ，由(1)可知,函数f x 在R 上单调递增，所以at 2+2t -2≥t -1，即不等式at 2+t -1≥0对一切t ∈1,4 恒成立,则a ≥1t 2-1t =1t -12 2-14，又1t ∈14,1 ，所以当1t =1时，1t 2-1t 取最大值，最大值为0，所以要使a ≥1t2-1t 恒成立，则a ≥0，所以a 的取值范围为[0,+∞).20.已知不等式mx 2-3x +b >4的解集为-∞,1 ∪2,+∞ ．(1)求m ，b 的值；(2)解关于x 的不等式ax 2+m -a x +a -b <a -5m a ∈R ．【答案】(1)m =1,b =6;(2)答案见解析.【详解】(1)由题设mx 2-3x +b -4>0的解集为-∞,1 ∪2,+∞ ，所以1,2是mx 2-3x +b -4=0的两个根，且m >0，Δ=9-4m (b -4)>0，所以3m =3b -4m=2⇒m =1b =6 ，满足Δ=9-4×(6-4)=1>0，故m =1,b =6.(2)由(1)知：ax 2+1-a x -1=(ax +1)(x -1)<0，当a =0，则x -1<0，即x <1，解集为(-∞,1)；当a ≠0，则a x +1a (x -1)<0，若a >0，则x +1a (x -1)<0，可得-1a <x <1，解集为-1a ,1 ；若a <0，则x +1a (x -1)>0，当-1a <1，即a <-1时，可得x <-1a 或x >1，解集为-∞,-1a∪(1,+∞)；当-1a =1，即a =-1时，可得x ≠1，解集为(-∞,1)∪(1,+∞)；当-1a >1，即-1<a <0时，可得x <1或x >-1a ，解集为(-∞,1)∪-1a ,+∞ ；21.在2021年的全国两会上，“碳达峰”“碳中和”被首次写入政府工作报告，也进一步成为网络热词．为了减少自身消费的碳排放，节省燃料．经多次实验得到某种型号的汽车每小时耗油量Q (单位：L )与速度v (单位：km/h )(40≤v ≤120)的数据关系：Q v =0.000025v 3-0.004v 2+0.25v 40≤v <100 0.00625v 2-1.101v +57.6100≤v ≤120．(1)王先生购买了一辆这种型号的汽车接送孩子上学，由于城市道路拥堵，每小时只能行驶40km ，王先生家距离学校路程为8km ，王先生早上开车送孩子到学校，晚上开车接回家，求王先生每天开车接送孩子的耗油量；(2)周末，王先生开车带全家到周边游玩，经过一段长度为100km 平坦的高速公路(匀速行驶)，这辆车应以什么速度在这段高速公路行驶才能使总耗油量最少？【答案】(1)2.08(L )(2)80km/h【详解】(1)王先生的汽车每小时耗油量为Q 40 =0.000025×403-0.004×402+0.25×40=5.2(L )，每天开车接送孩子的时间为840×2=0.4(h )，则王先生每天开车接送孩子的耗油量为5.2×0.4=2.08(L ).(2)设总油耗量为W ，当40≤v <100时，Q v =0.000025v 3-0.004v 2+0.25v ，∴W =100v×Q v =0.0025v 2-0.4v +25=0.0025(v -80)2+9，∴当v =80时，W 取得最小值为9，当100≤v ≤120时，Q v =0.00625v 2-1.101v +57.6，∴W =100v ×Q v =0.625v +5760v-110.1，令v 1,v 2∈[100,120]，且v 1<v 2，则W 1-W 2=0.625v 1+5760v 1-110.1-0.625v 2+5760v 2-110.1 =0.625(v 1-v 2)+57601v 1-1v 2 =(v 1-v 2)0.625v 1v 2-5760v 1v 2，当v 1,v 2∈[100,120]且v 1<v 2时，v 1-v 2<0,v 1v 2>0，0.625v 1v 2-5760>0.625×1002-5760=490>0，则W 1-W 2<0，可得W =0.625v +5760v-110.1在[100,120]上单调递增，∴当v =100时，W 取得最小值为10，综上，当40≤v ≤120时，W 的最小值为9，此时对应的v =80，所以，这辆车应以80km/h 速度行驶才能使总耗油量最少．22.设函数f x ，g x 具有如下性质：①定义域均为R ；②f x 为奇函数，g x 为偶函数；③f x +g x =e x (常数e 是自然对数的底数，e =2.71828⋯)．利用上述性质，解决以下问题：(1)求函数f x ，g x 的解析式；(2)证明：对任意实数x ，f x 2-g x 2为定值，并求出这个定值；(3)已知m ∈R ，记函数y =2m ⋅g 2x -4f x ，x ∈-1,0 的最小值为φm ，求φm ．【答案】(1)f x =e x -e -x 2,g x =e x +e -x 2(2)-1(3)φm =m 1e -e 2-21e -e +2m ,m ≤2e 2-e 22m ,m >2e 2-e2【详解】(1)由性质③知，f x +g x =e x ，所以f -x +g -x =e -x ，由性质②知，f -x =-f x ,g -x =g x ，所以-f x +g x =e -x ，解得f x =e x -e -x 2,g x =e x +e -x 2.(2)由(1)可得：f x 2-g x 2=e x -e -x 2 2-e x +e -x 2 2=e 2x +e -2x -24-e 2x +e -2x +24=-1(3)函数y =2m ⋅g 2x -4f x =m e 2x +e -2x -2e x -e -x ，设t =e x -e -x ，因为函数y =e x 、y =-e -x 均为R 上的增函数，故函数t 为R 上的增函数，当x ∈-1,0 时，t ∈1e -e ,0 ，t 2=e x -e -x 2=e 2x +e -2x -2，所以e 2x +e -2x =t 2+2，所以原函数即y =mt 2-2t +2m ，t ∈1e -e ,0，设h t =mt 2-2t +2m ，t ∈1e -e ,0，当m =0时，h t =-2t 在t ∈1e -e ,0上单调递减，此时h t min =h 0 =0．当m ≠0时，函数h t 的对称轴为t =1m ，当1m >0时，即m >0时，h t 开口向上，在1e-e ,0 上单调递减，此时h t min =h 0 =2m ，当1m <12e -e 2时，即0>m >2e 2-e 2时，函数开口向下，此时h t min =h 0 =2m ，当12e -e 2≤1m <0时，即m ≤2e 2-e 2时，函数开口向下，此时h t min =h 1e -e =m 1e -e 2-21e-e +2m ，综上所述，φm =m 1e -e 2-21e -e +2m ,m ≤2e 2-e 22m ,m >2e 2-e 2.。
北师大版高一数学上期中试题及答案
（满分120分考试90分钟）年级 高一 学科 数学 （期中试卷）宝鸡市石油中学 齐宗锁一、选择题（本大题共12小题，每小题4分，共48分。
在每小题的四个选项中， 只有一项是符合题目要求的。
）1、设集合}35|),{(},64|),{(-==+-==x y y x B x y y x A ，则B A = （ ）A ．{1，2}B ．{x =1，y =2}C ．{（1，2）}D ．（1，2）2、已知函数)(x f 是定义在[]5,1a -上的偶函数，则a 的值是 （ ）A ．0 B.1 C.6 D.-6 3、若01a a >≠且，则函数1x y a-=的图象一定过点 （ ）A ．（０,１）B ．（０,-1）C ．(１,０) Ｄ．（１,１）4．若1)(+=x x f ，则=-)2(f 1（ ）A 、3B 、2C 、1D 、35．下列四个图像中，是函数图像的是 （ ）A 、（1）B 、（1）、（3）、（4）C 、（1）、（2）、（3）D 、（3）、（4）6、下列函数中既是奇函数，又在区间（0，+∞）上单调递增的是 ( )A ．2y x =-B ．()12xy g =C ．1y x x=+D ． ||x e y =7、若方程2ax 2－x －1=0在（0，1）内恰好有一个解，则a 的取值范围是 （ ）A ．a <-1B ．a >1C ．－1<a <1D ．0≤a <18、已知函数⎩⎨⎧=x x x f 3log )(2)0()0(≤>x x ，则)]41([f f 的值是 （ ）A.91B.41 C. 4 D. 99．为了得到函数13()3x y =⨯的图象，可以把函数1()3xy =的图象 （ ）A ．向左平移3个单位长度B ．向右平移3个单位长度C ．向左平移1个单位长度D ．向右平移1个单位长度10．．设a =log 0.34，b =log 43，c =0.3 –2，则a 、b 、c 的大小关系为 （ ）（1）（2）（3）（4）A ．b ＜a ＜cB ．a ＜c ＜bC ．c ＜b ＜aD ．a ＜b ＜c 11、函数)1lg(+=x y 的图象是 （ ）12、函数)32(log )(221--=x x x f 的单调递增区间是 （ ）A ．（－∞，1）B ．（－∞，－1）C ．（3，+∞）D ．（1，+∞） 二、填空题（本大题共4小题，每小题5分，共20分）13、已知集合{}{}0)1(,12=-=+==x x x B t x x A ，则=⋃B A 。
2022-2023学年北京师范大学附属中学高一上学期期中数学试卷含详解
北京师大附中2022—2023学年（上）高一期中考试数学试卷考生须知1．本试卷有三道大题，共6页．考试时长120分钟，满分150分．2．考生务必将答案填写答题纸上，在试卷上作答无效．3．考试结束后，考生应将答题纸交回．一、选择题共10小题，每小题4分，共40分．在每小题列出的四个选项中，选出符合题目要求的一项．1.若集合{}|52x x A =-<<，{}|33x x B =-<<，则A ⋂B =A.{}|32x x -<< B.{}|52x x -<<C.{}|33x x -<< D.{}|53x x -<<2.已知集合{}10A x x =+≤，{|}B x x a =≥，若A B ⋃=R ，则实数a 的值可以为（）A.2B.1C.0D.2-3.函数2()23f x x x =-+-在闭区间[0,3]上的最大值和最小值分别是（）A.0，–2B.–2，–6C.–2，–3D.–3，–64.下列函数值中，在区间(0,)+∞上不是．．单调函数的是（）A.y x= B.2y x = C.y x = D.1y x =-5.如果()f x 是定义在R 上的奇函数，那么下列函数中，一定是偶函数的是A.()y x f x =+ B.()y x f x =⋅C.2()y x f x =+ D.2()y x f x =⋅6.“0a b >>”是“22a b >”的（）A.充分而不必要条件B.必要而不充分条件C .充分必要条件D.既不充分也不必要条件7.函数3()1f x x x =+-的零点所在的一个区间是（）A.(2,1)-- B.(1,0)- C.(0,1) D.(1,2)8.下列函数中，满足f （2x ）＝2f （x ）的是（）A.f （x ）＝（x +2）2B.f （x ）＝x +1C.()4f x x=D.f （x ）＝x ﹣|x |9.已知3()4f x ax bx =+-，若(2)6f =，则(2)f -=（）A.-14B.14C.-6D.1010.根据统计，一名工人组装第x 件某产品所用的时间(单位：分钟)为f (x)＝,x A x A <≥(A ，c 为常数)．已知工人组装第4件产品用时30分钟，组装第A 件产品用时15分钟，那么c 和A 的值分别是A.75,25B.75,16C.60,25D.60,16二、填空题共10小题，每小题4分，共40分．11.函数()f x =_________．12.已知集合M ＝{0，1，2，3}，N ＝{x |x ＝2a ，a ∈M }，则集合M ∩N ＝_____．13.设函数()1,0,x f x x ∈⎧=⎨∉⎩Q Q ，其中Q是有理数集，则(f f ⎡⎤⎣⎦的值为_____________．14.设a 为常数，函数23 ()2f x x x =-+，若()f x a +为偶函数，则=a ___________．15.设函数f (x )＝1,12,1x x x x ⎧>⎪⎨⎪--≤⎩则f (f (2))＝________，函数f (x )的值域是________．16.若0x >，则1()49f x x x=+的最小值为___________；取到最小值时，x =___________．17.已知1x >，且1x y -=，则1x y+的最小值是___________；取到最小值时，x =__________．18.若函数()2f x x x a =-+为偶函数，则实数=a ________，函数()f x 的单调递增区间是___________.19.已知函数()f x 是定义在R 上的奇函数，当0x >时，2()2f x x ax a =-+，其中a ∈R ．①1()2f -=_______；②若()f x 的值域是R ，则a 的取值范围是_______．20.已知集合{1,2,,}U n = ，*N n ∈.设集合A 同时满足下列三个条件：①A U ⊆；②若x A ∈，则2x A ∉；③若U x A ∈ð，则2U x A ∉ð.（1）当4n =时，一个满足条件的集合A 是__________；（写出一个即可）（2）当10n =时，满足条件的集合A 的个数为_________.三、解答题共6小题，共70分.解答应写出文字说明，演算步骤或证明过程.21.已知全集U =R ，集合(){}20P x x x =-≥，{}26M x a x a =<<+（1）求集合U C P（2）若U C P M ⊆，求实数a 的取值范围.22.已知函数()12x f x x +=+．（1）求f [f （1）]的值；（2）若f （x ）＞1，求x 的取值范围；（3）判断函数在（-2，+∞）上的单调性，并用定义加以证明．23.已知函数2()f x a x=-，()g x x a =-（a ∈R ）.（1）当1a =时，解关于x 的不等式()0f x ≥；（2）判断函数()()y f x g x =-的奇偶性，并证明；（3）若()()0f x g x +≥在(0,)+∞上恒成立，求a 的取值范围.24.定义域为R 的函数()f x 满足：对任意的,m n ∈R 有()()()f m n f m f n +=⋅，且当0x >时，有0()1<<f x ，1(2)4f =.（1）求出(0)f 的值，并证明：()0f x >在R 上恒成立；（2）证明：()f x 在R 上是减函数；（3）若存在正数x 使不等式()24()()f x f ax f x <成立，求实数a 的取值范围．25.已知函数21()2f x ax x c =-+（a 、c ∈R ），满足(1)0f =，且x ∈R 时，()0f x ≥恒成立．（1）求a 、c 的值；（2）若函数()()g x f x mx =-在区间[,2]m m +上有最小值–5，请求出实数m 的值．26.已知集合{1,2,3,,}n A n =⋅⋅⋅，集合(,)n m M 为集合n A 的m 元子集，且(,)n m M 中元素均为孤立元素．孤立元素的定义为：当x A ∈，1x A -∉且1x A +∉时，则称x 为集合A 中的孤立元素．（1）列出所有符合题意的集合(4,2)M ；（2）设()k m 为集合(8,)m M 的所有可能的集合个数，求()k m 的最大值，并说明理由；（3）在集合(2,)m m M 的所有可能集合中，存在元素在所有可能的集合中出现的次数最少，求出这样的元素并指出其出现次数，并说明理由．北京师大附中2022—2023学年（上）高一期中考试数学试卷考生须知1．本试卷有三道大题，共6页．考试时长120分钟，满分150分．2．考生务必将答案填写答题纸上，在试卷上作答无效．3．考试结束后，考生应将答题纸交回．一、选择题共10小题，每小题4分，共40分．在每小题列出的四个选项中，选出符合题目要求的一项．1.若集合{}|52x x A =-<<，{}|33x x B =-<<，则A ⋂B =A.{}|32x x -<< B.{}|52x x -<<C.{}|33x x -<<D.{}|53x x -<<【答案】A【详解】在数轴上将集合A ，B 表示出来，如图所示，由交集的定义可得，A B ⋂为图中阴影部分，即{}32x x -<<，故选A.考点：集合的交集运算.2.已知集合{}10A x x =+≤，{|}B x x a =≥，若A B ⋃=R ，则实数a 的值可以为（）A.2B.1C.0D.2-【答案】D【分析】由题意可得{|1}A x x =≤-，根据A B ⋃=R ，即可得出1a ≤-，从而求出结果．【详解】{|},1{|}A x x B x x a =≤-=≥ ，且A B ⋃=R ，1a ∴≤-，∴a 的值可以为2-．故选：D ．【点睛】考查描述法表示集合的定义，以及并集的定义及运算．3.函数2()23f x x x =-+-在闭区间[0,3]上的最大值和最小值分别是（）A.0，–2B.–2，–6C.–2，–3D.–3，–6【答案】B【分析】根据二次函数的单调性求最值.【详解】()f x 的对称轴为1x =，开口向下，所以()f x 在[)0,1上单调递增，在(]1,3上单调递减，则()()max 12f x f ==-，又()03f =-，()36f =-，所以()min 6f x =-.故选：B.4.下列函数值中，在区间(0,)+∞上不是．．单调函数的是（）A.y x =B.2y x = C.y x = D.1y x =-【答案】D【分析】结合一次函数，二次函数，幂函数的性质可进行判断．【详解】由一次函数的性质可知，y x =在区间(0,)+∞上单调递增；由二次函数的性质可知，2y x =在区间(0,)+∞上单调递增；由幂函数的性质可知，y x =+(0,)+∞上单调递增；结合一次函数的性质可知，1y x =-在()0,1上单调递减，在()1,+∞上单调递增．故选：D ．【点睛】本题主要考查了基本初等函数的单调性的判断，属于基础试卷．5.如果()f x 是定义在R 上的奇函数，那么下列函数中，一定是偶函数的是A.()y x f x =+ B.()y x f x =⋅C.2()y x f x =+ D.2()y x f x =⋅【答案】B【详解】试卷分析：由题意得，因为函数()f x 是定义在R 上的奇函数，所以()()f x f x -=-，设()()g x xf x =，则()()()()()g x x f x xf x g x -=--==，所以函数()g x 为偶函数，故选B ．考点：函数奇偶性的判定．6.“0a b >>”是“22a b >”的（）A.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件【答案】A【分析】根据充分不必要条件的概念判断即可.【详解】当0a b >>时，22a b >；当22a b >时，a b >，不一定0a b >>，所以“0a b >>”是“22a b >”的充分不必要条件.故选：A.7.函数3()1f x x x =+-的零点所在的一个区间是（）A.(2,1)-- B.(1,0)- C.(0,1)D.(1,2)【答案】C【分析】利用零点的存在性定理即可求解.【详解】因为3()1f x x x =+-在x ∈R 上单调递增，(2)110,(1)30f f -=-<-=-<，根据零点的唯一性定理知函数在(2,1)--上无零点，故A 错误;(1)30,(0)10f f -=-<=-<，根据零点的唯一性定理知函数在(1,0)-上无零点，故B 错误;(0)10,(1)10f f =-<=>，根据零点的唯一性定理知函数在(0,1)上有唯一零点，故C 正确;(1)10,(2)90f f =>=>，根据零点的唯一性定理知函数在(1,2)上无零点，故D 错误;故选:C.8.下列函数中，满足f （2x ）＝2f （x ）的是（）A.f （x ）＝（x +2）2B.f （x ）＝x +1C.()4f x x= D.f （x ）＝x ﹣|x |【答案】D【分析】对每一个选项的函数逐一验证即得解.【详解】A.f （x ）＝（x +2）2,所以222(2)(22)484,2()288f x x x x f x x x =+=++=++，所以不满足满足f （2x ）＝2f （x ）；B.f （x ）＝x +1，所以(2)21,2()22,(2)2()f x x f x x f x f x =+=+∴≠；C.()4f x x =，所以428(2),2(),(2)2()2f x f x f x f x x x x===∴≠；D.f （x ）＝x ﹣|x |，所以(2)22||,2()22||f x x x f x x x =-=-，满足f （2x ）＝2f （x ）.故选D【点睛】本题主要考查求函数值，意在考查学生对这些知识的理解掌握水平.9.已知3()4f x ax bx =+-，若(2)6f =，则(2)f -=（）A.-14B.14C.-6D.10【答案】A【分析】先计算(2)+(2)f f -，再代入数值得结果.【详解】(2)+(2)8248248f f a b a b -=+----=-Q ，又(2)6f =，所以(2)14,f -=-故选：A10.根据统计，一名工人组装第x 件某产品所用的时间(单位：分钟)为f (x)＝,x A x A <≥(A ，c 为常数)．已知工人组装第4件产品用时30分钟，组装第A 件产品用时15分钟，那么c 和A 的值分别是A.75,25 B.75,16C.60,25D.60,16【答案】D【详解】由题意可得：f （A ）=15，所以而f （4）=30，可得出2=30=4，可得A=16从而=60故答案为D二、填空题共10小题，每小题4分，共40分．11.函数()f x =_________．【答案】()2,∞+【分析】根据函数表达式可得20x ->，解不等式即可.【详解】由()f x =20x ->，解得2x >，所以函数的定义域为()2,∞+.故答案为：()2,∞+12.已知集合M ＝{0，1，2，3}，N ＝{x |x ＝2a ，a ∈M }，则集合M ∩N ＝_____．【答案】{0，2}【分析】先求出集合N ，再求M ∩N.【详解】∵M ＝{0，1，2，3}，N ＝{0，2，4，6}，∴M ∩N ＝{0，2}．故答案为{0，2}【点睛】本题主要考查集合的交集运算，意在考查学生对这些知识的理解掌握水平.13.设函数()1,0,x f x x ∈⎧=⎨∉⎩QQ ，其中Q是有理数集，则(f f ⎡⎤⎣⎦的值为_____________．【答案】1【分析】利用分段函数的定义即可求解.【详解】因为Q ,所以(f ，又因为0∈Q ,所以(0)=1f ，所以(=(0) 1.f f f ⎡⎤=⎣⎦故答案为:1.14.设a 为常数，函数23 ()2f x x x =-+，若()f x a +为偶函数，则=a ___________．【答案】1【分析】先得到()22()2223f x a x a x a a +=+-+-+，利用函数为偶函数，得到()()f x a f x a -+=+，列出方程，得到220a -=，求出a .【详解】23 ()2f x x x =-+，()()()222()232223f x a x a x a x a x a a +=+-++=+-+-+，因为()f x a +为偶函数，所以()()f x a f x a -+=+，故()()()()222222232223x a x a a x a x a a -+--+-+=+-+-+，故220a -=，解得：1a =.故答案为：115.设函数f (x )＝1,12,1x x x x ⎧>⎪⎨⎪--≤⎩则f (f (2))＝________，函数f (x )的值域是________．【答案】①.－52②.[－3，＋∞)【分析】由内层依次代入函数，即可得出结果.分段求出各段的值域，再求并集即为答案.【详解】∵f (2)＝12，∴f (f (2))＝12f ⎛⎫⎪⎝⎭＝－12－2＝－52.当x ＞1时，f (x )∈(0,1)，当x ≤1时，f (x )∈[－3，＋∞)，∴f (x )∈[－3，＋∞)．【点睛】本题考查分段函数的函数值与值域.属于基础题.复合函数的函数值求法：由内层依次代入计算.分段函数的值域问题：分段求出各段的值域，再求并集.16.若0x >，则1()49f x x x =+的最小值为___________；取到最小值时，x =___________．【答案】①.43②.16【分析】利用基本不等式计算可得.【详解】解：因为0x >，所以14()493f x x x =+≥=，当且仅当149x x =，即16x =时取等号，故答案为：43；1617.已知1x >，且1x y -=，则1x y+的最小值是___________；取到最小值时，x =__________．【答案】①.3②.2【分析】依题意可得1x y =+，且0y >，再利用基本不等式求出1x y+的最小值，即可得解.【详解】解：∵1x >，且1x y -=，∴1x y =+，且0y >，所以11113x y y y +=++≥+=，当且仅当1y =，即2x =时取等号.故答案为：3；218.若函数()2f x x x a =-+为偶函数，则实数=a ________，函数()f x 的单调递增区间是___________.【答案】①.0②.1,02⎛⎫-⎪⎝⎭、1,2⎛⎫+∞ ⎪⎝⎭【分析】由偶函数的定义得出x a x a +=-，等式两边平方可求得实数a 的值，求出函数()f x 在()0,∞+上的增区间和减区间，利用偶函数的基本性质可得出函数()f x 的单调递增区间.【详解】函数()2f x x x a =-+的定义域为R ，且该函数为偶函数，则()()f x f x -=，即()22x x a x x a ---+=-+，所以，x a x a -=+，等式x a x a -=+两边平方可得222222x ax a x ax a -+=++，可知0ax =对任意的x R ∈恒成立，所以，0a =，则()2f x x x =-.当0x >时，()2f x x x =-，则函数()f x 在()0,∞+上的减区间为10,2⎛⎫ ⎪⎝⎭，增区间为1,2⎛⎫+∞⎪⎝⎭.由于函数()f x 为偶函数，因此，函数()f x 的单调递增区间为1,02⎛⎫-⎪⎝⎭、1,2⎛⎫+∞ ⎪⎝⎭.故答案为：0；1,02⎛⎫-⎪⎝⎭、1,2⎛⎫+∞ ⎪⎝⎭.【点睛】求函数的单调区间：首先应注意函数的单调区间是其定义域的子集；其次掌握一次函数、二次函数等基本初等函数的单调区间．求函数单调区间的常用方法：根据定义、利用图象、单调函数的性质.19.已知函数()f x 是定义在R 上的奇函数，当0x >时，2()2f x x ax a =-+，其中a ∈R ．①1()2f -=_______；②若()f x 的值域是R ，则a 的取值范围是_______．【答案】①.14-②.(,0][1,)-∞⋃+∞【分析】①利用奇函数的定义，计算即可得到所求的值；②由()f x 的图象关于原点对称，以及二次函数的图象与x 轴的交点，由判别式不小于0，解不等式即可得到答案.【详解】①由题意，函数()f x 是定义在R 上的奇函数，当0x >时，()22f x x ax a =-+，则211111()([()2]22224f f a a -=-=--⨯+=-；②若函数()f x 的值域为R ，由函数的图象关于原点对称，可得当0x >时，函数()22f x x ax a =-+的图象与x 轴有交点，则2(2)40a a ∆=-≥，解得0a ≤或1a ≥，即实数a 的取值范围是(,0][1,)-∞⋃+∞.【点睛】本题主要考查了函数的奇偶性的应用，及函数的值域的应用，其中解答中根据函数的奇偶性和合理利用二次函数的图象与性质求解是解答的关键，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，属于中档试卷.20.已知集合{1,2,,}U n = ，*N n ∈.设集合A 同时满足下列三个条件：①A U ⊆；②若x A ∈，则2x A ∉；③若U x A ∈ð，则2U x A ∉ð.（1）当4n =时，一个满足条件的集合A 是__________；（写出一个即可）（2）当10n =时，满足条件的集合A 的个数为_________.【答案】①.{}1,4；（{}1,4，{}1,3,4，{}2，{}2,3任写一个即可）②.32【分析】（1）4n =时，集合{}1,2,3,4U =，则由题意可得1，4同属于集合A ，此时2属于A 的补集，或2属于集合A ，1，4同属于集合A 的补集，元素3与集合A 的关系不确定，从而可求出集合A ，（2）当10n =时，集合{}1,2,3,4,5,6,7,8,9,10U =，1，4必须同属于A ，此时2,8属于A 的补集；或1，4必须同属于A 的补集，此时2,8属于A ；当3A ∈时，则6A ∉；则3U A ∉ð，则6A ∈；当5A ∈时，则10A ∉；当5U A ∉ð，则10A ∈；而元素7，9没有限制，从而可得满足条件的集合A【详解】（1）4n =时，集合{}1,2,3,4U =，由①A U ⊆；②若x A ∈，则2x A ∉；③若U x A ∈ð，则2U x A ∉ð，可知：当1A ∈时，则2A ∉，即2U A ∈ð，则4U A ∉ð，即4A ∈，但元素3与集合A 的关系不确定，故{}1,4A =或{}1,3,4A =；当2A ∈时，则4A ∉，1A ∉，元素3与集合A 的关系不确定，故{}2A =，或{}2,3A =．综上，{}1,4A =，或{}1,3,4A =，或{}2A =，或{}2,3A =．（2）当10n =时，集合{}1,2,3,4,5,6,7,8,9,10U =，由①A U ⊆；②若x A ∈，则2x A ∉；③U x A ∈ð，则2U x A ∉ð，可知：当1A ∈时，则2A ∉，即2U A ∈ð，则4U A ∉ð，即4A ∈，8U A ∈ð，1，4必须同属于A ，此时2,8属于A 的补集；或1，4必须同属于A 的补集，此时2,8属于A ；此时1248，，，的放置有2种；当3A ∈时，则6A ∉；则3U A ∉ð，则6A ∈；此时36，的放置有2种；当5A ∈时，则10A ∉；当5U A ∉ð，则10A ∈；此时510，的放置有2种；而元素7，9没有限制，此时79，的放置，各有2种；所以集合A 可能为{}{}{}{}1,4,2,8,1,4,7,9,2,8,7,9，{}{}{}{}1,4,3,2,8,3,1,4,6,2,8,6，{}{}{}{}1,4,5,2,8,5,1,4,10,2,8,10，{}{}{}{}1,4,7,2,8,7,1,4,3,7,1,4,3,9，{}{}{}{}1,4,9,2,8,9,2,8,3,7,2,8,3,9，{}{}{}{}1,4,67,1,4,6,9,2,8,6,7,2,8,6,9,，{}{}{}{}1,4,5,7,1,4,5,9,2,8,5,7,2,8,5,9，{}{}{}{}1,4,10,7,1,4,10,9,2,8,10,7,2,8,10,9所以满足条件的集合A 共有32个．故答案为：{}1,4，（{}1,4，{}1,3,4，{}2，{}2,3任写一个即可）；32三、解答题共6小题，共70分.解答应写出文字说明，演算步骤或证明过程.21.已知全集U =R ，集合(){}20P x x x =-≥，{}26M x a x a =<<+（1）求集合U C P（2）若U C P M ⊆，求实数a 的取值范围.【答案】（1）{}|02x x <<；（2）[]2,0-【分析】（1）分析可得，P 是不等式的解集，由不等式的解法，容易解得P ，进而可得U P ð可得答案．（2）根据U P M ⊆ð，利用区间端点值建立不等关系，最后解不等式组即可求实数a 的取值范围．【详解】解：（1）因为全集U =R ，集合(){}20P x x x =-≥，所以{|2P x x =≥或0}x ≤所以{|(2)0}U P x x x =-<ð，即集合{}|02U P x x =<<ð（2）因为U P M ⊆ð，所以0262a a ⎧⎨+⎩解得02.a a ⎧⎨-⎩所以[]2,0a ∈-【点睛】本题考查集合间的交、并、补的混合运算，这类题目一般与不等式、方程联系，难度不大，注意正确求解与分析集合间的关系即可，属于基础题．22.已知函数()12x f x x +=+．（1）求f [f （1）]的值；（2）若f （x ）＞1，求x 的取值范围；（3）判断函数在（-2，+∞）上的单调性，并用定义加以证明．【答案】（1）58（2）（-∞，-2）（3）增函数，证明见解析【分析】（1）可以求出()213f =，然后代入x =23即可求出f [f （1）]的值；（2）根据f （x ）＞1即可得出112x x ++，化简然后解分式不等式即可；（3）分离常数得出()112f x x =-+，从而可看出f （x ）在（-2，+∞）上是增函数，根据增函数的定义证明：设任意的x 1＞x 2＞-2，然后作差，通分，得出()()()()12121222x x f x f x x x --=++，然后说明f （x 1）＞f （x 2）即可得出f （x ）在（-2，+∞）上是增函数．【详解】（1）f [f （1）]=2125323823f +⎛⎫== ⎪⎝⎭+；（2）由f （x ）＞1得，112x x ++＞，化简得，102x +＜，∴x ＜-2，∴x 的取值范围为（-∞，-2）；（3）()11122x f x x x +==-++，f （x ）在（-2，+∞）上是增函数，证明如下：设x 1＞x 2＞-2，则：()()12211122f x f x x x -=-++=()()121222x x x x -++，∵x 1＞x 2＞-2，∴x 1-x 2＞0，x 1+2＞0，x 2+2＞0，∴()()1212022x x x x -++＞，∴f （x 1）＞f （x 2），∴f （x ）在（-2，+∞）上是增函数．【点睛】本题考查了已知函数求值的方法，分式不等式的解法，分离常数法的运用，增函数的定义，考查了计算能力和推理能力，属于基础题．23.已知函数2()f x a x=-，()g x x a =-（a ∈R ）.（1）当1a =时，解关于x 的不等式()0f x ≥；（2）判断函数()()y f x g x =-的奇偶性，并证明；（3）若()()0f x g x +≥在(0,)+∞上恒成立，求a 的取值范围.【答案】（1）02x <≤（2）奇函数，证明见解析（3）a ≤【分析】(1)解分式不等式即可；(2)利用奇函数的定义证明;(3)利用基本不等式求出最小值即可求解.【小问1详解】当1a =时,22()10x f x x x-=-=≥,即(2)00x x x -≥⎧⎨≠⎩,解得02x <≤.【小问2详解】依题意，2()()y f x g x x x =-=-，判断函数为奇函数，证明如下：令2()h x x x =-，定义域为(),0(0,)-∞⋃+∞，因为22()()()h x x x h x x x-=-+=--=-，所以函数为奇函数.【小问3详解】由题可知220x a x+-≥在(0,)+∞上恒成立，即22x a x +≥在(0,)+∞上恒成立，因为2x x +≥，当且仅当2x x =，x =所以要使22x a x +≥在(0,)+∞上恒成立，则min22x a x ⎛⎫+≥ ⎪⎝⎭，即2a ≥，解得a ≤.24.定义域为R 的函数()f x 满足：对任意的,m n ∈R 有()()()f m n f m f n +=⋅，且当0x >时，有0()1<<f x ，1(2)4f =.（1）求出(0)f 的值，并证明：()0f x >在R 上恒成立；（2）证明：()f x 在R 上是减函数；（3）若存在正数x 使不等式()24()()f x f ax f x<成立，求实数a 的取值范围．【答案】（1）(0)1f =,证明见解析（2）证明见解析.（3）1a >-【分析】(1)令0,1m n ==即可求解(0)1f =;(2)根据题设令121,m x n x x ==-即可证明;(3)利用题设条件和单调性求解.【小问1详解】令0,1m n ==,则有(1)(0)(1)f f f =⋅,因为0(1)1f <<,所以(0)1f =.因为0x >时，有0()1<<f x ，令,==-m x n x ，则有()()()f x x f x f x -+=-⋅，即()()(0)1f x f x f -⋅==，所以1()1()f x f x -=>，即0x <时，()0f x >.综上，()0f x >在R 上恒成立.【小问2详解】1212,,,x x x x ∀∈<R 则有210x x ->，所以()2101f x x <-<，因为[]2121121()()()()f x f x x x f x f x x =+-=⋅-，所以2211()()1()=-<f x f x x f x ，所以12()()f x f x >，所以()f x 在R 上是减函数.【小问3详解】由（1）可知1(2)=4(2)f f -=，所以()4()()=(2)()()(2)()21f x f ax f f x f ax f x f ax f a x -=-+=-++⎡⎤⎣⎦，所以存在正数x 使不等式()221()f a x f x -++<⎡⎤⎣⎦成立，因为()f x 在R 上是减函数，所以存在正数x 使不等式()221a x x -++>成立，即存在正数x 使不等式21a x x >+-成立，所以min21a x x ⎛⎫>+- ⎪⎝⎭，因为0x >，所以211x x +-≥-，当且仅当2x x=，x =时等号成立,所以1a >.25.已知函数21()2f x ax x c =-+（a 、c ∈R ），满足(1)0f =，且x ∈R 时，()0f x ≥恒成立．（1）求a 、c 的值；（2）若函数()()g x f x mx =-在区间[,2]m m +上有最小值–5，请求出实数m 的值．【答案】（1）14a c ==.（2）3m =-或1m =-+【分析】（1）讨论0a =时不满足题意，0a ≠时根据题意(1)0f =得12a c +=，又可列出201()402a ac >⎧⎪⎨--≤⎪⎩解得21()04a -≤，又21()04a -≥，即可求解.（2）由（1）知2111()424f x x x =-+，得2111()()()424g x f x mx x m x =-=-++，然后求出对称轴21x m =+，并分三种情况讨论二次函数在指定区间[,2]m m +上的单调性即可.【小问1详解】当0a =时，1()2f x x c =-+.由(1)0f =得：102c -+=，即12c =，11().22f x x ∴=-+显然1x >时，()0f x <这与条件相矛盾，不符合题意.0a ∴≠，函数21()2f x ax x c =-+是二次函数.由于对一切x R ∈，都有()0f x ≥，于是由二次函数的性质可得：201(402a ac >⎧⎪⎨--≤⎪⎩，即01016a ac >⎧⎪⎨≥>⎪⎩由(1)0f =得12a c +=，即12c a =-，代入01016a ac >⎧⎪⎨≥>⎪⎩得11()216a a -≥.整理得2110216a a -+≤，即21(04a -≤，而21()04a -≥，14a ∴=.14c ∴=.14a c ∴==【小问2详解】14a c == ,2111()424f x x x ∴=-+，2111()()()424g x f x mx x m x ∴=-=-++，该函数图像开口向上，且对称轴为21x m =+，假设存在实数m 使函数2111()()()424g x f x mx x m x =-=-++，在区间[],2m m +上有最小值5-.①当1m <-时，21m m +<，函数()g x 在区间[],2m m +上是递增的，()5,g m ∴=-即2111()5424m m m -++=-解得3m =-或73m =71,3>- 73m ∴=舍去.②当11m -≤<时，212m m m ≤+<+函数()g x 在区间[],21.m m +上是递减的，而在区间[]21,2m m ++上是递增的，(21)5g m ∴+=-即()()211121215424m m m ⎛⎫+-+++=- ⎪⎝⎭.解得12m =--或12m =-+.③当m 1≥时，212m m +≥+，函数()g x 在区间[],2m m +上是递减的，(2)5,g m ∴+=-即2111(2)()(2)5424m m m +-+++=-解得1m =--1m =-+综上可得，当3m =-或1m =-+时，函数函数()()g x f x mx =-（在区间[,2]m m +上有最小值–5.26.已知集合{1,2,3,,}n A n =⋅⋅⋅，集合(,)n m M 为集合n A 的m 元子集，且(,)n m M 中元素均为孤立元素．孤立元素的定义为：当x A ∈，1x A -∉且1x A +∉时，则称x 为集合A 中的孤立元素．（1）列出所有符合题意的集合(4,2)M ；（2）设()k m 为集合(8,)m M 的所有可能的集合个数，求()k m 的最大值，并说明理由；（3）在集合(2,)m m M 的所有可能集合中，存在元素在所有可能的集合中出现的次数最少，求出这样的元素并指出其出现次数，并说明理由．【答案】（1）{1,3},{1,4},{2,4}（2）当2m =时，()k m 的最大值为21，理由见解析.（3）2和21m -在所有可能的集合中只能出现一次，理由见解析.【分析】（1）列出4{1,2,3,4}A =的二元子集，观察符合要求的子集.（2）转化为元素不相邻问题，采用插空法求解，可推导公式为81(C )mm k m -+=；（3）取{1,3,5,7,,23,21}m m ⋅⋅⋅--，{1,3,5,7,,23,2}m m ⋅⋅⋅-，{1,4,6,8,,22,2}m m ⋅⋅⋅-，{2,4,6,8,,22,2}m m ⋅⋅⋅-说明除2和21m -外的其它元素在所有集合中至少出现两次，再证明2只能出现一次.【小问1详解】4{1,2,3,4}A =，所有的二元子集共有6个：{1,2},{1,3},{1,4},{2,3},{2,4},{3,4}，其中符合要求的有3个：{1,3},{1,4},{2,4}【小问2详解】将{1,2,3,,8}n A =⋅⋅⋅中元素从小到大排列一队，共有8个位置，由孤立元素的定义知(8,)m M 中任何两个元素都不能为相邻的整数，因此可转化为两个元素不相邻问题，可用插空法解决：第一步：将这m 个元素对应的位置取出来，剩下8m -个位置；第二步：将这m 个位置插入，保证它们互不相邻，只要从81m -+个空隙中选出m 个空放入，共有81C m m -+种方法.因此，81(C )mm k m -+=18C )8(1k ==，27C )21(2k ==，36C )20(3k ==，45C )5(4k ==，当5m ≥时，任取出的m 个数都会有相邻的整数，不是孤立元素，故(8,)m M 不存在.综上：当2m =时，()k m 的最大值为21.【小问3详解】2{1,2,3,,2}m A m =⋅⋅⋅取(2,)m m M 中两个集合：{1,3,5,7,,23,21}m m ⋅⋅⋅--，{1,3,5,7,,23,2}m m ⋅⋅⋅-在这两个集合中除21m -外的奇数出现2次；再取(2,)m m M 中两个集合：{1,4,6,8,,22,2}m m ⋅⋅⋅-，{2,4,6,8,,22,2}m m ⋅⋅⋅-，这两个集合中除2以外的偶数都出现了两次；若(2,)m m M 中有2存在，则这样集合只有{2,4,6,2}m ⋅⋅⋅，一个，因为大于2的第一个数必须为4，否则元素个数就不会凑够m 个，以此类推，后面元素分别为6,2m ⋅⋅⋅，8，，故含有2的集合只有{2,4,6,2}m ⋅⋅⋅，一个.所以2在所有可能的集合中只能出现一次.由位置对称性知：21m -在所有可能的集合中只能出现一次.综上：只有2和21m -在所有可能的集合中只能出现一次.【点睛】插空法，是用来解决某些元素不相邻的排列组合题，即不邻问题.在解决对于某几个元素要求不相邻的问题时，先将其它元素排好，再将指定的不相邻的元素插入已排好元素的间隙或两端位置，从而将问题解决的策略.用这种方法解题思路清晰、简便易懂.除了插空法，还有其他解排列问题的方法，如：插板法，用于处理分组问题；捆绑法，用于处理相邻问题.。
北师大版英语高一上学期期中试卷及解答参考
北师大版英语高一上学期期中复习试卷及解答参考一、听力第一节（本大题有5小题，每小题1.5分，共7.5分）1、What are the speakers mainly discussing?A. The weather forecast for the weekend.B. The importance of wearing a coat in cold weather.C. How to stay warm during the winter season.Answer: AExplanation: The conversation starts with one speaker mentioning the cold weather forecast for the weekend, indicating that the topic of discussion is the weather.2、How does the woman feel about the movie she just watched?A. She enjoyed it very much.B. She found it too predictable.C. She thought it was not as good as expected.Answer: BExplanation: The woman expresses her disappointment with the movie, stating that it was predictable, which implies that she did not enjoy it as much as she had hoped.3、You hear a conversation between two students discussing their weekend plans.A)They are planning to go hiking.B)They are planning to go shopping.C)They are planning to go to a concert.Answer: BExplanation: The conversation indicates that they will go shopping this weekend, as one student mentions, “I’m going to the mall this weekend.”4、You hear a news report about a new scientific discovery.A)The discovery is about a new type of plant.B)The discovery is about a new animal species.C)The discovery is about a new medical treatment.Answer: CExplanation: The news report specifically mentions that the discovery is related to a new medical treatment, which is being tested in clinical trials.5、You will hear a short conversation between two students, Tom and Lucy. Listen carefully and answer the following question.What is the price of the book Tom wants to buy?A)$10B)$20C)$30D)$40Answer: B)$20Explanation: In the conversation, Lucy asks Tom if he needs any help. Tom responds, “Yeah, I need to buy this book, and it costs$20.” Therefore, thecorrect answer is B)$20.二、听力第二节（本大题有15小题，每小题1.5分，共22.5分）1、What are the speakers mainly discussing?A) The advantages of online learning.B) The challenges of studying abroad.C) The importance of language skills in the job market.Answer: BExplanation: The conversation revolves around the challenges that the speakers are facing due to the high cost of studying abroad, indicating that they are discussing the challenges of studying abroad, not the advantages of online learning or the importance of language skills in the job market.2、Why does the man mention the need for a laptop?A) Because it’s a requirement for the new course.B) Because he needs it for his online classes.C) Because he’s planning to travel soon.Answer: BExplanation: In the conversation, the man specifically mentions the need for a laptop because he will be taking online classes, which suggests that the reason for the laptop is directly related to his studies, not a new course requirement or an upcoming trip.3、You will hear a conversation between two students discussing their weekendplans. Listen to the conversation and answer the question.What does the student suggest they do on Saturday afternoon?A. Go to the movies.B. Visit a museum.C. Go hiking.Answer: BExplanation: In the conversation, one student mentions, “I was thinking we could visit the art museum on Saturday afternoon,” which indicates that visiting a museum is the suggested activity.4、You will hear a short passage about the importance of exercise. Listen to the passage and answer the question.What is the main reason given for exercising regularly?A. To improve physical appearance.B. To reduce stress levels.C. To enhance academic performance.Answer: BExplanation: The passage emphasizes the benefits of regular exercise, particularly in reducing stress and improving overall well-being. The speaker mentions, “Exercise is a great way to reduce stress and improve mental health,” which aligns with option B.5.You are listening to a conversation between two students, Alex and Emily, discussing their weekend plans.Alex: Hey Emily, are you going anywhere this weekend?Emily: Yeah, I’m planning to go hiking in the mountains. How about you, Alex?A lex: That sounds fun! I think I’ll just stay in the city and visit the museum. Emily: Oh, that’s cool. What are you most excited about seeing there? Alex: The new art exhibit. I’ve heard it’s really impressive.Emily: I bet it is. I hope you enjoy your visit. Maybe we can meet up on Sunday afternoon?Question: What is Alex planning to do this weekend?A) Go hiking.B) Visit the mountains.C) Stay in the city and visit the museum.D) Go to the art exhibit.Answer: C) Stay in the city and visit the museum.Ex planation: In the conversation, Alex says, “I think I’ll just stay in the city and visit the museum,” indicating his plan for the weekend.6.You are listening to a news report about a local community event.News Anchor: And now, let’s go to our community r eporter, Sarah, who is at the City Park for the annual Earth Day Festival.Sarah: Good afternoon, everyone. The City Park is bustling with activity as thousands of people have gathered for the Earth Day Festival. The event kicked off with a parade of eco-friendly vehicles and ended with a tree planting ceremony. There are also several workshops and booths offering information onsustainable living and environmental conservation. I spoke with some of the attendees who shared their thoughts on the importance of Earth Day.Attendee 1: It’s so important to raise awareness about the en vironment.I brought my whole family today.Attendee 2: I think events like this really inspire people to make small changes in their daily lives.Sarah: Definitely. It’s great to see so many people coming together for a cause they care about.Question: What is the main purpose of the Earth Day Festival?A) To promote tourism in the city.B) To celebrate Earth Day and raise awareness about environmental issues.C) To provide entertainment for the local community.D) To promote local businesses.Answer: B) To celebrate Earth Day and raise awareness about environmental issues.Explanation: The news report clearly states that the Earth Day Festival is an event to celebrate Earth Day and raise awareness about environmental issues, as mentioned by the news anchor and the attendees.7.You’re listening to a conversation between two students discussing their weekend plans.A. What did the man do last weekend?B. What does the woman plan to do this weekend?C. Why does the woman need to study more?D. How does the woman feel about the upcoming test?Answer: BExplanation: The woman says, “I’m planning to go hiking this weekend,” which indicates her plans for the weekend. The other options are not supported by the dialogue.8.Listen to a short dialogue between a teacher and a student about a missing book.A. What happened to the book?B. Where did the student leave the book?C. Why doesn’t the student have the book?D. How does the teacher respond to the student’s explanation?Answer: CExplanation: Th e student explains, “I’m sorry, I lost it,” which is why they do not have the book. The other options are not mentioned in the dialogue, and the teacher’s response does not confirm the location of the book or the exact situation.9.You hear:W: I heard you’re planning a trip to Beijing this summer. Are you excited? M: Yeah, I can’t wait. I’ve always wanted to visit the Great Wall.Question: What is the man planning to do this summer?A) Visit the Great WallB) Go on a tripC) Travel to ChinaD) Learn ChineseAnswer: A) Visit the Great WallExplanation: The man explicitly states, “I’ve always wanted to visit the Great Wall,” which indicates his plan for the summer is to visit the Great Wall in Beijing.10.You hear:W: I’m thinking of studying abroad, but I’m not sure which country would be best for me.M: Well, if you’re interested in technology and innovation, you should consider going to the United States. They have some of the best universities in the world.Question: What advice does the man give to the woman?A) To study in ChinaB) To focus on technology and innovationC) To choose a country with the best universitiesD) To go to the United StatesAnswer: D) To go to the United StatesExplanation: The man suggests that the woman should consider going to the United States because it has “some of the best universities in the world,” which is related to her interest in technology and innovation.11.You will hear a conversation between two students discussing theirweekend plans. Listen and answer the question.What are the students planning to do on Saturday afternoon?A. Go shopping.B. Visit a museum.C. Have a picnic.Answer: BExplanation: In the conversation, the student mentions that they are planning to visit the local museum on Saturday afternoon. Therefore, the correct answer is B.12.You will hear a short dialogue between a teacher and a student about a homework assignment. Listen and answer the question.What subject is the homework assignment for?A. Math.B. Science.C. History.Answer: AExplanation: The student asks the teacher about the homework assignment, and the teacher responds by saying, “It’s for your math class.” This indicates that the homework is for mathematics, making the correct answer A.13.You will hear a short conversation between two students discussing their plans for the weekend. Listen carefully and choose the best answer to the question.Question: What are the students planning to do this weekend?A. Go to a movie.B. Visit a museum.C. Go hiking.D. Stay home and relax.Answer: B. Visit a museum.Explanation: In the conversation, the students discuss visiting the art museum as their plan for this weekend.14.You will hear a news report about a recent environmental project in the city. Listen carefully and answer the question.Question: What is the main goal of the environmental project?A. To reduce air pollution.B. To plant more trees in the city.C. To promote recycling.D. To create more public parks.Answer: B. To plant more trees in the city.Explanation: The news report mentions that the main goal of the project is to plant more trees in the city to improve the environment and reduce air pollution.15.A) 4B) 5C) 6D) 7Answer: B) 5Explanation: In this question, you need to listen for the specific number mentioned in the dialogue. The correct answer is 5, as it is clearly stated by one of the speakers. The other options are either not mentioned or are incorrect numbers related to the topic.三、阅读第一节（第1题7.5分，其余每题10分，总37.5分）第一题Reading PassageOn a cold November morning in 1922, British archaeologist Howard Carter stood in front of the sealed door of a small, unassuming tomb in the Valley of the Kings in Egypt. Little did he know that behind this door lay one of the greatest mysteries of ancient Egypt: the resting place of the legendary Pharaoh Tutankhamun. For over 3,000 years, Tutankhamun’s tomb had been hidden from the world, preserved in perfect condition by the sands of time.The discovery of Tutankhamun’s tomb was a turning point in the stu dy of ancient Egypt. Carter’s meticulous excavation revealed an astonishing wealth of artifacts, including furniture, jewelry, and the pharaoh’s own tomb. The most famous find, however, was the golden sarcophagus containing Tutankhamun’s mummy. This remarkable discovery provided invaluable insights into the life and times of ancient Egypt.1、Why did Howard Carter stand in front of the sealed door of a small tomb in the Valley of the Kings?A. He was searching for a new source of revenue.B. He had received a tip about the existence of a hidden tomb.C. He was conducting a routine archaeological survey.D. He was looking for a new challenge in his career.2、What did the discovery of Tutankhamun’s tomb mean for the study of ancient Egypt?A. It provided a new source of income for Egyptologists.B. It led to the development of new archaeological techniques.C. It allowed historians to better understand the life and times of ancient Egyptians.D. It sparked a renewed interest in the study of ancient Egypt.3、What was the most famous find in Tutankhamun’s tomb?A. The pharaoh’s chariot.B. The golden sarcophagus.C. The tomb’s entrance.D. The pharaoh’s mummy.4、What did the discovery of Tutankhamun’s tomb provide to historians?A. A new source of revenue.B. A better understanding of ancient Egyptian culture.C. A new source of inspiration for writers.D. A new perspective on the ancient Egyptian civilization.Answers:1、B2、C3、B4、B第二题Read the following passage and answer the questions that follow.In the small town of Willow Creek, there was a tradition that every autumn, the townspeople would gather to celebrate the Harvest Festival. This festival was not just a time for joy and festivities; it was also a way to honor the hard work of the farmers and to give thanks for the bountiful harvest.The festival began with a parade through the town, led by the mayor and the local school children, dressed in colorful costumes. The parade was followed by a variety of activities, including a pumpkin carving contest, a pie eating contest, and a traditional folk music performance.One of the highlights of the festival was the Harvest Market, where local farmers sold their fresh produce, homemade jams, and baked goods. The market was a bustling place, with people of all ages coming to buy and sell.1、What is the main purpose of the Harvest Festival in Willow Creek?A. To celebrate the fall season.B. To honor the farmers and give thanks for the harvest.C. To promote tourism in the town.D. To educate children about history.2、Who leads the parade during the Harvest Festival?A. The town’s firefighters.B. The mayor and the local school children.C. The town’s police officers.D. The local chefs.3、What is one of the activities that takes place during the Harvest Festival?A. A basketball tournament.B. A fishing competition.C. A pumpkin carving contest.D. A talent show.4、What is the Harvest Market known for?A. It is a place to buy and sell antiques.B. It is a place to learn about local wildlife.C. It is a place to purchase fresh produce and homemade goods.D. It is a place to watch traditional dances.Answers:1、B. To honor the farmers and give thanks for the harvest.2、B. The mayor and the local school children.3、C. A pumpkin carving contest.4、C. It is a place to purchase fresh produce and homemade goods.Section III. ReadingPassage:Title: The Power of ReadingReading has been a source of knowledge, inspiration, and entertainment for countless generations. It is a bridge that connects us to the past, a window that opens up new worlds, and a mirror that reflects our own thoughts and emotions. In today’s fast-paced world, where technology dominates every aspect of our lives, the importance of reading cannot be overstated.For students, especially those in high school, reading goes beyond the pages of textbooks. It fosters critical thinking, enhances vocabulary, and improves writing skills. The act of delving into a good book can be a transformative experience, taking readers on a journey of self-discovery and personal growth.Moreover, reading is a form of escape. It allows us to step away from the stresses and pressures of daily life and immerse ourselves in another reality. Whether it’s the thrill of a mystery novel, the romance of a love story, or the wisdom of a philosophical text, books offer a diverse range of emotions and perspectives.The benefits of reading extend beyond the individual. It is a powerful tool for building empathy and understanding across cultures and societies. By reading about different people, places, and experiences, we broaden our horizons and become more tolerant and accepting of diversity.In conclusion, the power of reading is undeniable. It is a fundamental partof our education, a source of personal growth, and a means of connecting with others. Let us embrace the joy of reading and continue to nourish our minds and souls with the wealth of knowledge and inspiration it offers.Third Question: Reading Comprehension1.What is the main idea of the passage?Answer: The power and importance of reading.2.How does reading contribute to students’ development?Answer: Reading fosters critical thinking, enhances vocabulary, and improves writing skills in students.3.What does the author mean when they say “reading is a form of escape”? Answer: The author means that reading allows us to step away from daily stresses and immerse ourselves in another reality, providing a sense of escape.4.What are some of the broader benefits of reading mentioned in the passage? Answer: Some of the broader benefits of reading mentioned in the passage include building empathy and understanding across cultures and societies, broadening horizons, and becoming more tolerant and accepting of diversity.第四题Reading Comprehension (Section A)Read the following passage and answer the questions that follow. Choose the best answer from the options given.Passage:The Great Wall of China, one of the greatest wonders of the world, was listed as a World Heritage by UNESCO in 1987. Just like a gigantic dragon, the GreatWall winds up and down across deserts, grasslands, mountains, and plateaus, stretching approximately 21,196 kilometers from east to west of China. With a history of more than 2000 years, some parts of the Great Wall are now in ruins or have disappeared. However, it is still one of the most appealing attractions all around the world owing to its historical significance.Originally built in different periods by various states to defend their borders against invasions, the Great Wall was later connected and fortified during the Ming Dynasty (1368-1644) to form a comprehensive defense system. The construction of the wall involved millions of workers and took over 2,000 years to complete. It is not only a symbol of the Chinese nation but also a representation of the wisdom and tenacity of ancient Chinese people. Today, the Great Wall serves as a reminder of the past and a source of inspiration for future generations, attracting millions of visitors each year who come to marvel at its grandeur and learn about its rich history.1.What is the total length of the Great Wall of China as mentioned in the passage?•A) 21,196 meters•B) 21,196 kilometers•C) 21,196 feet•D) 21,196 milesAnswer: B) 21,196 kilometers2.In which year did the Great Wall of China get listed as a World Heritage by UNESCO?•A) 1985•B) 1987•C) 1990•D) 1995Answer: B) 19873.During which dynasty was the Great Wall significantly connected and fortified to forma comprehensive defense system?•A) Tang Dynasty•B) Song Dynasty•C) Yuan Dynasty•D) Ming DynastyAnswer: D) Ming Dynasty4.Which of the following statements is NOT true according to the passage?•A) The Great Wall was built to protect the borders from invaders.•B) Some sections of the Great Wall are in ruins or have vanished.•C) The construction of the Great Wall spanned over 2,000 years.•D) The Great Wall no longer attracts tourists due to its deteriorated state.Answer: D) The Great Wall no longer attracts tourists due to itsdeteriorated state.四、阅读第二节（12.5分）Reading SectionPassage:The Great Wall of China is one of the most remarkable architectural wondersin the world. Stretching over 13,000 miles, it was built over several centuries by various dynasties to protect the Chinese empire from invasions. The Wall is a symbol of China’s strength and resilience, and it has become an iconic representation of the country.Today, the Great Wall is a popular tourist destination, attracting millions of visitors each year. It’s not just the length and height of the Wall that fascinates tourists, but also the intricate carvings and the diverse landscapes it traverses. The Wall winds through mountains, deserts, and grasslands, offering a unique blend of natural beauty and historical significance.One of the most famous sections of the Wall is the Mutianyu section, located in the western part of Beijing. This section is known for its well-preserved structures and stunning views. It was built in the 6th century and has been restored several times over the years.Question:What is the primary reason for the construction of the Great Wall of China?A. To provide a scenic view for tourists.B. To attract foreign investors to China.C. To protect the Chinese empire from invasions.D. To promote tourism in China.Answer: C. To protect the Chinese empire from invasions.五、语言运用第一节 _ 完形填空（15分）Title: A Journey of Self-DiscoveryAs a freshman in high school, I found myself standing at a crossroads, unsure of the path ahead. The first semester was filled with excitement and anxiety in equal measure. Every day seemed like a new adventure, yet the pressure to excel academically and socially weighed heavily on my shoulders. It was during this time that I embarked on a journey of 1 discovery, a journey that taught me invaluable lessons about myself.One of the most significant challenges I faced was 2 with my peers. Coming from a different background, I felt like an outsider at times. I struggled to find common ground and feared being judged for my differences. However, through participating in various extracurricular activities and opening up to my classmates, I gradually learned to 3 the beauty in diversity and embrace my uniqueness.Another turning point came when I was assigned to a difficult math course. Math had always been a 4 subject for me, and the thought of struggling through it filled me with dread. But instead of giving up, I decided to face my fears head-on. With the help of my teacher and countless hours of 5, I managed to not only pass the course but also develop a deeper appreciation for the subject. This experience taught me resilience and the importance of perseverance.Moreover, the semester brought about a renewed sense of 6towards learning. As I delved deeper into my studies, I discovered a passion for literature thatI never knew existed. Reading books from diverse cultures and time periods broadened my horizons and sparked a curiosity within me. I realized that education was not just about acquiring knowledge but also about fostering a lifelong love for learning.Through these experiences, I came to understand that life is full of 7 and uncertainties. But it is precisely these challenges that shape us into the individuals we become. I learned to embrace the unknown, to have courage in the face of adversity, and to cherish every moment of growth and self-discovery.Cloze Test Questions (with Options and Answer Key)1.A. personalB. culturalC. scientificD. socialAnswer: A2.A. fitting inB. standing outC. keeping upD. breaking awayAnswer: A3.A. recognizeB. createC. ignoreD. reject Answer: A4.A. fascinatingB. challengingC. relaxingD. enjoyable Answer: B5.A. practiceB. leisureC. debateD. research Answer: A6.A. enthusiasmB. skepticismC. boredomD. apathy Answer: A7.A. certaintiesB. opportunitiesC. obstaclesD. routines Answer: CNote: The passage and questions are designed to test students’ understanding of context, vocabulary, and logical reasoning while engaging in a meaningful topic of self-discovery and growth.六、语言运用第二节 _ 语法填空（15分）For the Grammar Fill-in-the-Blanks SectionDirections: Read the following passage. Fill in each blank with a proper word given below. There are ten blanks followed by four choices marked A, B, C and D. Choose the most appropriate answer for each blank.Text:The Arctic is one of the coldest places on Earth. However, it’s also home to a variety of wildlife that has adapted to live in such harsh conditions. Polar bears, ____(1)____ example, have thick fur and a layer of fat to keep them warm. They spend much of their time hunting seals on the ice. In summer, when the ice melts, polar bears often go onto land and may not eat ____(2)____ several months until the sea freezes over again. Another animal that lives here is the arctic fox, which changes its coat color from brown in summer to white in winter ____(3)____ blend into the snow. Sadly, global warming is causing the Arctic ice ____(4)____ (melt), putting these animals’ habitats at risk.The people who live in the Arctic also face many challenges. The Inuit, ____(5)____ traditionally hunt and fish for a living, must now adapt to changing conditions. For centuries, they ____(6)____ (depend) on the predictability ofthe seasons and the migration patterns of animals. Now, however, they find that the ice is too thin for travel ____(7)____ foot or by dog sled, disrupting their way of life. Despite these difficulties, the Inuit remain resilient and continue to preserve their culture and traditions.Scientists warn that if we do not take action soon, the Arctic might change forever, affecting not just the local inhabitants but the entire planet. It’s hoped that by working together, we can protect this unique environment and ensure it ____(8)____ (survive) for future generations.(1)A. for B. with C. as D. by(2)A. to B. for C. at D. in(3)A. to B. so as C. in order to D. for(4)A. melt B. melting C. melted D. to melt(5)A. whom B. whose C. who D. that(6)A. depend B. depended C. have depended D. had depended(7)A. in B. on C. by D. with(8)A. survive B. survived C. will survive D. has survivedNow let’s provide the answers to this exercise.The correct answers for the grammarfill-in-the-blanks exercise are as follows:1.A. for2.B. for3.A. to4.D. to melt5.C. who6.C. have depended7.B. on8.C. will survive七、写作第一节 _ 应用文写作（15分）Section VII. WritingPart A. Application WritingTask: Write a letter to a friend, expressing your excitement about a recent event you both attended. Include details about the event, your personal feelings, and a suggestion for a future activity.Example:Dear [Friend’s Na me],I hope this letter finds you well! I just wanted to share with you my excitement about the concert we both attended last weekend. It was absolutely fantastic!The venue was beautiful, with a stunning view of the city skyline. The artist, [Artist’s Name], performed live, and the music was simply amazing. The crowd was lively and energetic, and I couldn’t help but feel a sense of joy and unity throughout the evening.I was particularly impressed by [specific part of the concert, e.g., the artist’s performance of a particular song]. The passion and energy he brought to the stage were truly inspiring. It was a night to remember, and I’m so gladwe were able to experience it together.Now, I have a suggestion for our next outing. How about we plan a picnic at the park? We could bring some delicious food, enjoy the beautiful weather, and perhaps even bring along a portable speaker to play some of our favorite music. It would be a great way to relax and spend quality time together.Looking forward to hearing from you soon!Best regards,[Your Name]Analysis:•The letter is well-structured, starting with a greeting and a clear introduction to the topic.•The writer expresses excitement and shares personal feelings about the concert, making the letter more engaging.•Specific details about the concert are provided, such as the venue, the artist, and the crowd, which helps to paint a picture for the reader.• A suggestion for a fu ture activity is offered, showing the writer’s thoughtfulness and desire to continue enjoying time together.•The conclusion is friendly and inviting, encouraging the friend to respond.八、写作第二节 _ 读后续写（25分）Title: “A Surprise Encounter in the Library”。
陕西省西安市西安高新第一中学2024-2025学年高一上学期期中考试英语试题
陕西省西安市西安高新第一中学2024-2025学年高一上学期期中考试英语试题一、阅读理解As the excitement of the Summer Olympics attracts audiences worldwide, it presents a golden opportunity to inspire our younger generation not just in sports but in the fields of Science, Technology, Engineering, and Mathematics (STEM). Here, we explore a series of innovative STEM challenges designed for both lower elementary and upper elementary students, aimed at sparking their curiosity and fostering a love for learning.Surfing STEM ChallengeThe Surfing STEM Challenge invites students to explore the fascinating world of magnetic forces (磁力) by keeping a surfer suspended in mid-air. Using a magnetic stick. students guide the surfer’s track, mirroring the fluid movements of surfing waves.Basketball STEM challengeThe Basketball STEM challenge attracts students by tasking them with inventing a basketball game tailored for one-handed play, making use of materials at hand. The activity highlights the importance of creativity, engineering capability, and collaborative effort, all within the vibrant atmosphere of the Summer Olympics.Skateboarding STEM ActivityThe Skateboarding STEM Activity challenges students to craft a small skateboard, along with constructing a skatepark to demonstrate their fingerboarding skills. It’s an engaging way to connect with the dynamic world of skateboarding in the Summer Olympics, encouraging teamwork, innovation, and hands-on learning.Golf STEM ChallengeThe Golf STEM Challenge tasks students with the exciting project of designing and building a mini golf course, along with creating their own golf club. This innovative approach to learning not only brings the spirit of the Summer Olympics into the classroom but also fosters teamwork, problem-solving, and a deeper understanding of the scientific concepts behind the game of golf.Click Here to find out more information.1．What is the main goal of the STEM challenges described in the passage?A．To train students for future careers in STEM.B．To encourage students’ curiosity and love for learning.C．To help students understand complex scientific theories.D．To promote the importance of sports and physical activity.2．Which two STEM challenges encourage students to build something physical?A．Surfing STEM Challenge and Golf STEM Challenge.B．Surfing STEM Challenge and Basketball STEM Challenge.C．Skateboarding STEM Activity and Golf STEM Challenge.D．Basketball STEM Challenge and Skateboarding STEM Activity.3．Where is this passage most likely to be found?A．A sports magazine.B．A travel blog.C．A scientific report.D．An educational website.In medical school and throughout his career as a neonatologist, William Cashore was often asked to read and correct others’ work. Little did they know that he was a spelling champion, witha trophy(奖品) at home to prove it. Cashore won the Scripps National Spelling Bee in 1954 at age14. Now 84, he’s the oldest living champi on of the contest, which dates back to 1925. As contestants from this year’s competition headed home, he reflected on his experience and the effect it had on him.Cashore credits his parents for helping him prepare for his trip to Washington, D．C．, for the spelling bee. His mother was an elementary school teacher and his father was a lab technician with a talent for taking words apart and putting them back together.When the field narrowed to two competitors, the other boy misspelled “uncinate”, which means bent like a hook. Cashore spelled it correctly, then clinched the title with the word “transept”, an architectural term for the transverse part of a cross- shaped church.Cashore, who was given $500 and an encyclopedia set, enjoyed a brief turn as a celebrity. He didn’t br ag about his accomplishment after returning to Pennsylvania, but the experience quietly shaped him in multiple ways. “It gave me much more self-confidence and also gave me asense that it’s very important to try to get things as correct as possible,” he said. “I’ve always been that way, and I still feel that way. If people are careless about spelling and writing, you wonder if they’ re careless about their thinking.”“Preparing for a spelling bee today requires more concentration and technique than it did decades ago,” Cashore said. “The vocabulary of the words are far, far more technical. The English language, in the meantime, has imported a great many words from foreign languages which were not part of the English language when I was in eighth grade,” he added.4．What can we learn about Cashore from the first two paragraphs?A．He is the best ever spelling champion.B．He prepared alone for the spelling bee.C．He used his spelling ability to help others.D．He preferred to show off his spelling talent. 5．What does the word “clinched” underlined in paragraph 3 refer to?A．Won.B．Declined.C．Removed.D．Gave.6．How did the spelling bee competition affect Cashore?A．It helped him start his career.B．It strengthened his confidence.C．It limited his way of thinking.D．It brought him lasting popularity.7．What should we do to prepare for today’s spelling bee according to Cashore?A．Learn multiple languages.B．Go abroad for language study.C．Be a remarkable technician.D．Try to improve relevant skills.What are the most important factors that distinguish NBA legends from the rest? When scoping out talent, scouts (星探) look for two main things: outstanding previous performance and “unexploited potential”. There is an assumption that physical stature and athleticism (e.g., arm span, flexibility, swiftness, etc) are markers of unexploited potential, and will therefore continue predicting success in the NBA．But is this assumption correct?A new study sheds some new light on the issue. Unsurprisingly, measures of physical makeup and athleticism clearly influence selection into the NBA draft. But they did not distinguish productive NBA players from unproductive players after taking into account prior performance. In fact, the only variables that predicted NBA success were youth, college performance, and college quality. Of course, the researchers didn’t analyze every possible measure of skill, such as passing, defense and shooting. Also, many external factors could also have playeda role, such as coaching and team social dynamics. But none of them may be as important as the biggest elephant in the room. Let’s turn to Michael Jordan, the basketball legend.A lot has been made of Jordan’s “cut” from his school basketball team in Grade Two in high school. In reality, Jordan wasn’t cut, but just placed elsewhere: on the junior team. The school team really needed size, and at the tryouts Jordan was 5feet 10 inches. He didn’t have his big growth shoot until the following year, when he was placed on the school team. What’s more, at the tryouts, Jordan didn’t really stand out on any of the characteristics the coaches were looking out for.Well, except one: “If Jordan distinguished himself at all during the tryout, it was through his extreme effort.”What truly saddens me is that professionals judge the “natural” performer who demonstrates early evidence of ability to be more talented, and more hirable than the “striver”, who shows early evidence of high motivation and perseverance. Maybe it’s time to completely rethink this whole belief of unexploited potential. Perhaps we should think about early ability as merely potential that was realized earlier than others. Nothing more, nothing less. 8．According to the passage, what do team managers generally value in players?A．Athletic skills and training experiences.B．Motivation and team social dynamics.C．Physical makeup and athletic competence.D．Early performance and professional spirit.9．What does “the biggest elephant in the room” in paragraph 2 refer to?A．The most efficient training program.B．The most essential quality as a player.C．The greatest NBA legend Michael Jordan.D．The most obvious factor but one that is ignored.10．Why did Jordan fail to make the school team as a Grade Two student?A．He lacked talent for the sport.B．He was too short to win the coach’s favor.C．He was needed more by the school’s junior team.D．He didn’t make enough efforts to impress the coach.11．What message does the writer want to convey in the passage?A．Super efforts can make up for the lack of athleticism.B．The concept of “unexploited potential” doesn’t make sense.C．Those who show early evidence of ability are not truly talented.D．Passion and dedication are as worth considering as athletic characteristics.A new study published in the journal Science Advances finds that AI enhances creativity by boosting the novelty of story ideas as well as the “usefulness” of stories — their ability to engage the target audience and potential for publication. It finds that AI “professionalizes” stories, making them more enjoyable, more likely to have plot twists, better written and less boring.In a study, 300 participants were tasked with writing a short, eight-sentence “micro story” for a target audience of young adults. They were divided into three groups: one group was allowed no Ad help, a second group could use ChatGPT to provide a single three-sentence starting idea, and writers in the third group could choose from up to five AI-generated ideas for their inspiration.They then asked 600 people to judge how good the stories were, assessing them for novelty and “usefulness”. They found that writers with the most access to AI experienced the greatest gains to their creativity, their stories scoring 8.1% higher for novelty and 9% higher for novelty compared with stories written without AI. Writers who used up to five AI-generated ideas also scored higher for emotional characteristics, producing stories that were better written, more enjoyable, less boring and funnier.The researchers evaluated the writers’ creativity using a Divergent Association Task (DA T) and found that more creative writers — those with the highest DAT scores — benefited least from generative AI ideas.Less creative writers conversely saw a greater increase in creativity: access to five AI ideas improved novelty by 10.7% and usefulness by 11.5% compared with those who used no AI ideas. Their stories were judged to be up. to 26.6% better written, up to 22.6% more enjoyable and up to 15.2% less boring.Anil Doshi, Assistant Professor at the UCL School said, “While these results point to an increase in individual creativity, there is risk of losing collective novelty. If the publishing industry were to use more generative AI-inspired stories, our findings suggest that the stories would become less unique and more similar to each other. That is not encouraged in writing.”12．What were some participants assigned to do in the study?A．Help targeted young writers.B．Improve some less creative works.C．Create a mini story for young men.D．Use AI to judge the quality of stories. 13．Who benefited most from the AI-assisted writing in the study?A．Those using AI most.B．Those of most creativity.C．Those with the help of ChatGPT.D．Those writing the most.14．What might be Anil Doshi’s attitude toward the AI-assisted writing?A．Optimistic.B．Objective.C．Doubtful.D．Critical. 15．What can be the best title for the passage?A．AI Is Being Used to Create Novel Stories NowadaysB．AI Is Catching on But Threatening the Future of WritersC．AI Can Boost Creativity in Writing But Has Its DisadvantagesD．AI Helps Improve Writing Skills But Is Limited to Young WritersPeru is one of the most popular countries in South America, mostly thanks to its Wonder of the World, Machu Picchu. There is much more to see and do in Peru if you’re willing to get out there and explore. 16 You don’t need a lot of money to visit here. Here are a few ways to cut down your costs in Peru.17 They are the cheapest accommodation you can find and offer personalized service to guests. try to stay in these as often as possible.Take public transportation. Embrace public transportation to get around — it’s super affordable so skip the taxis. You’ll save a fortune.Travel off-season. For a low-cost trip, the best times to visit Peru are the months of April and May or September and October. 18Take the cheap buses. They cost around 2-10 PEN for a ride. 19 But there is always a door person whom you can ask if the bus is going to your location. There are not always marked bus stops, so look for gathering crowds.Go on a free walking tour. 20 Free Walking Tour Peru has tours that can guide you around both Lima and Cusco. Just remember to tip your guide at the end!Bring a water bottle. The tap water here isn’t safe to drink so bring a reusable water bottlewith a filter to save money and reduce your plastic use.A．Stay at family-run hotels.B．After all, it’s safe to travel alone in Peru.C．Prices are usually cheaper during these periods.D．Best of all, traveling around Peru is inexpensive.E．Peru has a ton of fancy hotels to stay throughout thé country.F．They are a bit confusing, as they don’t necessarily have a schedule.G．This is a great way to learn the history behind the places you are seeing.二、完形填空I could feel the excitement rising in me as I held the beautifully wrapped present in my hands. Unable to 21 my curiosity any longer, I 22 it open. It was a brand-new laptop. I had scored straight A for my examination and this was my 23 .Soon I became glued to my new-found 24 friend. I began to spend increasing time on it and was 25 to the glowing screen before me. During weekends, I could 26 spend a few hours on my laptop. It came to a point where I chose to 27 myself in my room, refusing to come out and engage in family 28 . I preferred to stay connected to the people I befriended in cyberspace (网络空间) although I had never met them. Gradually, my sleep was 29 to only scattered naps, and my grades at school also began 30 .It was about this time that my parents started 31 their disapproval. I was asked to reduce my usage of the internet. Yet, however hard I tried, I was unable to pull myself away and concentrate on my lessons. I was 32 for just another click and another look at my message board. Then one day, my father 33 my laptop. I complained but in vain.Over time, I realized that Dad and Mom had always wanted the best for me. That was when I decided to take the big step — to 34 my internet addiction. Quitting was harder than I expected.But with dogged determination, I knew that I would 35 a new leaf. 21．A．keep B．show C．contain D．improve 22．A．tore B．cut C．broke D．sliced23．A．order B．reward C．inspiration D．strategy 24．A．electronic B．physical C．virtual D．authentic 25．A．committed B．exposed C．devoted D．addicted 26．A．hardly B．easily C．officially D．regretfully 27．A．trap B．adapt C．lock D．shelter 28．A．conflicts B．discussions C．investments D．experiments 29．A．instructed B．adjusted C．limited D．reduced 30．A．rising B．advancing C．declining D．shifting 31．A．voicing B．sharing C．challenging D．exchanging 32．A．responsible B．grateful C．ready D．desperate 33．A．gave away B．took away C．threw away D．swept away 34．A．ignore B．admit C．monitor D．overcome 35．A．turn over B．pick up C．rely on D．attend to三、语法填空阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。
2025届陕西省西安高新第一中学化学高一第一学期期中监测模拟试题含解析
2025届陕西省西安高新第一中学化学高一第一学期期中监测模拟试题注意事项：1．答卷前，考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。
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一、选择题(共包括22个小题。
每小题均只有一个符合题意的选项）1、下列物质在水溶液中的电离方程式书写正确的是A．KClO3＝K++3O2—+Cl5+B．NaHSO4＝ Na++HSO4—C．H2SO4＝H++ SO42—D．Al2(SO4)3＝ 2Al3++3SO42—2、在下列状态下，属于能够导电的电解质是（）A．氯化钠晶体B．液态氯化氢C．硝酸钾溶液D．熔融氢氧化钠3、引爆H2和Cl2组成的混合气体，相同状况下，测得反应前后体积不变。
下列的分析正确的是A．混合气中H2和Cl2的体积相等B．混合气中H2体积大于Cl2体积C．混合气中H2的体积小于Cl2的体积D．以上情况均有可能4、下列实验操作不能达到实验目的是A．A B．B C．C D．D5、下列属于电解质的是A．氯化钠B．酒精C．氯气D．铁6、实验室加热高锰酸钾制取氧气，用排水法收集满氧气后，下一步应采取的操作及原因是A．先熄灭酒精灯，以节约燃烧B．先把导管移出水面，再熄灭酒精灯以免水倒吸引起试管炸裂C．先撤酒精灯，再撤导气管，防止水倒吸D．检验收集的气体是否纯净7、研究表明：多种海产品如虾、蟹、牡蛎等，体内含有+5价的砷（As）元素，但它对人体是无毒的，砒霜的成分是As2O3，属剧毒物质，专家忠告：吃饭时不要同时大量食用海鲜和青菜，否则容易中毒，并给出了一个公式：大量海鲜+大量维生素C = 砒霜。
2021-2022学年陕西省西安市雁塔区高新一中高一(上)期中数学试卷(附答案详解)
2021-2022学年陕西省西安市雁塔区高新一中高一（上）期中数学试卷一、单选题（本大题共10小题，共40.0分）1. 已知集合A ={x|0<x <6}，集合B ={x|3≤x ≤8}，则A ∪B =( )A. {x|0<x ≤3}B. {x|6<x ≤8}C. {y|0<y ≤8}D. 以上答寮都不对2. 下列函数中，在定义域上为奇函数，并且在区间(0,+∞)上是减函数的是( )A. f(x)=x 3B. f(x)=1xC. f(x)=(12)xD. f(x)=3−x3. 若函数y =x 2−2x −4的定义域为[0,m]，值域为[−5,−4]，则m 取值范围是( )A. [0,1]B. (1,2]C. [1,2]D. [0,2]4. 下列区间中，包含函数f(x)=log 12x +1x 的零点的是( )A. (3,4)B. (2,3)C. (1,2)D. (0,1)5. 已知f(x)=(m 2−2m −7) xm−23是幂函数，且在(0,+∞)上单调递增，则满足f(a −1)>1的实数a 的范围为( )A. (−∞,0)B. (2,+∞)C. (0,2)D. (−∞,0)∪(2,+∞)6. 函数f(x)=log 2(3−2x −x 2)的单调递增区间是( )A. (−∞,−1)B. (−3,−1)C. (−1,+∞)D. (−1,1)7. 设函数f(x)=e |x|，则使得f(2x −1)<f(x)成立的x 的取值范围是( )A. (13,1) B. (−∞,13)∪(1,+∞) C. (−13,13)D. (−∞,13)∪(13,+∞)8. 设a =(√2)3，b =(12)−3，c =log 23，则a ，b ，c 的大小关系为( )A. a <b <cB. b <a <cC. b <c <aD. c <a <b9. 若函数f(x)={(a −1)x −2a,x <2log a x,x ≥2在R 上单调递减，则实数a 的取值范围是( )A. (0,1)B. (0,√22]C. [√22,1)D. (1,+∞)10. 若函数f(x)=log a x −x +a(a >0且a ≠1)有两个零点，则实数a 的取值范围是二、单空题（本大题共4小题，共16.0分）11.已知集合A={(x,y)|2x−y=4}，B={(x,y)|x+y=5}，则A∩B中元素个数为______．12.计算求值：8−13+lg5+lg2+e ln2+1lg0.01=______．413.设f(x)=ax5+bx3+cx+7(其中a、b、c为常数，x∈R)，若f(−2021)=−17，则f(2021)=______．14.已知函数f(x)是定义在R上的周期为2的奇函数，且当0≤x<1时，f(x)=2x+a，f(1)=0，则f(−3)+f(14−log27)=______．三、解答题（本大题共7小题，共64.0分）15.已知集合A={x|x2+2x−3<0}，B={y|y<−1或y>3}，C={x|−2<x<m+1}，其中m>−3．(1)求A∩B；(2)若(A∪B)∩C=C，求实数m的取值范围．16.已知二次函数f(x)=ax2+bx+c的图象过点(0,3)，且不等式ax2+bx+c<0的解集为{x|1<x<3}．(1)求f(x)的解析式；(2)若g(x)=f(x)−(2t−4)x在区间[−1,2]上有最小值2，求实数t的值．17.已知定义域为R的函数f(x)=1−2x．1+2x(1)判断f(x)的奇偶性；(2)试判断函数f(x)=1−2x在R上的单调性，并用函数单调性的定义证明；1+2x(3)若对于任意t∈R，不等式f(t2−2t)+f(t2−k)<0恒成立，求实数k的取值范围．−4．18.已知函数f(x)=x+3x(1)求关于x的不等式xf(x)<(m−3)(x−1)(m∈R)的解集；(2)若关于x的方程f(a x)−k⋅a−x−k=0(a>0,a≠1)有两个不相等的实数根，求实数k的取值范围．19.已知函数f(x)=log2x，g(x)=ax2−4x+1．(Ⅰ)若函数y=f(g(x))的值域为R，求实数a的取值范围,2]，都存在t∈[−1,1]使得不(Ⅱ)函数ℎ(x)=f2(x)−f(x2)，若对于任意的x∈[12等式ℎ(x)>k⋅2t−2成立，求实数k的取值范围20.已知函数f(x)=3|x|+2+x+93|x|+1的最大值为M，最小值为m，则求M+m的值．21.已知f(x)为偶函数，g(x)为奇函数，且满足f(x)−g(x)=21−x．(1)求f(x)，g(x)；(2)若方程mf(x)=[g(x)]2+2m+9有解，求实数m的取值范围；(3)若ℎ(x)=|12[f(x)+g(x)]−1|，且方程[ℎ(x)]2−(2k+12)ℎ(x)+k=0有三个解，求实数k的取值范围．答案和解析1.【答案】C【解析】解：∵集合A ={x|0<x <6}，集合B ={x|3≤x ≤8}， ∴A ∪B ={y|0<y ≤8}． 故选：C ．利用并集定义直接求解．本题考查集合的运算，考查并集定义、不等式性质等基础知识，考查运算求解能力，是基础题．2.【答案】B【解析】解：函数f(x)=x 3的定义域为R ，为奇函数，在区间(0,+∞)上是增函数，故选项A 错误；函数f(x)=1x 为奇函数，且在区间(0,+∞)上是减函数，故选项B 正确；函数f(x)=(12)x ，因为f(−x)=(12)−x ≠−f(x)，所以函数为非奇非偶函数，故选项C 错误；函数f(x)=3−x ，因为f(−x)≠−f(x)，所以函数为非奇非偶函数，故选项D 错误． 故选：B ．利用奇偶性的定义判断是否为奇函数，然后由基本初等函数的单调性判断函数的单调性，即可得到答案．本题考查了函数奇偶性与单调性的判断，解题的关键是掌握奇偶性的定义以及基本初等函数的性质，考查了逻辑推理能力，属于基础题．3.【答案】C【解析】解：函数y =x 2−2x −4=(x −1)2−5 当x =1时y =−5 当x =0或2时，函数y =−4 则函数的定义域为[0,2]解得：0≤m≤2所以求得：m的范围为[1,2]或[0,1]进一步结合四个选项m的范围为[1,2]故选：C首先把函数转化为：函数y=x2−2x−4=(x−1)2−5，进一步当x=1时y=−5，当x=0或2是函数y=−4，则函数的定义域为[0,2]，最后确定参数的范围．本题考查的知识要点：二次函数顶点式与一般式的互化，根据值域确定定义域，及求参数的范围．4.【答案】C【解析】解：因为函数f(x)在(0,+∞)上单调递减，f(1)=1>0，f(2)=−1+12<0，所以f(x)的零点在(1,2)内，故选：C．利用零点存在性定理即可求解．本题考查了零点存在性定理，属于基础题．5.【答案】D【解析】解：∵f(x)=(m2−2m−7) x m−23是幂函数，且在(0,+∞)上单调递增，∴{m2−2m−7=1 m−23>0，解得m=4，∴f(x)=x23=√x23，定义域为R，且是偶函数，∵f(x)在(0,+∞)上单调递增，∴f(x)在(−∞,0)上单调递减，又∵f(−1)=f(1)=1，f(0)=0，∴由f(a−1)>1可得：a−1<−1或a−1>1，解得a<0或a>2，∴实数a的范围为(−∞,0)∪(2,+∞)，奇偶性，根据函数的单调性和奇偶性求出满足f(a−1)>1的实数a的范围即可．本题主要考查了幂函数的定义和性质，是基础题．6.【答案】B【解析】解：由3−2x−x2>0，得x2+2x−3<0，解得−3<x<1．令t=−x2−2x+3，其对称轴方程为x=−1，函数t=−x2−2x+3在(−3,−1)上为增函数，且外层函数y=log2t是定义域内的增函数，∴函数f(x)=log2(3−2x−x2)的单调递增区间是(−3,−1)．故选：B．由对数函数的真数大于0求解函数的定义域，再求出内层函数的增区间，即可得到原函数的增区间．本题主要考查了复合函数的单调性以及单调区间的求法．对应复合函数的单调性，一要注意先确定函数的定义域，二要利用复合函数与内层函数和外层函数单调性之间的关系进行判断，判断的依据是“同增异减”，是中档题．7.【答案】A【解析】解：f(x)是偶函数，且在[0,+∞)上是增函数，∴由f(2x−1)<f(x)得，f(|2x−1|)<f(|x|)，∴|2x−1|<|x|，<x<1，∴(2x−1)2<x2，解得13,1)．∴x的取值范围是(13故选：A．根据f(x)的解析式即可判断出f(x)的奇偶性与单调性，将不等式f(2x−1)<f(x)合理转化，即可解出x的范围．本题主要考查函数奇偶性与单调性的综合，考查指数函数的性质，属于基础题．8.【答案】D【解析】解：∵a =(√2)3=2√2∈(2,3)，b =(12)−3=23=8，c =log 23<log 24=2， ∴c <a <b ， 故选：D ．利用有理指数幂和对数的运算性质分别求得a ，b ，c 的范围(或值)得答案． 本题考查对数值的大小比较，考查有理指数幂与对数的运算性质，是基础题．9.【答案】C【解析】解：根据题意，函数f(x)={(a −1)x −2a,x <2log a x,x ≥2在R 上单调递减，必有{a −1<00<a <12(a −1)−2a ≥log a 2，化简可得{0<a <1log a 2≤−2，解可得√22≤a <1，即a 的取值范围是[√22,1)；故选：C ．根据题意，由函数的单调性的性质列出不等式组，求解可得a 的取值范围，即可得答案． 本题考查函数单调性的应用，关键是掌握函数单调性的定义．10.【答案】B【解析】解：令f(x)=0，有log a x =x −a ，①当a >1时，函数y =log a x 单增，函数y =x −a 相当于函数y =x 向下至少移动了1个单位，故函数y =log a x 与y =x −a 的图象有两个交点；②当0<a <1时，函数y =log a x 与y =x −a 的图象显然仅有一个交点， 综上，a >1． 故选：B ．只需函数y =log a x 与y =x −a 图象的交点个数为2，分a >1及0<a <1即可作出结论． 本题考查函数零点与方程根的关系，考查对数函数的图象及性质，属于基础题．11.【答案】1【解析】解：∵集合A ={(x,y)|2x −y =4}，B ={(x,y)|x +y =5}， ∴A ∩B ={(x,y)|{2x −y =4x +y =5}={(3,2)}，∴A ∩B 中元素个数为1． 故答案为：1．利用交集定义直接求解．本题考查集合的运算，考查交集定义等基础知识，考查运算求解能力，是基础题．12.【答案】3【解析】解：8−13+lg5+lg2+e ln2+14lg0.01=(23)−13+lg5×2+2+14lg10−2=2−1+lg10+2+14×(−2) =12+3−12 =3． 故答案为：3．由指数与对数的运算性质求解即可．本题主要考查指数与对数的运算性质，考查运算求解能力，属于基础题．13.【答案】31【解析】解：∵f(x)=ax 5+bx 3+cx +7(其中a ，b ，c 为常数，x ∈R)，f(−2021)=−17， ∴f(2021)=a ⋅20215+b ⋅20213+c ⋅2021+7， f(−2021)=a(−2021)5+b(−2021)3+c(−2021)+7， ∴f(2021)+f(−2021)=14，∴f(2021)−17=14， ∴f(2021)=14+17=31． 故答案为：31．由已知得f(2021)=a ⋅20215+b ⋅20213+c ⋅2021+7，f(−2021)=a(−2021)5+b(−2021)3+c(−2021)+7，由此能求出f(2021)．本题主要考查函数的求值问题，考查函数的性质的应用，属于基础题．14.【答案】−34【解析】 【分析】本题考查函数的奇偶性与周期性的综合应用，考查对数函数的性质，注意求出a 的值，属于中档题．根据题意，由奇函数的性质可得f(0)=20+a =0，解可得a 的值，即可得函数的解析式，结合函数的奇偶性与周期性求出f(−3)与f(14−log 27)的值，计算即可得答案． 【解答】解：根据题意，函数f(x)是定义在R 上的奇函数且当0≤x <1时，f(x)=2x +a ， 则f(0)=20+a =0，解可得a =−1， 则f(x)=2x −1，函数f(x)是定义在R 上周期为2的奇函数且f(1)=0， 则f(−3)=f(1)=0，f(14−log 27)=f(2−log 27)，又由2<log 27<3，则−1<2−log 27<0，则有0<log 27−2<1， 则f(log 27−2)=f(log 274)=2log 274−1=34，则f(14−log 27)=−f(log 27−2)=−34； 则f(−3)+f(14−log 27)=0−34=−34． 故答案为：−34．15.【答案】解：(1)∵集合A ={x|x 2+2x −3<0}={x|−3<x <1}，B ={y|y <−1或y >3}， ∴A ∩B ={x|−3<x <−1}；(2)A ∪B ={x|x <1或x >3}，C ={x|−2<x <m +1}，其中m >−3． (A ∪B)∩C =C ， ∴C ⊆(A ∪B)， ∵m >−3，∴C ≠⌀， ∴m +1≤1，解得m ≤0， ∴实数m 的取值范围是(−3,0]．【解析】(1)求出集合A ，利用交集定义能求出A ∩B ；(2)求出A ∪B ={x|x <1或x >3}，由(A ∪B)∩C =C ，得C ⊆(A ∪B)，由m >−3，得C ≠⌀，从而m +1≤1，由此能求出实数m 的取值范围．本题考查集合的运算，考查并集、并集定义、不等式性质等基础知识，考查运算求解能力，是基础题．16.【答案】解：(1)由f(0)=3，得c =3，又1和3是方程ax 2+bx +c =0的两根， ∴c a =3，−b a=4.解得a =1，b =−4， ∴f(x)=x 2−4x +3．(2)g(x)=f(x)−(2t −4)x =x 2−2tx +3，x ∈[−1,2].开口向上且对称轴为x =t ， 当t ≤−1时，g(x)在[−1,2]上为增函数，g(x)min =g(−1)=2t +4=2，解得t =−1，符合题意；当−1<t <2时，g(x)在[−1,t]上为减函数，g(x)在[t,2]上为增函数，g(x)min =g(t)=−t 2+3=2，解得t =±1，其中t =−1舍去；当t ≥2时，g(x)在[−1,2]上为减函数，g(x)min =g(2)=7−4t =2，解得t =54，不符合题意．综上可得，t =1或t =−1．【解析】(1)根据题意，分析可得c =3，又由一元二次不等式的解知，1和3是方程ax 2+bx +c =0的两根且f(0)=c ，利用根与系数的关系即可求参数，写出解析式．(2)根据题意，由二次函数的开口及对称轴，结合其在闭区间上的最小值，讨论t ≤−1、−1<t <2、t ≥2情况下，求符合条件的t 值即可．本题考查二次函数的性质以及应用，涉及函数单调性的性质和应用，属于基础题． 17.【答案】解：(1)函数f(x)=1−2x 1+2x 的定义域为R ，f(−x)=1−2−x 1+2−x =2x −12x +1=−f(x)， 可得f(x)为奇函数；(2)函数f(x)=1−2x 1+2x=−1+22x +1在R 上为减函数． 证明：设x 1，x 2∈R ，且x 1<x 2，f(x 1)−f(x 2)=22x 1+1−22x 2+1=2(2x 2−2x 1)(2x 1+1)(2x 2+1)， 由x 1<x 2，可得0<2x 1<2x 2，所以2(2x 2−2x 1)(2x 1+1)(2x 2+1)>0，即f(x 1)>f(x 2)， 所以f(x)在R 上为减函数；(3)对于任意t ∈R ，不等式f(t 2−2t)+f(t 2−k)<0恒成立，可得f(t 2−2t)<−f(t 2−k)=f(k −t 2)，因为f(x)在R 上为减函数，可得t 2−2t >k −t 2，即k <2t 2−2t 恒成立， 由2t 2−2t =2(t −12)2−12≥−12，所以k <−12，即k 的取值范围是(−∞,−12).【解析】(1)由函数的奇偶性的定义可得结论；(2)f(x)在R 上为减函数，运用单调性的定义证明，注意作差、变形和下结论等步骤；(3)由f(x)的奇偶性和单调性，可得t 2−2t >k −t 2，即k <2t 2−2t 恒成立，再由二次函数的最值求法，可得所求范围．本题考查函数的奇偶性和单调性的判断和运用，以及不等式恒成立问题解法，考查转化思想和运算能力、推理能力，属于中档题．18.【答案】解：(1)由xf(x)<(m −3)(x −1)(m ∈R)，得x 2+3−4x <(m −3)(x −1)，即x 2−(m +1)x +m <0．∴(x −1)(x −m)<0．若m <1，则m <x <1；若m =1，则不等式无解；若m >1，则1<x <m ．∴当m <1时，不等式的解集为(m,1)；当m =1时，不等式无解；当m >1时，不等式的解集为(1,m)．(2)方程f(a x )−k ⋅a −x −k =0(a >0,a ≠1)可化为a 2x −(k +4)⋅a x +3−k =0． 令a x =t ，则t >0，故方程t 2−(k +4)t +3−k =0有两个不相等的正根，记两个正根分别为t 1，t 2， ∴{△=(k +4)2−4(3−k)>0t 1+t 2=k +4>0t 1t 2=3−k >0，解得−6+4√2<k <3．∴实数k 的取值范围是(−6+4√2,3)．【解析】(1)由xf(x)<(m −3)(x −1)(m ∈R)，x 2−(m +1)x +m <0，即(x −1)(x −m)<0，然后对m 分类求解得答案；(2)方程f(a x)−k⋅a−x−k=0(a>0,a≠1)可化为a2x−(k+4)⋅a x+3−k=0，令a x=t，则t>0，可得方程t2−(k+4)t+3−k=0有两个不相等的正根，再由一元二次方程的判别式及根与系数的关系列不等式组求解．本题考查函数零点与方程根的关系，训练了一元二次不等式的解法，考查一元二次方程根的分布与系数间的关系，是中档题．19.【答案】解：(Ⅰ)a<0时，内层函数有最大值，故函数值不可能取到全体正数，不符合题意；a=0时，内层函数是一次函数，内层函数值可以取遍全体正数，值域是R，符合题意；a>0时，要使内层函数的函数值可以取遍全体正数，只需要函数最小值小于等于0，故只需△≥0，解得a∈(0,4]．综上得a∈[0,4]．(Ⅱ)由题意可得k⋅2t<ℎ(x)+2=log22x−2log2x+2在x∈[12,2]恒成立，则k⋅2t<ℎ(x)min+2=1在t∈[−1,1]有解，即k<12t在t∈[−1,1]有解，∴k<(12t)max=2，综上，实数k的取值范围k<2．【解析】(Ⅰ)根据对数的性质，当真数取遍全体正数时，对数函数的值域是R，问题转化为内层函数的值域包括全体正数，再根据参数a的取值范围的不同，分三种情况讨论即可解答本题；(Ⅱ)不等式成立求参数取值范围的问题通常转化为最值问题求解，所以先求ℎ(x)min，再转化求关于t的函数的最值，从而得出答案．本题考查对数函数的性质，全称命题及特称命题的逻辑关系，转化化归的思想，本题的疑点是第一问函数值域是R的转化，难点是第二问中对于全称命题及特称命题逻辑关系的理解及正确转化．20.【答案】解：f(x)=3|x|+2+x+93|x|+1=9(3|x|+1)+x3|x|+1=9+x3|x|+1．设g(x)=x3|x|+1，∵g(x)的定义域为R，关于原点对称，且g(−x)=−g(x)，∴g(x)为奇函数，则g(x)max+g(x)min=0．∵M=f(x)max=g(x)max+9，m=f(x)min=g(x)min+9，∴M+m=18．【解析】把已知函数解析式变形，结合奇函数的对称性即可求解．本题考查函数的最值及其几何意义，考查奇函数的对称性及应用，是中档题．21.【答案】解：(1)根据题意，∵f(x)是偶函数，g(x)是奇函数，且f(x)−g(x)=21−x①，∴f(−x)=f(x)，g(−x)=−g(x)，∴f(−x)+g(−x)=21+x，即f(x)+g(x)=21+x②；由①+②解得f(x)=2x+2−x，①−②解得g(x)=2x−2−x；(2)方程mf(x)=[g(x)]2+2m+9有解，则m(2x+2−x)=22x+2−2x−2+2m+9=(2x+2−x)2+2m+5有解，、令t=2x+2−x≥2，当且仅当x=0时取等号，∴mt=t2+2m+5在[2,+∞)有解，即m(t−2)=t2+5，当t=2时，不成立，当t>2时，m= t2+5t−2=(t−2)2+4(t−2)+9t−2=(t−2)+9t−2+4≥2√(t−2)⋅9t−2+4=6+4=10，当且仅当t=5时取等号，故m的取值范围为[10,+∞)；(3)ℎ(x)=|12[f(x)+g(x)]−1|=|2x−1|∈[0,+∞)，令ℎ(x)=a，则a∈[0,+∞)，函数ℎ(x)的图象，如图所示为：∵方程[ℎ(x)]2−(2k+12)ℎ(x)+k=0有三个解，∴a2−(2k+12)a+k=0有两个根，且0<a1<1<a2，或者a1=0，0<a2<1，或者0<a1<1，a2=1当a1=0，0<a2<1，有k=0，a2−12a=0，解得a2=12满足题意，则|2x−1|=0，解得x=0，则|2x−1|=12，存在两个x值满足，当a2=1时，记p(a)=a2−(2k+12)a+k，∴{ϕ(0)=k>0ϕ(1)=12−k≤0，解得k≥12，故k的取值范围为[12,+∞)．【解析】将方程进行等价转化，利用换元法转化为一元二次方程，结合一元二次方程根的分别进行求解即可．本题考查函数的零点与方程的根的关系，及函数的性质，考查学生的运算能力，属于难题．。
2021-2022年高一数学上学期期中试题高新部
2021-2022年高一数学上学期期中试题高新部一、选择题（12题，每小题5分，共60分）1．若集合A ＝{(1,2)，(3,4)}，则集合A 中元素的个数是( )A ．1B ．2C ．3D ．4 2．把集合{x |x 2－3x ＋2＝0}用列举法表示为( )A ．{x ＝1，x ＝2}B ．{x |x ＝1，x ＝2}C ．{x 2－3x ＋2＝0}D ．{1,2}3．下列集合的表示方法正确的是( )A ．第二、四象限内的点集可表示为{(x ，y )|xy ≤0，x ∈R ，y ∈R}B ．不等式x －1<4的解集为{x <5}C ．{全体整数}D ．实数集可表示为R 4．若函数f (x )＝⎩⎨⎧ x 2－1，x ≤1log 2x ，x ＞1，则f [f (－3)]等于( )A ．1B ．2C ．0D ．35．定义在R 上的偶函数f (x )，在x >0时是增函数，则( )A ．f (3)<f (－4)<f (－π)B．f(－π)<f(－4)<f(3)C．f(3)<f(－π)<f(－4)D．f(－4)<f(－π)<f(3)6．已知集合M＝{(x，y)|x＋y＝2}，N＝{(x，y)|x－y＝4}，那么集合M∩N为( ) A．x＝3，y＝－1B．(3，－1)C．{3，－1}D．{(3，－1)}7．设集合A＝{5,2a}，集合B＝{a，b}，若A∩B＝{2}，则a＋b等于( ) A．1B．2C．3D．48．设集合A＝{a，b}，B＝{a＋1,5}，若A∩B＝{2}，则A∪B等于( ) A．{1,2} B．{1,5}C．{2,5} D．{1,2,5}9．如图所示的Venn图中，若A＝{x|0≤x≤2}，B＝{x|x＞1}，则阴影部分表示的集合为( )A．{x|0＜x＜2}B．{x|1＜x≤2}C．{x|0≤x≤1，或x≥2}D．{x|0≤x≤1，或x＞2}10．已知集合M中的元素a，b，c是△ABC的三边，则△ABC一定不是( )A．锐角三角形B．钝角三角形C．直角三角形D．等腰三角形11．下面有三个命题：①集合N中最小的数是1；②若－a∉N，则a∈N；③若a ∈N，b∈N，则a＋b的最小值是2.其中正确命题的个数是( )A．0个B．1个C．2个D．3个12．下列正确的命题的个数有( )①1∈N；②2∈N*；③12∈Q；④2＋2∉R；⑤42∉Z.A．1个B．2个C．3个D．4个二、填空题(20分，每题5分)13.已知集合P中元素x满足：x∈N，且2＜x＜a，又集合P中恰有三个元素，则整数a＝__________.14.集合P中含有两个元素分别为1和4，集合Q中含有两个元素1和a2，若P与Q相等，则a＝__________.15．设集合A是由1，k2为元素组成的集合，则实数k的取值范围是________．16．由实数t，|t|，t2，－t，t3所构成的集合M中最多含有________个元素．三、解答题（6小题，满分70分，）17．设P，Q为两个数集，P中含有0,2,5三个元素，Q中含有1,2,6三个元素，定义集合P＋Q中的元素是a＋b，其中a∈P，b∈Q，求P＋Q中元素的个数．(10分)18.若集合A＝{x|x2＋x－6＝0}，B＝{x|x2＋x＋a＝0}，且B⊆A，求实数a的取值范围．(12分)19.已知A＝{x|x2－ax＋a2－19＝0}，B＝{x|x2－5x＋6＝0}，C＝{x|x2＋2x－8＝0}，且∅(A∩B)，A∩C＝∅，求a的值．(12分)20．设集合A＝{－2}，B＝{x|ax＋1＝0，a∈R}，若A∩B＝B，求a的值．21．设全集I＝R，已知集合M＝{x|(x＋3)2≤0}，N＝{x|x2＋x－6＝0}．(1)求(∁I M)∩N；(2)记集合A＝(∁I M)∩N，已知集合B＝{x|a－1≤x≤5－a，a∈R}，若B∪A＝A，求实数a的取值范围．22．集合A＝{x|－1<x<1}，B＝{x|x<a}．(1)若A∩B＝∅，求a的取值范围；(2)若A∪B＝{x|x<1}，求a的取值范围．．答案及解析1.【解析】由列举法可知，A中含有(1,2)，(3,4)两个元素．【答案】B2.【解析】解方程x2－3x＋2＝0得x＝1或x＝2，所以集合{x|x2－3x＋2＝0}用列举法可表示为{1,2}．【答案】D3.【解析】选项A中应是xy<0；选项B的本意是想用描述法表示，但不符合描述法的规范格式，缺少了竖线和竖线前面的代表元素x；选项C的“{}”与“全体”意思重复．【答案】D4.答案：D解析：f(－3)＝(－3)2－1＝8，所以f[f(－3)]＝f(8)＝log28＝3.解析：∵f (x )在R 上是偶函数，∴f (－π)＝f (π)，f (－4)＝f (4)，而3<π<4且f (x )在(0，＋∞)上是增函数．∴f (3)<f (π)<f (4)，即f (3)<f (－π)<f (－4)．6．D [M 、N 中的元素是平面上的点，M ∩N 是集合，并且其中元素也是点，解⎩⎨⎧ x ＋y ＝2，x －y ＝4，得⎩⎨⎧ x ＝3，y ＝－1.]7．C [依题意，由A ∩B ＝{2}知2a ＝2，所以，a ＝1，b ＝2，a ＋b ＝3，故选C.]8.解析：选D ∵A ∩B ＝{2}，∴2∈A,2∈B ，∴a ＋1＝2，∴a ＝1，b ＝2，即A ＝{1,2}，B ＝{2,5}．∴A ∪B ＝{1,2,5}．9.解析：选D 因为A ∩B ＝{x |1＜x ≤2}，A ∪B ＝{x |x ≥0}，阴影部分为A ∪B 中除去A ∩B 的部分，即为{x |0≤x ≤1，或x ＞2}．10.【解析】 因为集合中元素具有互异性，所以a ，b ，c 互不相等，因此选D.11.【解析】因为自然数集中最小的数是0，而不是1，所以①错；对于②，取a ＝2，则－2∉N，2∉N，所以②错；对于③，a＝0，b＝0时，a＋b取得最小值是0，而不是2，所以③错．【答案】A12.【解析】∵1是自然数，∴1∈N，故①正确；∵2不是正整数，∴2∉N*，故②不正确；∵12是有理数，∴12∈Q，故③正确；∵2＋2是实数，∴2＋2∈R，所以④不正确；∵42＝2是整数，∴42∈Z，故⑤不正确．【答案】B13.解析：∵x∈N，且2＜x＜a，集合P中恰有三个元素，∴x的值为3,4,5.又∵a ∈N，∴a＝6.答案：614.解析：由题意，得a2＝4，a＝±2.答案：±215.【解析】∵1∈A，k2∈A，结合集合中元素的性质可知k2≠1，解得k≠±1.【答案】k≠±116.【解析】由于|t|至少与t和－t中的一个相等，故集合M中至多有4个元素．17.解析：当a ＝0时，由b ∈Q 可得a ＋b 的值为1,2,6；当a ＝2时，由b ∈Q 可得a ＋b 的值为3,4,8；当a ＝5时，由b ∈Q 可得a ＋b 的值为6,7,11.由集合元素的互异性可知，P ＋Q 中的元素为1,2,3,4,6,7,8,11，共8个．18.解 A ＝{－3,2}．对于x 2＋x ＋a ＝0，①当Δ＝1－4a <0，即a >14时，B ＝∅，B ⊆A 成立； ②当Δ＝1－4a ＝0，即a ＝14时， B ＝⎩⎪⎨⎪⎧⎭⎪⎬⎪⎫－12， B ⊆A 不成立；③当Δ＝1－4a >0，即a <14时，若B ⊆A 成立， 则B ＝{－3,2}，∴a ＝－3×2＝－6.综上：a 的取值范围为a >14或a ＝－6.19.解：B ＝{x |x 2－5x ＋6＝0}＝{x |(x －2)(x －3)＝0}＝{2,3}，C ＝{x |x 2＋2x －8＝0}＝{x |(x －2)(x ＋4)＝0}＝{2，－4}，∵A ∩B ≠∅，A ∩C ＝∅，∴3∈A ，将x ＝3代入x 2－ax ＋a 2－19＝0得：a 2－3a －10＝0，解得a ＝5或－2.当a ＝5时，A ＝{x |x 2－5x ＋6＝0}＝{2,3}与A ∩C ＝∅矛盾；当a ＝－2时，A ＝{x |x 2＋2x －15＝0}＝{3，－5}符合题意．综上a ＝－2.20．解 ∵A ∩B ＝B ，∴B ⊆A .∵A ＝{－2}≠∅，∴B ＝∅或B ≠∅.当B ＝∅时，方程ax ＋1＝0无解，此时a ＝0.当B ≠∅时，此时a ≠0，则B ＝{－1a}， ∴－1a ∈A ，即有－1a ＝－2，得a ＝12. 综上，得a ＝0或a ＝12.21.解：(1)∵M ＝{x |(x ＋3)2≤0}＝{－3}，N ＝{x |x 2＋x －6＝0}＝{－3, 2}，∴∁I M ＝{x |x ∈R 且x ≠－3}，∴(∁I M )∩N ＝{2}．(2)A ＝(∁I M )∩N ＝{2}，∵A ∪B ＝A ，∴B ⊆A ，∴B ＝∅或B ＝{2}，当B ＝∅时，a －1>5－a ，∴a >3；当B ＝{2}时，⎩⎨⎧ a －1＝2，5－a ＝2，解得a ＝3，综上所述，所求a 的取值范围为{a |a ≥3}．22.解：(1)如下图所示，A ＝{x |－1<x <1}，B ＝{x |x <a }，且A ∩B ＝∅，∴数轴上的点x ＝a 在x ＝－1的左侧(含点x ＝－1)，∴a ≤－1，即a 的取值范围为{a |a ≤－1}．(2)如下图所示，A ＝{x |－1<x <1}，B ＝{x |x <a }，且A ∪B ＝{x |x <1}，∴数轴上的点x ＝a 在x ＝－1和x ＝1之间(含点x ＝1，但不含点x ＝－1)， ∴－1<a ≤1，即a 的取值范围为{a |－1<a ≤1}H C39437 9A0D 騍20938 51CA 凊39607 9AB7 骷C&SK-22639 586F 塯m x。
北师大版高中数学必修一上学期期中考试高一年级数学试卷.doc
吉安一中2009－2010学年度上学期期中考试高一年级数学试卷班级 学号 姓名 命题教师：刘志乐、罗飞兰 说明：1．考试时间：120分钟 ， 试卷满分：150分2．本试卷分公共题、A 类题、B 类题。
宏志班、春蕾班、实验班、重点班的学生做公共题和A 题，其它班级的学生做公共题和B 类题。
3．要求将所有答案填写在答题卷上，考试结束时只交答题卷。
一、选择题（本大题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一个选项是符合题目要求的）1．下列关系式正确的是 （ ）A ．Q ∈2B ．{}{}x x x 222== C ．{}{}a b b a ,,= D ．{}2005∅∈2．定义集合运算：A ※B ＝｛t|t ＝xy , x ∈A ，y ∈B }，设A ＝｛1，2｝，B ＝｛0，2｝， 则集合A ※B 的所有元素之和为（ ） A ．6B ．3C ．2D ．03．下列函数中既是奇函数，又在区间（0，+∞）上单调递增的是 （ ） A ．2y x =-B ． 1y x x=+C ．()12xy g = D ． ||x ey =4．已知)(x f 的图象恒过（1，1）点，则)4(-x f 的图象恒过（ ） A ．（－3，1）B ．（5，1）C ．（1，－3）D ．（1，5）5．20xx +=在下列哪个区间内有实数解（ ）A ．[]2,1--B ． []0,1C ．[]1,2D ．[]1,0-6．已知函数22(1)() (12)2 (2)x x f x x x x x +≤-⎧⎪=-<<⎨⎪≥⎩, 若f (a )=3 ，则a 的值为（ ）A.3 B. －3 C. ±3 D. 以上均不对7．设a >1，实数x ，y 满足()xf x a =，则函数()f x 的图象形状大致是（ ）8．若lg lg 0,()()a b a b f x x g x x +===则函数与在第一象限内的图象关于（ ）对称A . 直线x y =B ．x 轴C ．y 轴D ．原点 9．若函数y=f (x )的定义域是[2，4]，则y=f (x log 21)的定义域是（ ）A ．[21，1] B ．[4，16] C ．[41,161] D ．[2，4]10．（A 类）对任意实数x 规定y 取14,1,(5)2x x x -+-三个值中的最小值，则函数y （ ） A ．有最大值2，最小值1， B ．有最大值2，无最小值， C ．有最大值1，无最小值， D ．无最大值，无最小值。
2021-2022学年陕西省西安市高新第一中学高一上学期期中数学试题(解析)
2021-2022学年陕西省西安市高新第一中学高一上学期期中数学试题一、单选题1．已知集合A x|0x6，集合B x|3x8，则A B（）A．x|0x3C．y|0y8【答案】C【分析】根据并集运算直接求解即可.【详解】B．x|6x8 D．以上答案都不对A x|0x6,B x|3x8,A B(0,8),故选：C2．下列函数中，在定义域上为奇函数，并且在区间0,上是减函数的是（）3 A．f(x)x 【答案】B1B．f(x)x1C．f(x)2xD．f(x)3x【分析】根据定义分析奇偶性，利用常见的基本函数的性质可得出单调性.【详解】A选项，f(1)1，f(2)8，故在0,上并不递减，不符合题意；C选项，由指数函数性质可知，f(x)(0,)(x R)，故f(x)f(x)0，不是奇函数；D选项，f(x)3x，f(x)3x，故f(x)f(x)6，不是奇函数；综上ACD选项错B选项，f(x)误；1定义域为{x|x0}，且f(x)f(x)0，是奇函数，且显然0, x上递减，B选项正确.故选：B.23．f x x2x4定义域为0,m，值域为5,4，则m的取值范围是（）A．1【答案】CB．1,C．1,2D．1,2【分析】由二次函数的性质知f(x)开口向上且顶点为(1,5)，且f(0)f(2)4，结合闭区间对应值域即可确定m的范围.【详解】由f(x)(x1)25，其开口向上且顶点为(1,5)，当f(x)4时，可得x 0或x 2，因为f(x)定义域为0,m 对应值域为5,4，所以1m 2.故选：C.4．函数f x log 1x21的零点所在区间是（）xA ．3,4C ．1,2【答案】CB ．2,3D ．0,1【分析】根据函数解析式可得f x 在(0,)上单调递减，再根据零点存在性定理判断零点所在区间；【详解】解：函数f x log 1x 21在(0,)上单调递减，且xf 110,f 2110，f x 的零点在1,2内.2故选：C 5．已知f xm 2m 7x2m 23是幂函数，且在0,上单调递增，则满足f a 11的实数a 的范国为（）A ．,0【答案】D【分析】由幂函数的定义求得m 的可能取值，再由单调性确定m 的值，得函数解析式，结合奇偶性求解.【详解】由题意m 22m 71，解得m4或m 2，又f(x)在0,上单调递增，所以2B ．2,D ．,0C ．0,22,m 20，m2，3所以m4，f(x)x 3，易知f(x)是偶函数，所以由f a 11得a 11，解得a 0或a2.故选：D.26．函数f x log 232x x 的单调递增区间是（）A ．,1B ．3,1C ．1,D ．1,1【分析】根据复合函数的单调性，即可求解.【详解】函数的定义域需满足32x x 20，解得：3x 1，函数分为内外层函数，y log 2t ，t 32x x 2x 14，定义域内，内层函数在区间3,1是增函数，在区间1,1是减函数，根据复合函数的单调性可知，函数的单调递增区间是3,1.故选：B7．设函数f(x)e |x|，则使得f(2x 1)f(x)成立的x 的取值范围是（）1A ．,13111B ．,(1,)C ．,33311D ．,,332【答案】A【解析】首先判断出函数为偶函数，再判断出函数的单调性，根据单调性可得2x 1x ，解绝对值不等式即可求解.【详解】f(x)e |x|，则fxexe f x ，函数为偶函数，x当x 0时，f(x)e x ，所以函数在0,单调递增，所以函数在,0上单调递减，若f(2x 1)f(x)，则2x 1x ，即3x 24x 10，解得x 1，1所以不等式的解集为,1.313故选：A 8．设a12，b，c log 23，则a 、b 、c 的大小关系为（）233A ．a b c 【答案】DB ．b a cC ．b c aD ．c a b【分析】利用指数函数和对数函数的单调性可得出a 、b 、c 三个数的大小关系.1【详解】b232223323a ，且a22log 24log 23c ，故c a b .32(a1)x2a,x29．若函数f(x)在R上单调递减，则实数a的取值范围是（）log x,x2a20, A．2【答案】B2,1B．2C．(0,1)D．[1,)a10【分析】要使函数f x是减函数，须满足0a1，解出不等式组即可.2(a1)2a log2aa10(a1)x2a,x2【详解】若函数f(x)在R上单调递减，则0a1，log x,x2a2(a1)2a log2a得2a1. 2故选：B.10．若函数f x logax x a（a0且a1）有两个零点，则实数a的取值范围是（）A．0,1【答案】B【解析】将函数f x loga x x a（a0且a1）有两个零点转化为函数y logax和函数y x a图象有两个交点问题，然后分类讨论a1和0a1进行判断即可得出答案.【详解】当函数f x logax x a（a0且a1）有两个零点，则若f x0，loga x x a有两个根，则得函数y logax和函数y x a图象有两个交点，①当a1 B．1,C．1,e D．e,时，因为函数y logax的图象恒过点(1,0)，而直线y x a的图象恒过点M a,0，由于a1，则点M a,0一定在点(1,0)的右侧，如图所示：x与y x a的图象有两个交点，即a1时函数由图象可知，当a1时，函数y logax x a有两个零点；f x logax与y x a的图象仅有一个交点，如图所示：②当0a1时，函数y logax x a有且仅有一个零点；此时函数f x loga综上可得a1满足题意.故选：B.【点睛】本题考查了分类讨论思想和数形结合思想的应用，考查了函数零点个数的问题，属于中档题.二、填空题2x y4,B x,y∣x y5，则A B中元素个数为11．已知集合A x,y∣__________．【答案】1【分析】利用交集的定义直接求解.2x y4，B x,y∣x y5，【详解】∵集合A x,y∣2x y4∴A B x,y3,2，x y5∴A B中元素个数为1.故答案为：1.112．计算求值8lg5lg2eln2lg0.01=___________.413【答案】3【分析】根据指数幂与对数的运算性质，准确运算，即可求解.11133【详解】由8lg5lg2e lg0.01(2)(lg5lg2)eln2lg0.01441111lg102(2)123.2422ln213故答案为：3.13．设f(x)ax 5bx 3cx 7（其中a b c 为常数，x R ），若f(2021)17.则f 2021___________.【答案】3120215b?20213c?20217，【分析】由已知得f 2021a?f2021a2021b 2021c 20217，由此能求出f 2021．53【详解】f(x)ax 5bx 3cx 7（其中a ，b ，c 为常数，x R )，f202117，f 2021a?20215b?20213c ?20217f2021a2021b 2021c 2021753f 2021f 202114，f 20211714f 2021141731．故答案为：31．x 14．已知函数f x 是定义在R 上的周期为2的奇函数，且当0x 1时，f x2a ，f 10，则f3f 4log 27______.3【答案】4【分析】根据题意，由奇函数的性质可得f(0)20a0，解可得a 的值，即可得函数的解析式，结合函数的奇偶性与周期性求出f (3)与f(4log 27)的值，计算即可得答案．【详解】解：根据题意，函数f(x)是定义在R 上的奇函数且当0x 1时，f(x)2x a ，则f(0)20a0，解可得a1，则f(x)2x 1，函数f(x)是定义在R 上周期为2的奇函数且f 10，则f 3f 10，f(4log 27)f(2log 27)，又由2log 273，则12log 270，则有0log 2721，7log 2734f(log 72)f(log )21，则2244则f(4log 27)f(log 272)；则f(3)f(4log 27)0；3故答案为：．4343434三、解答题∣x 22x 30,B {y ∣y1或y 3},C{x ∣2x m 1}，其中15．已知集合Ax m 3．(1)求A B ；(2)若AB CC ，求实数m 的取值范围．【答案】(1)A B{x |3x1}；(2)(3，0]．【分析】（1）求出集合A ，利用交集定义能求出A B ；（2）求出A B{x |x 1或x3}，由(AB )CC ，得C(AB )，由m 3，得C，从而m 11，由此能求出实数m 的取值范围．(1)解：集合A {x |x 22x 30}{x |3x 1}，B {y |y1或y 3}，AB{x |3x1}.(2)解：A B{x |x 1或x3}，C{x |2x m1}，其中m 3．因为(AB )C C ，C(A B )，m3，C ，m11，解得m 0，实数m的取值范围是(3，0]．16．已知二次函数f(x)ax2bx c的图象过点(0,3)，且不等式ax2bx c0的解集为{x|1＜x＜3}．(1)求f(x)的解析式；(2)若g(x)f(x)(2t4)x在区间[1,2]上有最小值2，求实数t的值．2【答案】(1)f x x4x3；(2)t1或t1.1和3是方程ax2+bx+c 【分析】（1）根据题意得f(0)c＝3，又由一元二次不等式的解知，＝0的两根，利用根与系数的关系即可求参数，写出解析式．（2）由二次函数的开口及对称轴，结合其在闭区间上的最小值，讨论t≤1、1＜t ＜2、t≥2情况下求符合条件的t值即可．(1)由f(0)3，得c＝3，又1和3是方程ax2+bx+c＝0的两根，∴bc3，4．解得a＝1，b＝4，a a∴f(x)x24x3．(2)g(x)f(x)(2t4)x＝x22tx3，x∈[1,2]．开口向上且对称轴为x＝t，1、当t≤1时，g(x)在[1,2]上为增函数，g(x)min＝g(1)＝2t+4＝2，解得t＝1，符合题意；2、当1＜t＜2时，g(x)在[1,t]上为减函数，g(x)在[t,2]上为增函数，g(x)ming t t232，解得t＝±1，其中t＝1舍去；3、当t≥2时，g(x)在[1,2]上为减函数，g(x)min＝g(2)＝74t＝2，解得t 合题意．综上，t＝1或t＝1．12x17．已知定义域为R的函数f x．12x 5，不符4(1)判断f x的奇偶性12x(2)试判断函数f x在R上的单调性，并用函数单调性的定义证明；12x22(3)若对于任意t R，不等式f t2t f t k0恒成立，求实数k的取值范围．【答案】(1)奇函数(2)减函数,证明见解析1-)(3)(-，2【分析】（1）由函数的奇偶性的定义可得结论；（2）f(x)在R 上为减函数，运用单调性的定义证明，注意作差、变形和下结论等步骤；（3）由f(x)的奇偶性和单调性，可得t 22t k t 2，即k 2t 22t 恒成立，再由二次函数的最值求法，可得所求范围．(1)12x 函数f(x)的定义域为R ，12x12x 2x 1f(x)x f(x)，x1221可得f(x)为奇函数；(2)12x2函数f(x)在R 上为减函数．112x2x 1222(2x 22x 1)证明：设x 1，x 2R ，且x 1x 2，f(x 1)f(x 2)x 1x 2，2121(2x 11)(2x 21)2(2x 22x 1)0，即f(x 1)f(x 2)，由x1x 2，可得022，所以x 1(21)(2x 21)x 1x 2所以f(x)在R 上为减函数；(3)对于任意t R ，不等式f(t 22t )f(t 2k)0恒成立，可得f(t 22t )f(t 2k)f(k t 2)，因为f(x)在R 上为减函数，可得t 22t k t 2，即k 2t 22t 恒成立，由2t 22t 2(t )212121，211所以k，即k 的取值范围是(,)．2218．已知函数f xx 34．x(1)求关于x 的不等式xf x m 3x 1m R 的解集；x x(2)若关于x 的方程f a k ak 0(a0,a 1)有两个不相等的实数根，求实数k 的取值范围．【答案】(1)答案见解析；(2)642,3【分析】（1）根据题意得x 1x m 0，再分m 1，m1，m 1三种情况讨论求解即可；2（2）令a xt ,t 0，进而根据题意将问题转化为t 4k t 3k 0有两个不相等的正实数根，再结合二次方程根的分布求解即可.(1)解：因为f x x34，x2所以xf xx 4x 3x 1x 3，所以原不等式等价于x 1x 3m3x 1，即x 1x m0，所以当m 1时，不等式的解集为m ,1；当m1时，不等式的解集为，当m 1时，不等式的解集为1x m .综上，当m1时，不等式的解集为m ,1；当m1时，不等式无解；当m 1时，不等式的解集为1,m .(2)x x解：因为f a a 34，a xx x所以方程f a k a k 0(a0,a 1)有两个不相等的实数根等价于a x3k4k0有两个不相等的实数根，a x令a x t ,t 0，x则a 3k 4k0有两个不相等的实数根等价于t 24k t 3k 0有两个不相xa 等的正实数根，记作t 1,t 2.Δ4k 243k所以t 1t 24k 0，解得642k 3.tt 3k 012所以实数k 的取值范围是642,3219．已知函数f xlog 2x ，g xax 4x 1.（1）若函数y f g x 的定义域为R ，求实数a 的取值范围；1（2）函数h(x)f 2(x)f(x 2)，若对于任意的x,2，都存在t 1,1使得不等式2h(x)k 2t 2成立，求实数k 的取值范围.【答案】（1）a 4；（2）k 2.a2a 0a0【分析】（1）由题意可得x R ，恒成立，讨论或，列出ax 4x 100即可求解.1t2（2）由题意可得k 2h(x)2log 2x 2log 2x 2在x,2恒成立，只需21k 2th xmin21在t 1,1有解，进而化为kt2即可求解.2max【详解】（1）由题可知：对任意的x R ，ax 24x 10恒成立.当a 0时，不合题意；a 0a当a0时，由，可得164a 0解得a 4，综上，a 4；1t2由题意可得k 2h(x)2log 2x 2log 2x 2在x,2恒成立，（2）2t 则k 2h xmin2在t 1,1有解，2令y h x 2log 2x 2log 2x 2，1由x,2，则log 2x1,1，2则y log 22x 2log 2x 2log 2x 111,5，所以y min 1，t 所以k 2h xmin21在t1,1有解，21t 1,1有解，t 在21kt2，综上，实数k 的取值范围k2.2max即k<20．已知函数f x 【答案】183x2x 9的最大值为M ，最小值为m ，求M m 的值．x31【分析】对函数f(x)的解析式进行化简，构造奇函数，利用奇函数的性质进行求解即可.【详解】解：根据题意，f x 设g x 3x2x 993x 9x 9，xx x 313131x ，x 31x xx x g(x)，所以函数g xx 因为g xx是奇函数，313131所以g xmaxg(x)min0，所以M mg(x)max 9g(x)min918，所以Mm 181x 21．已知f x 为偶函数，g x 为奇函数，且满足f x g x2.x （1）求f x 、g x ；（2）若方程mf x g x2m9有解，求实数m 的取值范围；（3）若h x211f xg x1h x 2k，且方程h x k0有三个解，求222实数k 的取值范围.1x x x x 【答案】（1）f x22，g x 22；（2）10,；（3）0,+.2【分析】（1）由已知条件可得出f x 、g x 的等式组，由此可解得这两个函数的解析式；2（2）令t 2x 2x2，分析可知函数F tt mt 2m5在2,上有零点，分m m 2、2两种情况讨论，结合二次函数的零点分布可得出关于实数m 的不等式，22综合可得出实数m 的取值范围；（3）作出函数h x 的图象，分析可知方程h x1有两个不等的实根，从而方程2h x2k 有且只有一个根，数形结合可求得实数k 的取值范围.1x 【详解】（1）因为f x 为偶函数，g x 为奇函数，由已知可得fxg x2，即f xg x21x x x 1xf x 22f xg x 2，所以，xx；1x，解得g x22f xg x2x xx xm 22442m7，g x2m9（2）由mf x 可得2令t 2x 2x22x 2x2，当且仅当x0时，等号成立，则t 24x4x2，故有t 2mt 2m50，其中t 2，2令F tt mt 2m 5，其中t 2，则函数F t 在2,上有零点，①当m2时，即当m4时，则F t 在2,上单调递增，所以，F tF 290，2不合乎题意；②当m2时，即当m4时，则有m 28m 200，解得m 10，此时m 10.2综上所述，实数m 的取值范围是10,；12x ,x 01x f x g x （3）h x 1212x 1,x 0，作出函数h x 的图象如下图所2示：211h x 由2k2h x k0可得h x2h x 2k0，由图可知，方程h x 1有两个不等的实根，21.2由题意可知，方程h x2k 有且只有一个根，故2k 0或2k 1，解得k0或k1因此，实数k 的取值范围是0,+.2。
高新一中高一语文期中试卷
一、选择题（每题2分，共20分）1. 下列词语中，字形、字音、字义完全正确的一项是（）A. 惊愕（jīng è）拈轻怕重（niān qīng pà zhòng）B. 翩跹（piān xiān）情不自禁（qíng bù zì jīn）C. 赏心悦目（shǎng xīn yuè mù）鸡鸣狗盗（jī míng gǒu dào）D. 摇头晃脑（yáo tóu huàng nǎo）雕虫小技（diāo chóng xiǎo jì）2. 下列句子中，没有语病的一项是（）A. 通过这次活动，使同学们受到了很大的启发和教育。
B. 为了提高学生的综合素质，学校决定开设一些选修课程。
C. 随着科技的不断发展，人们对未来的生活充满了期待。
D. 这本书的内容丰富，插图精美，深受广大读者的喜爱。
3. 下列各句中，没有使用修辞手法的一项是（）A. 花开两朵，各表一枝。
B. 雨后的天空，像一块洗过的蓝宝石。
C. 他的笑声如银铃般清脆悦耳。
D. 她的眼神里闪烁着坚定的光芒。
4. 下列词语中，没有错别字的一项是（）A. 惊慌失措（jīng huāng shī cuò）B. 鸡飞蛋打（jī fēi dàn dǎ）C. 美轮美奂（měi lún měi huàn）D. 震天动地（zhèn tiān dòng dì）5. 下列各句中，句式不同的一项是（）A. 他不仅学习好，而且乐于助人。
B. 我们一定要努力学习，争取早日成才。
C. 雨中的湖面，像一面镜子。
D. 我看着他的背影，心里充满了感慨。
二、填空题（每空2分，共20分）6. 下列句子出自哪部文学作品？（）《桃花源记》《离骚》《岳阳楼记》《醉翁亭记》7. “落红不是无情物，化作春泥更护花”出自哪位诗人的作品？（）8. 下列词语中，加点的字在词语中的含义相同的一项是（）A. 谈笑风生、风和日丽B. 惊天动地、地久天长C. 落花流水、水落石出D. 惊涛骇浪、浪迹天涯9. 下列各句中，加点词的用法相同的一项是（）A. 他对这个问题一无所知。
北师大版高一数学上学期期中试卷及答案
高一数学上学期期中试卷(必修1)命题人：宝鸡石油中学 沈涛考试说明：本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（填空题、解答题）两部分，第Ⅰ卷1至2页，第Ⅱ卷3至6页，满分150分，考试时间120分钟。
注意事项：1．答第I 卷前，考生务必将自己的姓名、班级、学校用蓝、黑墨水钢笔或圆珠笔、签字笔写在答卷上。
2．第I 卷每小题得出答案后，请将答案填写在答题卷相应表格指定位置上。
答在第Ⅰ卷上不得分。
3．考试结束，考生只需将第Ⅱ卷（含答卷）交回。
第Ⅰ卷（选择题共60分）一、选择题（本大题共12小题，每小题5分，共60分．在每小题给出的四个选项中，只有一个是符合题目要求的．）1.（改编自北师大版必修一第9页习题1-2A 组第2题第2问）集合2{|60} ,M x x x =--=则以下正确的是（ ）. {2} . 2 . 3 . 3A M B M C M D M -∈-⊂∈∈－2．如图，U 是全集，M 、P 是U 的子集，则阴影部分所表示的集合是A ．()U M P I ðB ．M P IC ．()U M P I ðD ．()()U U M P I 痧3．下列各组函数中，表示同一函数的是A ．1y =，0y x =B ．y x = , 2x y x=C ．y x =，ln x y e =D ．||y x = ，2()y x =4．函数()x f x a =在[0,1]上的最大值与最小值之和为3，则a 的值是A ． 12B ．2C ．3D ．325．二次函数2()23f x x bx =+-()b R ∈零点的个数是 A ．0B ．1C ．2D ．46．如图的曲线是幂函数n y x =在第一象限内的图象。
已知n 分别取1-，l ，12，2四个值，与曲线1C 、2C 、3C 、4C 相应的n 依次为 A ．2，1，12，1- B ．2，1-，1，12C ．12，1，2，1-D ．1-，1，2，127．已知0.70.70.7log 0.8,log 0.9,log 1.1a b c ===，那么A ．a b c <<B ．a c b <<C ．c b a <<D ．c a b <<8．我市2008年底人口总数约为100万，经统计近年来我市的年人口增长率约为10％，预计到2011年底我市人口总数将达到（ ）万人（精确到0．1）．A ．121B ．133．1C ．133．2D ．146．49．根据表格中的数据，则方程20x e x --=的一个根所在的区间可为x1-0 123x e12x + 1 23 4 5A ．(1,0)-B ．(0,1)C ．(1,2)D ．(2,3)10．某公司为了适应市场需求对产品结构做了重大调整,调整后初期利润增长迅速，后来增长越来越慢，若要建立恰当的函数模型来反映该公司调整后利润y 与时间x 的关系，可选用A ．一次函数B ．二次函数C ．指数型函数D ．对数型函数11．若1ab =（其中1,1a b ≠≠），则函数()log a f x x =与函数()log b g x x =的图象A ．关于x 轴对称B ．关于y 轴对称C ．关于原点对称D ．关于直线y x =对称12．（必修1第三章习题3-5 B 组第3题改编）关于函数xxx f +-=11lg )(，有下列三个命题：①对于任意)1,1(-∈x ，都有0)()(=-+x f x f ； ②)(x f 在)1,1(-上是减函数；③对于任意)1,1(,21-∈x x ，都有)1()()(212121x x x x f x f x f ++=+；其中正确命题的个数是 （ ）A ．0B ．1C ．2D ．3第Ⅱ卷（非选择题共90分）二、填空题（本大题共4小题，每小题4分，共16分）13．已知25(1)()21(1)x x f x x x +>⎧=⎨+≤⎩，则[(1)]f f = 。
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—第一学期期中 届高一数学试题 一、选择题：（本大题共12小题，每小题4分，共48分。
只有一项符合题目要求） 1．已知集合{}
0)1(=-=x x x A ，那么 （ ）
A ．A ∈0
B. A ∉1
C. A ∈-1
D. A ∉0
2．设a 为非零实数，则
a
a
的值组成的集合是 （
）
A.1
B.{
}1 C.1±
D.{}1,1-
3．若B A f →:能构成映射，下列说法正确的有 （ ）
⑴A 中的任一元素在B 中必须有像且唯一；
⑵B 中的多个元素可以在A 中有相同的原像； ⑶B 中的元素可以在A 中无原像； ⑷像的集合就是集合B A.1个 B.2个 C.3个 D.4个
4．设3log 2
1=a ，2
.0)31(=b ，31
2=c ，则c b a ,,的大小顺序是（
）
A. c b a <<
B. a b c <<
C. b a c <<
D. c a b <<
5．函数()()1log 32++-=x x x f 的定义域是
（
）
A.[)3,1-
B.()3,1-
C.(]3,1-
D.[]3,1-
6．函数()5222
+-+=x a x y 在区间()+∞,4上是增函数，则实数a 的取值范围是（ ） A.(]2,-∞-
B. [)+∞-,2
C.(]6,-∞-
D.[)+∞-,6
7．在同一坐标系中，函数x y 3log =与x
y 1
log 3=的图像之间的关系是 （ ）
A ．关于y 轴对称
B.关于x 轴对称
C.关于原点对称
D.关于直线y=x 对称
8．已知函数()x f y =是R 上的偶函数，且()1+x f 在[)+∞-,1上是减函数，若()()2-≥f a f ，则a 的取值范围是（ ） A. (]2,∞-
B. [)+∞,2
C.(][)+∞⋃∞-,22,
D.[]2,2-
9．已知{}2,log |2<==x x y y A ，⎭
⎬⎫⎩
⎨⎧
<==2,)2
1(|x y y B x
，则B A ⋂=（
）
A.φ
B.⎪⎭
⎫ ⎝⎛1,41
C.⎪⎭
⎫ ⎝⎛41,0
D.⎪⎭
⎫ ⎝⎛∞-41,
10．已知函数()x f 是R 上的增函数，()()1,3,1,0B A -是其图像上的两点，那么()1<x f 的
解集的补集是 （ ）
A. ()3,1-
B. ()3,0
C. (][)+∞⋃∞-,30,
D. (][)+∞⋃-∞-,31,
11．函数()x f 定义域是R ，且对任意R y x ∈,，()()()y f x f y x f +=+恒成立，则下列选项中不恒成立的是
（
）
A.()00=f
B.()()122f f =
C.()12
1
21f f =
⎪⎭
⎫ ⎝⎛ D.()()0<-x f x f
12．已知函数()()132
+-+=x m mx x f 的图像与x 轴的交点至少有一个在原点右侧，则实数m 的取值范围是
（
）
A. (]1,0
B.()1,0
C.()1,∞-
D.(]1,∞-
二．填空题（本大题共4小题，每小题4分，共16分）
13．函数()12
+-=x x x f 在定义域[]2,0上的值域为 。
14．已知()()
()
⎩⎨
⎧>-≤+=0506
2x x
x x x f ，若()10=x f ，则x= 。
15．已知函数()x f y =是奇函数，当0≥x 时，()13-=x
x f ，则()2-f = 。
16．函
数
()()1,0log ≠>=a a x x f a ，若()2010
201021=⋅⋅⋅x x x f ，则
()()
()2
2010
2
22
1x f x f x f +⋅⋅⋅++= 。
三．解答题（本大题共5小题，共56分，解答应写出文字说明，证明过程或演算过程） 17．（本小题满分10分）求函数()()4lg 3
1
++-+-=x x x x f 的定义域。
18．（本小题满分10分）
⑴画出函数()4-=x x y 的图像。
⑵利用图像回答：当k 为何值时，方程
()k x x =-4有一解？有两解？有三解？
19．（本小题满分12分）设{
}
04|2
=+=x x x A ，
(){}
0112|22=-+++=a x a x x B 且
B B A =⋂,求实数a 的取值范围。
20（本小题满分12分）一片森林原来面积为a ，计划每年砍伐一些树，且每年砍伐面积的百分比相等，当砍伐到面积一半时，所用的时间是10年。
为保护生态环境，森林面积至少要保留原面积的
4
1，已知到今年为止，森林剩余面积为原来的22
⑴到今年为止，该森林已砍伐了多少年？
⑵今后最多还能砍伐多少年？
21．（本小题满分12分）已知()a
b
x f x x ++-=+122是定义域为R 的奇函数。
⑴求()x f 的解析式；并指出函数()x f 的增减性。
⑵若对于实数[]2010,3∈x ，不等式()
09292
<--+⎪⎭
⎫ ⎝⎛+t t f x x f 恒成立，求实数t 的取值范围。