厦门双十中学2015-2016学年（上）期中考试高一数学试题（2015-11-10 上午08：00-10：00）第Ⅰ卷（选择题 共50分）一、选择题：本大题共10小题，每小题5分，共50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.请把答案填涂在答题卷的相应位置.1． 设全集U 是实数集R ，{}{}1,02M x x N x x =<=<<都是U 的子集，则图中阴影部分所表示的集合是A ．{}12x x ≤<B ．{}01x x <<C ．{}0x x ≤D ．{}2x x <2． 下列函数中与x y =相等的是A ．2)(x y =B ．2x y =C ．x y 2log 2=D ．x y 2log 2=3． 若函数()(2)()xf x x x a =-+是奇函数，则a =A ． 2-B ．2C ．12-D ．124． 给定映射f ：()(),2,2x y x y x y →+-，在映射f 下，(3,1)-的原像为A ．(1,3)-B ．(5,5)C ．(3,1)-D ．(1,1)5． 已知函数2,0,()(1),0.x x f x f x x ⎧>=⎨-+≤⎩则(3)f -的值为A ．1B ．1-C ．0D ．9-6． 已知,k b ∈R ，则一次函数y kx b =+与反比例函数kby x=在同一坐标系中的图象可以是7． 已知()f x 是定义在R 上的偶函数，且在(],0-∞上是增函数，设()4log 7a f =，)3(log 2f b =，()0.60.2c f =，则,,a b c 的大小关系是8． 已知函数32()21f x x x x =+--，可用二分法计算其一个正数零点的近似值（精确度0.1）为参考数据：A ．1.5 D ．1.18759． 函数(13)2,1(),1xa x x f x a x -+≤⎧=⎨>⎩是R 上的减函数，则实数a 的取值范围为A ．1(,1)3B ．3[,1)4C ．13(,)34D ．13(,]3410．当实数k 变化时，对于方程||2||(21)(21)0x x k ----=的解的判断不正确．．．的是 A ．14k <-时，无解B ．14k =-时，有2个解 C ．104k -<≤时，有4个解 D ．0k >时，有2个解第Ⅱ卷（非选择题 共100分）二、填空题：本大题共6小题，每小题4分，共24分.请把答案填在答题卷的相应位置.11．函数()1f x x =-的定义域．．．为 ▲ . 12．已知3()2f x ax bx =+-，若(2015)7f =，则(2015)f -的值为 ▲ .13．已知全集U =R ，集合2{|0},{|320}A x x a B x x x =-≤=-+≤，且U A B = ðR ，则实数a 的取值范围是 ▲ .14．已知函数2()f x x ax b =++的零点是3-和1，则函数2()log ()g x ax b =+的零点是 ▲ . 15．若函数()6,2,2log ,2,a x x f x x x -+≤⎧=⎨+>⎩ （0a >，且1a ≠）的值域是[)4,+∞，则实数a 的取值范围是▲ .16．方程210x -=的解可视为函数y x =1y x=的图象交点的横坐标. 若方程490x ax +-=的各个实根12,,,(4)k x x x k ≤ 所对应的点9(,)(1,2,,)i ix i k x = 均在直线y x =的同三、解答题：本大题共6小题，每小题分数见旁注，共76分.解答应写出文字说明，证明过程或演算步骤.请在答题卷相应题目的答题区域内作答. 17．（本小题满分12分）（Ⅰ）求值：2lg5lg400⋅+；（Ⅱ）已知2log 3x =，求8822x xx x--++的值.18．（本小题满分12分）已知集合11{|132},{|24}4x A x a x a B x -=-<<+=<<. （Ⅰ）若1a =，求A B ；（Ⅱ）若A B =∅ ，求实数a 的取值范围. 19．（本小题满分12分）设函数1331()log (9)log ,2739x f x x x =⋅≤≤. （Ⅰ）设3log t x =，用t 表示()f x ，并指出t 的取值范围； （Ⅱ）求()f x 的最值，并指出取得最值时对应的x 的值.20．（本小题满分13分）小张周末自己驾车旅游，早上8点从家出发，驾车3 h 后到达景区停车场，期间由于交通等原因，小张的车所走的路程s （单位：km ）与离家的时间t （单位：h ）的函数关系式为s (t )＝－4t (t －13)． 由于景区内不能驾车，小张把车停在景区停车场．在景区玩到17点，小张开车从停车场以60 km/h 的速度沿原路返回．（Ⅰ）求这天小张的车所走的路程s （单位：km ）与离家时间t （单位：h ）的函数解析式； （Ⅱ）在距离小张家48 km 处有一加油站，求这天小张的车途经该加油站的时间．21．（本小题满分13分）已知函数2()1px q f x x +=+（,p q 为常数）是定义在(1,1)-上的奇函数，且1(1)2f =. （Ⅰ）求函数()f x 的解析式；（Ⅱ）判断并用定义证明()f x 在(1,1)-上的单调性； （Ⅲ）解关于x 的不等式(21)()0f x f x -+<.22．（本小题满分14分），其中a ∈R ．（Ⅰ）当1a =-时，在所给坐标系中作出()f x 的图象；（Ⅱ）对任意[1,2]x ∈，函数()f x 的图象恒在函数()14g x x =-+图象的下方，求实数a 的取值范围；（Ⅲ）若关于x 的方程()10f x +=在区间(1,0)-内有两个相异根，求实数a 的取值范围.一、选择题：本大题共10小题，每小题5分，共50分.二、填空题：本大题共6小题，每小题4分，共24分.11．【答案】21x x x ≤≠且 【解析】试题分析：根据题意，要使得函数()1f x x =-有意义，则要满足102,1420xx x x -≠⎧∴≤≠⎨-≥⎩且，故可知答案为{}21x x x ≤≠且.考点：函数定义域点评：解决的关键是根据分母不为零，偶次根式下为非负数，属于基础题。

12．【答案】11- 【解析】试题分析：显然(2015)(2015)4f f +-=-，所以(2015)4(2015)11f f -=--=- 考点：函数的奇偶性． 13．【答案】2a ≥ 14．【答案】2 15．【答案】【解析】当2x ≤，64x -+≥，要使函数()f x 的值域为[)4,+∞，只需1()2log (2)a f x x x =+>的值域包含于[)4,+∞，故1a >，所以1()2log 2a f x >+，所以2log 24a +≥，解得1a <≤【考点定位】分段函数求值域． 16．【答案】24a <-或24a >三、解答题：本大题共6小题，每小题分数见旁注，共76分. 17．（本小题满分12分） 【解析】 （Ⅰ）原式22lg 5(22lg 2)2)2lg 52lg 2lg 52(lg 2)2lg 52lg 2(lg 5lg 2)2lg 52lg 22=⋅++=+⋅+=+⋅+=+= ······················································································································· 6分 （Ⅱ）因为2log 3x =，所以23x=， ·············································································· 8分所以88(22)(414)17341491222299x x x x x x x xx x x x ------++-+==-+=-+=++. ························· 12分注意：公式化简正确，结果算错扣2分. 18．（本小题满分12分） 【解析】由11244x -<<可得212x -<-<，解得13x -<<，所以{|13}B x x =-<< ························ 4分 所以{|03}A B x x =<< . ··························································································· 6分（Ⅱ）若A =∅，则132a a -≥+，解得32a ≤-； ···································································· 8分 若A ≠∅，则32a >-，由于A B =∅ ，所以13a -≥或321a +≤-， ······························ 10分 解得312a -<≤-或4a ≥ ····························································································· 11分 综上，实数a 的取值范围是：(,1][4,)-∞-+∞ . ······························································ 12分注意：未讨论空集统一扣2分，未注意端点取等号统一扣1分，同一错误不重复扣分. 19．（本小题满分12分） 【解析】 （Ⅰ）设3log t x =，因为1279x ≤≤，所以32log 3x -≤≤，即23t -≤≤. ··························· 3分 此时，33332()log (9)log 3(log 2)(log 1)2x f x x x x t t =-⋅=-+⋅-=--+ 即2()2f x t t =--+，其中23t -≤≤ ············································································· 6分（Ⅱ）由（Ⅰ）可得，2219()2()24f x t t t =--+=-++ ···················································· 7分又23t -≤≤，函数22y t t =--+在1[2,)2--单调递增，在1(,3]2-单调递减， ······················ 8分所以当12t =-，即31log 2x =-，即x =时，()f x 取得最大值94； ································ 10分所以当3t =，即3log 3x =，即27x =时，()f x 取得最小值10-. ······································· 12分20．（本小题满分13分） 【解析】（Ⅰ）依题意得，当0≤t ≤3时，s (t )＝－4t (t －13)，∴s (3)＝－4×3×(3－13)＝120. ························································································ 2分 即小张家距离景点120 km ，小张的车在景点逗留时间为17－8－3＝6(h)． ·································································· 3分 ∴当3<t ≤9时，s (t )＝120， ··························································································· 4分 小张从景点回家所花时间为12060＝2(h)， ·········································································· 5分 ∴当9<t ≤11时，s (t )＝120＋60(t －9)＝60t －420. ································································ 7分 综上所述，这天小张的车所走的路程s (t )＝⎩⎪⎨⎪⎧－4t (t －13) 0≤t ≤3120 3<t ≤960t －420 9<t ≤11 ························································································· 8分（Ⅱ）当0≤t ≤3时，令－4t (t －13)＝48，得t 2－13t ＋12＝0，解得t ＝1或t ＝12(舍去)， ··········· 10分 当9<t ≤11时，令60t －420＝2×120－48＝192，解得t ＝515. ················································· 12分 答：小张这天途经该加油站的时间分别为9点和18时12分． ············································ 13分 21．（本小题满分13分） 【解析】（Ⅰ）依题意，(0)0,1(1),2f f =⎧⎪⎨=⎪⎩解得1,0p q ==，所以2()1x f x x =+. ········································ 3分 （Ⅱ）函数()f x 在(1,1)-上单调递增，证明如下： ···························································· 4分 任取1211x x -<<<，则12120,11x x x x -<-<<，从而 ····················································· 5分 12122212221221221212122212()()11(1)(1)(1)(1)()(1)(1)(1)f x f x x x x x x x x x x x x x x x x x -=-+++-+=++--=++< 所以12()()f x f x <， ··································································································· 7分 所以函数()f x 在(1,1)-上单调递增. ················································································ 8分 （Ⅲ）原不等式可化为：(21)()f x f x -<-，即(21)()f x f x -<- ······································ 9分 由（Ⅱ）可得，函数()f x 在(1,1)-上单调递增，所以1211,11,21,x x x x -<-<⎧⎪-<<⎨⎪-<-⎩········································································································· 11分 解得103x <<，即原不等式解集为1(0,)3. ········································································ 13分 22．（本小题满分14分）【解析】······································ 1分据此可作出图像如下：（Ⅱ）由题意，对任意[1,2]x ∈，()()f x g x <即()14f x x +<恒成立，只需()max ()14f x x +< ································································································································ 5分另一方面，222,()()32,()x ax x a f x x ax x a ⎧-+≤=⎨->⎩，即当0a ≥时，()f x 在(,)a -∞和(,)a +∞上均递增，∵2()f a a =，则()f x 在R 上递增， ··········· 6分 当0a <时，()f x 在(,)a -∞和······································ 7分 故()f x 在[1,2]x ∈上恒单调递增，从而()y f x x =+在[1,2]x ∈上也恒单调递增， ·················· 8分，解得04a <<，故实数a 的取值范围是(0,4). ························································································· 9分注意：本小题若未证明单调性，直接代端点并得出正确答案的统一给2分.（Ⅲ）记()()1F x f x =+，考虑()F x 在区间(1,0)-内有两个不同的零点即可.此时，2221,()()321,()x ax x a F x x ax x a ⎧-++≤=⎨-+>⎩，即 ·············· 10分 则由（Ⅱ）可知，当0a ≥时，()()1F x f x =+在R 上递增，方程()10f x +=在区间(1,0)-内至多有一个根，不符合要求，舍去；故0a <. ································································································· 11分 当x a ≤时，令()0F x =，可得1x a =（不符合x a ≤，舍去）或2x a =-，但21x a =<-，不在区间(1,0)-内，因此 ·············································· 12分当x a >时，2()321F x x ax =-+在区间(1,0)-内必有两个不同的零点，从而(1,0)(,)a -⊆+∞，所以210,34120,(0)0,(1)0,1a a f f a ⎧-<<⎪⎪∆=->⎪⎪>⎨⎪->⎪≤-⎪⎪⎩解得2a -<<···································································· 14分。