上海市高一上学期期中考试试卷数 学考试注意：试卷分第Ⅰ卷、第Ⅱ卷两部分。

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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，每小题给出的四个选项中，只有一项符合要求）1、已知{}{}2lg 0,(1)4,A x x B x x =>=-< 则A B =( ) {}.11A x x x <-≥或 {}.13B x x << {}.1C x x >- {}.3D x x >2、若函数23,12()3,25x x f x x x ⎧--≤≤=⎨-<≤⎩则方程()1f x =的解是( )A .4B 3C . 2D .43、方程43x e x =-+的根所在区间是( )A .1,04⎛⎫- ⎪⎝⎭B .10,4⎛⎫ ⎪⎝⎭C . 13,24⎛⎫ ⎪⎝⎭D . 11,42⎛⎫ ⎪⎝⎭4、下列函数中，与y x =是同一函数的是( )()1y = ()2log ;x a y a = ()log 3;a x y a = ()4y = ())5.y n N =∈*()().24A()().23B ()().12C ()().35D 5、若432a =，254b =，3log 0.2c =，则,,a b c 的大小关系是( ).Aa b c << .B c b a << .C b a c << .D c a b <<6、函数()2101x b y a a a +=+>≠且恒过定点()1,2，则b =( ).3A .3B - .2C - .1D7.已知函数()2221()1mm f x m m x --=--是幂函数，则m =( ) .0A .2B - .1C .2D8、函数2()23f x x ax =--在区间[]1,2上是单调函数，则( ).A (),1a ∈-∞ .B ()2,a ∈+∞ .C (][),12,a ∈-∞+∞ .D [)()1,22,a ∈+∞9、函数()21,x f x =-使()0f x ≤成立的x 的集合是( ).A {}0x x < .B {}=0x x .C {}1x x < .D {}1x x =10、函数lg x y x=的图像大致是( )11、若函数()212x x f x a+=-是奇函数，则使()3f x >成立的x 的取值范围是( ) .A ()1,1- .B (]1,1- .C [)0,1 .D ()0,112、某同学在研究函数()||1x f x x =+()x ∈R 时，分别给出下面几个结论： ①函数()f x 是奇函数； ②函数()f x 的值域为()1 1-，； ③函数()f x 在R 上是增函数； 其中正确结论的序号是( ).A ①②③ .B ①③ .C ②③ .D ①②第Ⅱ卷 非选择题（共90分）二、填空题（本大题共4小题，每小题5分）13、设,,a b R ∈集合{}1,,=0,,,b a b a b b a a ⎧⎫+-=⎨⎬⎩⎭则___________. 14、定义在R 上的函数()f x 是奇函数且每隔2个单位的函数值都相等,则()()()147f f f ++=_____________.15、已知集合{{}=,1,,,A B m A B A m ===则_____________.16、已知函数()f x 是定义在R 上的偶函数，当0x ≥时，2()2f x x x =-，如果函数()()g x f x m =-恰有4个零点，则实数m 的取值范围是 ．三、解答题（本大题6小题，共70分，解答应写出必要的文字说明，证明过程或演算步骤）17、(本小题10分)(1)已知函数()2(2)2x f x log ＝－，求()f x 的定义域；(2)解不等式2122x +⎛⎫> ⎪⎝⎭18、(本小题12分) 已知集合{}{}{}37,210,5.A x x B x x C x a x a =≤<=<<=-<<(1) 求A B 与()R A B .(2) 若(),A B C ⊆求实数a 的取值范围.19、(本小题12分)已知函数()()(10)x x f x ln a b a b >>>＝－．(1)求函数()f x 的定义域；(2)判断函数()f x 在定义域上的单调性，并说明理由；(3)当,a b 满足什么关系时,()f x 在[)2,+∞上恒取正值.20、(本小题12分)已知:函数()()()log 1log 1a a f x x x =+--（0a >且1a ≠）.(1)求函数()f x 的定义域;(2)判断函数()f x 的奇偶性,并加以证明;(3)设3a =,解不等式()0f x >.21、(本小题12分)求函数()()212log 43g x x x =++的单调区间(写出解答过程)22、(本小题12分)已知函数()f x 对一切实数,x y 都满足()()2)1(f x y f y x y x ＋＝＋＋＋,且()10f ＝.(1)求()0f 的值；(2)求()f x 的解析式；(3)当10,2x ⎡⎤∈⎢⎥⎣⎦时，()32f x x a <＋＋恒成立，求a 的范围．高一上学期期中考试答案一、选择题二、填空题三、解答题17、（本小题满分10分）解析：(1)由条件可知220,22x x ->∴<，······································2分 函数2x y =在R 上单调递增·····················································3分()1,,1x ∴<-∞即定义域为.···························································5分(2)2211112,222x x ++-⎛⎫⎛⎫⎛⎫ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭>∴> ··················································7分 又函数12x y ⎛⎫ ⎪⎝⎭=在R 上单调递减，21 3.x x ∴+<-∴<-，·····················9分 {}2123.2x x x +⎛⎫ ⎪⎝⎭∴><-不等式的解集为 ········································10分18．（本小题满分12分）解析：(1)由条件可知{}210x x A B B <<==································3分 又{}=37R A x x x <≥或.····························································4分(){}23710R A B x x x ∴=<<≤<或·············································6分(2) {}210x x C A B B <<⊆==··············································8分∴分析可知 10525a a a a ⎧⎪⎨⎪⎩≤-<-< ，解得10.a ≥ ········································10分 )10,.a ⎡⎣∴∈+∞ ·····································································12分19．（本小题满分12分）解析：(1)0,1x x x a a b b ⎛⎫ ⎪⎝⎭->∴>, ················································1分 又1a b>，0x ∴>·····································································2分∴定义域为()0,+∞.··································································3分 (2)函数在定义域上是单调递增函数.证明：1212,0x x x x ∀<<且，121210,,.x x x x a b a a b b >>>∴<> ················4分 1122x x x x a b a b ∴-<- ··································································6分()()1122ln ln x x x x a b a b ∴-<- ·························································7分()()12f x f x ∴<所以函数()f x 在定义域上是单调递增函数. ···································8分(3)要使得()f x 在)2,⎡⎣+∞上恒为正值，则()f x 在)2,⎡⎣+∞上的最小值必须大于0， 由(2)知()()22min 2ln()f x f a b ==- ················································9分22ln()0,a b ∴->22 1.a b ∴-> ··········································································11分∴()f x 在)2,⎡⎣+∞上恒为正值时，22 1.a b ->····································12分20．（本小题满分12分）解析：(1)由题意可知x 满足10,10x x ⎧⎨⎩+>-> 解得11x -<<，····································2分 ∴函数()f x 的定义域为()1,1- ··································································3分(2)()f x 是奇函数.证明：函数的定义域为()1,1-，关于原点对称，对于任意的()1,1x ∈-·················4分 ()log (1)log (1)a a f x x x -=-+++()log 1log (1)a a x x ⎡⎤⎣⎦=-+-- ······································································6分()f x =-所以函数()f x 是奇函数. ········································································8分(3)当3a =时，()330,log (1)log (1)f x x x >∴+>-3log y t =在定义域上是单调递增函数·····················································10分x ∴满足11,11x x x ⎧⎨⎩+>--<< 解得0 1.x << 所以不等式()0f x >的解集为{}01.x x << ················································12分21．（本小题满分12分）解析：由题意，2430,x x ++>解得3 1.x x <->-或()g x ∴的定义域为()(),31,.-∞--+∞ ·······················································3分 又243y x x =++在(),3-∞-上是单调递减函数，···········································5分 在()1,-+∞上是单调递增函数··································································7分 12log y x =在()0,+∞上是减函数·····························································9分 ∴由复合函数单调性可知：函数()()212log 43g x x x =++的单调递增区间是(),3-∞-，单调递减区间是()1,-+∞.······································································································12分22. （本小题满分12分）解析：(1)令()()()()()1,0,10111,012 2.x y f f f f ===++⨯∴=-=-则····························4分(2)()()0()01y f x f x x 令＝，则＝＋＋()2 2.f x x x ∴＝＋－···············································································6分(3)()232,1f x x a a x x +<+>-+由得； 设2211,1,2y x x y x x ⎛⎤ ⎥⎝⎦=-+=-+-∞则在上是减函数；·······································8分 所以2110,2y x x ⎡⎤⎢⎥⎣⎦=-+在上的范围为314y ≤≤；··········································10分1a ∴>.····························································································12分。