【月考试卷】福建省四校2017-2018学年高二下学期第二次联考试题(5月)数学(文)Word版含答案

“平和一中、华安一中、长泰一中、南靖一中”四校联考2017-2018学年第二学期第二次月考高二文科数学考试时间120分钟。

满分150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，只有一项是符合题目要求的。

1．已知集合{}29M x x =≤，{}2,3N =-，则M N =A. ∅B. {}2-C. {}3D. {}2,3- 2. 设,a b ∈R ，i 为虚数单位，且1i12ib a +=-，则b = A. 2- B. 1- C ．1 D. 23. 命题“x ∀∈R ,总有220x +>”的否定是A. “x ∀∉R ，总有220x +>”B. “x ∀∈R ，总有220x +≤”C. “x ∃∈R ，使得220x +>”D. “x ∃∈R ，使得220x +≤”4.“0a =”是“直线()1:110l ax a y -++=与直线2:210l x ay --=垂直”的 A. 充分不必要条件 B. 必要不充分条件 C. 充要条件 D. 既不充分也不必要条件5.若函数()21f x ax a =+-在区间)1,1(-上存在一个零点，则a 的取值范围是 A. 13a >B. 13a >或1-<aC. 113a -<< D. 1a <- 6．《孙子算经》是我国古代的数学名著，书中有如下问题：“今有五等诸侯，共分橘子六十颗，人别加三颗．问：五人各得几何？”其意思为：“有5个人分60个橘子，他们分得的橘子个数成公差为3的等差数列，问5人各得多少橘子．”根据这个问题，有下列3个说法：①得到橘子最多的人所得的橘子个数是15；②得到橘子最少的人所得的橘子个数是6；③得到橘子第三多的人所得的橘子个数是12．其中说法正确的个数是 A ．0 B ．1 C ．2 D ．37．已知213311,,ln 323a b c ⎛⎫⎛⎫=== ⎪ ⎪⎝⎭⎝⎭，则,,a b c 的大小关系为A. a b c >>B. a c b >>C. c a b >>D. c b a >>8. 若定义运算,,b a b a b a a b <⎧⊕=⎨⎩，≥，则函数()()()32log 1f x x x =+⊕-⎡⎤⎣⎦的值域是A. 把()na b +与()nab 类比，则有()nn na b a b +=+B. 把长方体与长方形类比，则有长方体的对角线平方等于其长宽高的平方和C. 把()log a x y +与()a b c +类比，则有()log log log a a a x y x y +=+D. 向量a ，b 的数量积运算与实数,a b 的运算ab a b =⋅类比，则有⋅=⋅a b a b 11．函数()ln ,0,cos ,0x x f x x x >⎧=⎨π⎩≤的图象上关于y轴对称的点共有A. 1对B. 2对C. 3对D. 4对 12.已知12,x x ∈R ，则()()212212e e x x x x -+-的最小值等于 A.12B.C. D. 2二、填空题：本大题共4小题，每小题5分，共20分。

13．设a ∈R ，若复数()()1i i a ++在复平面内对应的点位于实轴上，则a =_____． 14．已知f (x )为定义在R 上的奇函数，当x ≥0时，()2xf x m =+，则()3f -＝_____．15.二维空间中，圆的一维测度（周长）2l r =π，二维测度（面积）2S r =π；三维空间中，球的二维测度（表面积）24S r =π，三维测度（体积）343V r =π．应用合情推理，若四维空间中，“特级球”的三维测度312V r =π，则其四维测度W =_____．16.已知定义域为R 的函数()f x 的导函数为()f x '，且()()0f x f x '->，()12f =，则不等式()12ex f x ->的解集为_____．三、解答题：共70分。

解答应写出文字说明、证明过程或演算步骤。

第17～21题为必考题，每个试题考生都必须作答。

第22题、第23题为选考题，考生根据要求作答。

（一）必考题：共60分。

17．(12分)设函数(),0,2,0,x ax b x f x x +<⎧=⎨⎩≥ 且f (－2)＝3，f (－1)＝f (1).(1)求f (x )的解析式；(2)在如图所示的直角坐标系中画出()f x 的图象．18．(12分)设命题:p 实数x 满足()()30x a x a --<，其中0a >，命题:q 实数x 满足()()28240xx--≤．（1）若1a =，且p q ∧为真，求实数x 的取值范围； （2）若p ⌝是q ⌝的充分不必要条件，求实数a 的取值范围．19.（12分）已知函数x ax x x f ln 1)(2-++-=在1=x 处取得极值． (1)求)(x f ，并求函数)(x f 在点))2(,2(f 处的切线方程； (2) 求函数)(x f 的单调区间．20. （12分）“双十一网购狂欢节”源于淘宝商城（天猫）2009年11月11 日举办的促销活动，当时参与的商家数量和促销力度均有限，但营业额远超预想的效果，于是11月11日成为天猫举办大规模促销活动的固定日期.如今，中国的“双十一”已经从一个节日变成了全民狂欢的“电商购物日”.某淘宝电商分析近8年“双十一”期间的宣传费用x （单位：万元）和利润y （单位：十万元）之间的关系，得到下列数据：（1）请用相关系数r 说明y 与x 之间是否存在线性相关关系（当0.81r >时，说明y 与x 之间具有线性相关关系）；（2）根据（1）的判断结果，建立y 与x 之间的回归方程，并预测当24x =时，对应的利润ˆy为多少（ˆˆˆ,,ba y 精确到0.1）. 附参考公式：回归方程中ˆˆˆybx a =+中ˆb 和ˆa最小二乘估计分别为1221ˆni ii nii x ynx y bxnx ==-=-∑∑,ˆˆay bx =-，相关系数ni i x ynx yr -=∑参考数据：88211241,6i ii i i x yx ====≈=∑∑．21．（12分）已知函数()e xf x x a b =++（,a b ∈R ）在0x =处取得极值．（1）求()f x 的单调区间；（2）讨论()f x 的零点个数，并说明理由．（二）选考题：共10分。

请考生在第22、23两题中任选一题作答。

如果多做，则按所做的第一题计分。

22．[选修44-：坐标系与参数方程] (10分)在极坐标系中，曲线1C ：2cos ρθ=，曲线2C ：2sin 4cos ρθθ= .以极点为坐标原点，极轴为x 轴正半轴建立直角坐标系xOy ，曲线C的参数方程为12,2x t y ⎧=+⎪⎪⎨⎪=⎪⎩（t 为参数）.（1）求1C ，2C 的直角坐标方程；（2）C 与1C ，2C 交于不同四点，这四点在C 上的排列顺次为,,,P Q R S ，求PQ RS -的值．23．[选修45-：不等式选讲] (10分)已知函数()2f x x a x =-+-． （1）当=1a 时,解不等式()2f x …；（2）求证:()2f x a -≥d ．“平和、华安、长泰、南靖一中”四校联考2017-2018学年第二学期第二次月考高二文科数学参考答案评分说明：1．本解答给只出了一种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则。

2．解答右端所注分数，表示考生正确做到这一步应得的累加分数。

3．只给整数分数。

选择题和填空题不给中间分。

一、选择题：本大题考查基础知识和基本运算。

每小题5分，满分60分。

1．D ２．A ３．D ４．A ５．B ６．C７．D８．A９．C １0．B １1．C１2．D二、填空题：本大题考查基础知识和基本运算。

每小题5分，满分20分。

1３．1- 1４．7- 1５．43r π 1６．(),1-∞三、解答题：本大题共6小题，共70分。

解答应写出文字说明，证明过程或演算步骤。

17．本小题满分12分．解：(1)由f (－2)＝3，f (－1)＝f (1)得23,2,a b a b -+=⎧⎨-+=⎩ ··································································· 4分解得a ＝－1，b ＝1， ········································································································ 7分 所以()1,0,2,0.xx x f x x -+<⎧=⎨⎩≥ ······························································································· 8分 (2)f (x )的图象如图．································································································································ 12分 18．本小题满分12分．解：（1）当1a =时，由()()130x x --<，得{}13P x x =<<． ··········································· 2分由()()28240x x --≤，得2x4≤≤8，所以{}23Q x x =≤≤． ········································· 4分由p ∧q 为真，即p ，q 均为真命题，因此x 的取值范围是[)2,3． ·························································································· 6分 （2）若￢p 是￢q 的充分不必要条件，可得q 是p 的充分不必要条件，······································· 8分 由题意可得{}3P x a x a =<<，{}23Q x x =≤≤, ························································· 10分所以Q P ⊂，因此2a <且33a <，解得12a <<． ························································· 12分 19．本小题满分12分．解：(1)因为2()1ln f x x ax x =-++-，所以1()2(0)f x x a x x'=-+->． ······································ 1分 因为()f x 在1=x 处取得极值，所以(1)0f '=，即210a -+-=，解得所以3a =． ············································································································· 3分 因为1()23(0)f x x x x '=-+->，2ln 3)2(-=f ，3(2)2f '=-， 所以函数()f x 在点(2,(2))f 处的切线方程为36ln 22y x =-+-． ··········································· 6分 (2)由(1) 1()23(0)f x x x x'=-+->， 令()0f x '>，即1230x x -+->，解得121<<x ， 所以()f x 的单调递增区间为1(,1)2． ·················································································· 9分 令()0f x '<，即1230x x -+-<，解得210<<x 或1>x ， 所以()f x 的单调递减区间为)21,0(，),1(+∞．综上，()f x 的单调递减区间为1(0,)2和(1,)+∞，单调递增区间为1(,1)2． ································ 12分20．本小题满分12分．解：（1）由题意得6,4x y ==． ························································································ 2分又88211241,6i i ii i x y x ====≈=∑∑，所以882418640.990.818.256i i x yx yr --⨯⨯=≈≈>⨯∑， ······································· 5分 所以y 与x 之间具有线性相关关系. ···················································································· 6分因为81822218241864ˆ0.7356868i ii ii x yx ybxx ==--⨯⨯==≈-⨯-∑∑， ··································································· 8分（2）因为ˆˆ40.760.2a y bx =-≈-⨯=-， ········································································ 10分 所以回归直线方程为ˆ0.70.2yx =-， 当24x =时，ˆ0.70.20.7240.216.6yx =-=⨯-=， 即利润约为166万元. ······································································································ 12分 21．本小题满分12分．解：（1）因为()1e xf x a '=+， ································································································· 1分又()00f '=，即10a +=，解得1a =-． ·································································· 2分 令()0f x '＞，即1e 0x-＞，解得0x ＜；令()0f x '＜，即1e 0x-＜，解得0x ＞． ······································································ 4分所以()f x 的单调递增区间为(),0-∞，单调递减区间为()0,+∞． ····································· 5分 （2）由（Ⅰ）知()f x 在0x =处取得最大值1b -． ······························································ 6分 ①当10b -＜即b ＜1时，()0f x ＜，所以()f x 无零点． ······················································ 7分 ②当10b -=即b =1时，当且仅当0x =时，()0f x =，所以()f x 有一个零点． ······························································································· 8分③当10b -＞即b ＞1时，()()max 010f x f b ==-＞，因为0b -＜，且()ee 0bb f b b b ---=--+=-＜，又()f x 在(),0-∞上单调递增，所以()f x 在(),0-∞上有且只有一个零点． ························· 10分因为1b ＞，且()e 2e bbf b b b b =-+=-， 令()2e ,1xg x x x =-≥，则()2e 0xg x '=-＜，所以()g x 在[)1,+∞上单调递减，所以()()12e 0g x g =-≤＜， 所以()0f b ＜．又()f x 在()0,+∞上单调递减，所以()f x 在上()0+∞，有且只有一个零点．故当b ＞1时，()f x 有两个零点． ················································································· 12分22．本小题满分10分．解：（1）因为cos ,sin x y ρθρθ==， ··················································································· 1分 由2cos ρθ=得22cos ρρθ=， ···························································································· 2分 所以曲线1C 的直角坐标方程为()2211x y -+=． ····································································· 3分 由2sin 4cos ρθθ=得22sin 4cos ρθρθ=， ············································································· 4分 所以曲线2C 的直角坐标方程为: 24y x =． ············································································ 5分 （2）不妨设四个交点自下而上依次为,,,PQ R S ，它们对应的参数分别为1234,,,t t t t ． 把 12,2,x t y ⎧=+⎪⎪⎨⎪=⎪⎩代入24y x =，得234242t t ⎛⎫=+ ⎪⎝⎭，即238320t t --=, ·················································则()()21843324480∆=--⨯⨯-=＞，1483t t +=． ·································把 12,2,x t y ⎧=+⎪⎪⎨⎪=⎪⎩代入()2211x y -+=，得2212112t ⎫⎛⎫+-+=⎪ ⎪⎪⎝⎭⎝⎭，即20t t +=, ···········································分 则2=10∆＞，231t t +=-． ·································································································· 9分 所以21432314811|||||||()()||()||1|33PQ RS t t t t t t t t -=---=+-+=+=． ············································ 10分23．本小题满分10分．解：（1）当=1a 时,不等式()2f x ≥等价于不等式122x x -+-≥.当1x <时,不等式可化为122x x -+-≥,解得12x ≤，所以12x ≤． ··············································· 1分当1x ≤≤2时,不等式可化为122x x -+-≥,即21≥，这种情况无解． ··········································· 2分 当2x >时,不等式可化为122x x -+-≥,解得52x ≥，所以52x ≥． ··············································· 3分综上,当=1a 时,不等式()2f x ≥的解集为15,,.22⎛⎤⎡⎫-∞+∞ ⎪⎥⎢⎝⎦⎣⎭······················································· 5分（2）证明:()2f x x a x =-+-2a x x =-+- ················································································································ 7分2a x x -+-≥2.a =-所以不等式得证． ··········································································································· 10分。