厦门双十中学2014-2015学年（上）期中检测 高三数学（理科）试题（2014-11-13 08:00-10:00）【（试卷命题人：王成焱，审核人：张瑞炳）感谢高三数学（理科）备课组所有老师半学期的辛勤付出，你们辛苦了！当然，老师也在平时与同学们的交流中看到大家的不懈努力与对理想的执着与追求，在这阶段检测的时刻，让我们怀着感恩的心来证明自己吧！】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． 命题“对任意的x ∈R ，x 2＋1>0”的否定是（ ▲ ）A ．不存在x ∈R ，x 2＋1>0B ．存在x ∈R ，x 2＋1>0C ．存在x ∈R ，x 2＋1≤0D ．对任意的x ∈R ，x 2＋1≤02． 已知集合{}23,A a =，集合{}0,,1B b a =-，且{}1AB =，则A B =（ ▲ ）A ．{}0,1,3B ．{}0,1,2,3C ．{}1,2,4D ．{}0,1,2,3,43． “sin α≠sin β”是“α≠β”的（ ▲ ）A ．充分不必要条件B ．必要不充分条件C ．充分必要条件D ．既不充分也不必要条件4． 若a ，b ，c 为实数，且a <b <0，则下列命题正确的是（ ▲ ）A ．ac 2<bc 2B ．1a <1bC ．b a >a bD ．a 2>ab >b 25． 已知函数f (x )＝(x －a )(x －b )(其中a >b )的图像如下图所示，则函数g (x )＝a x ＋b 的图象是（ ▲ ）6． 设,x y 满足约束条件10,10,3,x y x y x -+≥⎧⎪+-≥⎨⎪≤⎩则23z x y =-的最小值是（ ▲ ）A ．3-B ．12C ．6-D ．12-7． 设斜率为2的直线l 过抛物线y 2＝ax (a ≠0)的焦点F ，且和y 轴交于点A . 若△OAF (O 为坐标原点)的面积为4，则抛物线的方程为（ ▲ ） A ．y 2＝4x B ．y 2＝8x C ．y 2＝±4x D ．y 2＝±8x8． 下列函数存在极值的是（ ▲ ）A. 2cos y x x =+B. ln x y e x =-C. 32331y x x x =++-D. 1ln y x x=-9． 定义：sin a b a b θ⨯=⋅⋅，其中θ为向量a与b的夹角，若2,5,6a b a b ==⋅=-，则a b ⨯=（ ▲ ） A ．6 B ．8 C ．－8 D ．8或－810．已知函数()f x 是定义在R 上的偶函数，对于任意x ∈R 都有(4)()(2)f x f x f +=+成立，当12,[0,2]x x ∈且12x x ≠时，都有1212()()0f x f x x x ->-. 给出下列命题：①函数()f x 一定是周期函数； ②函数()f x 在区间[6,4]--上为增函数；③直线4x =-是函数()f x 图像的一条对称轴； ④函数()f x 在区间[6,6]-上有且仅有4个零点.其中正确命题的个数是（ ▲ ） A ．1 B ．2 C ．3 D ．4第Ⅱ卷（非选择题 共100分）二、填空题：本大题共5小题，每小题4分，共20分.请把答案填在答题卷的相应位置.11．已知函数2,01()2,12x x f x x x ⎧<≤=⎨-<≤⎩，则20()f x dx ⎰等于 ▲ .12．已知双曲线C 1：x 2a 2－y 2b 2＝1(a >0，b >0)与双曲线C 2：y 216－x 24＝1有相同的渐近线，则C 1的离心率＝ ▲ . 13．已知2＋23＝223，3＋38＝338，4＋415＝4415，…，若7＋a b ＝7ab ，(a 、b 均为正实数)，则类比以上等式，可推测a 、b 的值，进而可得a ＋b ＝ ▲ .14．若定义在],[b a 上的函数13)(23+-=x x x f 的值域为]1,3[-，则a b -的最大值是 ▲ . 15．已知*(1,2,3,,,3,)i A i n n n N =≥∈是△AOB 所在的平面内的n 个相异点，且OB OA OB OA i ⋅=⋅. 给出下列命题： ①12n OA OA OA OA ====；的最小值不可能是OB ；③点12,,,,n A A A A 在一条直线上；④向量及i 在向量的方向上的投影必相等.其中正确命题的序号是 ▲ .（请填上所有正确命题的序号）三、解答题：本大题共6小题，每小题分数见旁注，共80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. 16．（本小题满分13分）已知全集U =R ，0m >，集合2{|120},{|3}A x x x B x x m =--<=-≤. (1)当2m =时，求()UAB ð；(2)命题p ：x A ∈，命题q ：x B ∈，若p 是q 的充分条件，求实数m 的取值范围. 17．（本小题满分13分）已知向量a ＝()3sin x ，－cos x ，b ＝()cos x ，cos x ，记函数f (x )＝a·b . (1)求f (x )的最小正周期及单调递增区间；(2)设△ABC 的内角A ，B ，C 的对边分别是a ，b ，c ，且c ＝3，f (C )＝12，若向量m ＝()1，sin A 与n ＝()2，sin B 共线，求a ，b 的值. 18．（本小题满分13分）平面直角坐标系中，点M 的坐标是，曲线1C 的参数方程为1cos ,sin ,x y αα=+⎧⎨=⎩（α为参数），在以坐标原点为极点、x 轴的非负半轴为极轴建立的极坐标系中，曲线2C 的极坐标方程为4sin ρθ=. (1)将曲线1C 和2C 化成普通方程，并求曲线1C 和2C 公共弦所在直线的极坐标方程；(2)若过点M ，倾斜角为3π的直线l 与曲线1C 交于A ，B 两点，求MA MB ⋅的值. 19．（本小题满分13分）经过多年的运作，“双十一”抢购活动已经演变成为整个电商行业的大型集体促销盛宴. 为迎接2014年“双十一”网购狂欢节，某厂家拟投入适当的广告费，对网上所售产品进行促销. 经调查测算，该促销产品在“双十一”的销售量p 万件与促销费用x 万元满足231p x =-+(其中a x ≤≤0，a 为正常数)．已知生产该产品还需投入成本102p +万元(不含促销费用)，产品的销售价格定为20(4)p+元／件，假定厂家的生产能力完全能满足市场的销售需求． (1)将该产品的利润y 万元表示为促销费用x 万元的函数；(2)促销费用投入多少万元时，厂家的利润最大？并求出最大利润的值. 20．（本小题满分14分）已知中心在坐标原点O ，焦点在x 轴上的椭圆C 的离心率为12，且经过点M (1，32)． (1)求椭圆C 的方程；(2)若F 是椭圆C 的右焦点，过F 的直线交椭圆C 于M 、N 两点，T 为直线x ＝4上任意一点，且T 不在x 轴上，(ⅰ)求FM →·FN →的取值范围；(ⅱ)若OT 平分线段MN ，证明：TF ⊥MN （其中O 为坐标原点）.21．（本小题满分14分）已知函数ln ()xf x x a=+(a ∈R )，曲线()y f x =在点(1,(1))f 处的切线方程为1y x =-. (1)求实数a 的值，并求()f x 的单调区间；(2)试比较20152014与20142015的大小，并说明理由；(3)是否存在k ∈Z ，使得()2kx f x >+对任意0x >恒成立？若存在，求出k 的最小值；若不存在，请说明理由.厦门双十中学2014-2015学年（上）期中检测 高三数学（理科）参考答案及评分标准（2014-11-13）17．（本小题满分13分） 【解析】 (1)依题意，21cos 2111()cos cos 22cos 2sin(2)22262x f x x x x x x x x π+=-=-=--=-- ··············································································································································································· 3分 所以最小正周期22T ππ==， ························································································································· 4分 令222,262k x k k Z πππππ-≤-≤+∈，解得,63k x k k Z ππππ-≤≤+∈，所以()f x 的单调递增区间是：[,],63k k k Z ππππ-+∈.············································································· 6分 (2)由11()sin(2)622f C C π=--=，得sin(2)16C π-=， ·········································································· 7分因为0C π<<，所以112666C πππ-<-<，所以262C ππ-=，解得3C π=， ···································· 8分因为向量m ＝()1，sin A 与n ＝()2，sin B 共线，所以sin 2sin B A =，由正弦定理得2b a =，…① ········ 9分在△ABC 中，由余弦定理得2222cos3c a b ab π=+-，即223a b ab +-=，……………………② ·············· 11分由①②，解得1,2a b ==. ································································································································ 13分 18．（本小题满分13分） 【解析】(1) 依题意，1C 的普通方程：22(1)1x y -+=，………………………………① ······································· 2分对2C ，24sin ρρθ=，所以224x y y +=，即22(2)4x y +-=，……② ······································ 4分 ①－②可得，20x y -=， ························································································································ 6分所以曲线1C 和2C 公共弦所在直线的极坐标方程为cos 2sin 0ρθρθ-=，1tan ()2R θρ=∈. ···· 7分 （注：本次考试，直线的极坐标方程若只写“cos 2sin 0ρθρθ-=”，或者“1tan 2θ=”均给分！）(2)解法一：依题意，直线l的参数方程为13,2,x t y ⎧=+⎪⎪⎨⎪=⎪⎩（t 为参数），点A 、B 分别对应参数12,t t ， ···················· 9分代入1C的方程：221(31))122t +-+=，整理得2560t t ++=，所以126t t =， ················· 12分 所以126MA MB t t ⋅==. ······························································································································ 13分解法二（注：了解即可！）：设曲线1C 的圆心为(1,0)C ，半径1r =，则由圆幂定理得2222()()(31)0)16MA MB MC r MC r MC r ⋅=+-=-=-+-=. ··········· 13分19．（本小题满分13分） 【解析】(1)由题意知，)210()204(p x p py +--+=， ···························································································· 3分 将231p x =-+代入化简得：x x y -+-=1416（0x a ≤≤）. ································································· 6分(2)13)1(14217)114(17=+⨯+-≤+++-=x x x x y ， ········································································· 8分 当且仅当1,114=+=+x x x 即时，上式取等号. ····························································································· 9分 当1a ≥时，促销费用投入1万元时，厂家的利润最大； ············································································ 10分当1a <时，)114(17+++-=x x y 在[]0,a 上单调递增， ······································································· 11分 所以x a =时，函数有最大值，即促销费用投入a 万元时，厂家的利润最大. ·········································· 12分 综上，当1a ≥时，促销费用投入1万元，厂家的利润最大；当1a <时，促销费用投入a 万元，厂家的利润最大. ······································································· 13分 20．（本小题满分14分） 【解析】(1)设椭圆C 的方程为22221(0)x y a b a b+=>>，则222221,2191,4,c e a a b a b c ⎧==⎪⎪⎪+=⎨⎪⎪=+⎪⎩解得224,3a b ==，所以椭圆22:143x y C +=. ································································ 4分 (2)(ⅰ)易得(1,0)F ， ··············································································································································· 5分①若直线l 斜率不存在，则1:=x l ，此时)23,1(M ，)23,1(-N ，FN FM ⋅＝49-； ···················· 6分②若直线l 斜率存在，设)1(:-=x k y l ，),(),,(2211y x N y x M ，则由⎪⎩⎪⎨⎧=+-=134)1(22y x x k y 消去y 得：01248)34(2222=-+-+k x k x k ······················································· 7分 ∴3482221+=+k k x x ，341242221+-=⋅k k x x ····························································································· 8分∴⋅),1(),1(2211y x y x -⋅-=]1)()[1(21212++-+=x x x x k ＝21149k +-- ···················· 9分∵02≥k ∴11102≤+<k ∴411432<+-≤k ∴493-<⋅≤-FM综上，FN FM ⋅的取值范围为]49,3[--． ························································································· 10分(ⅱ)线段MN 的中点为Q ，则由(ⅰ)可得，2122243,(1)24343Q Q Q x x k kx y k x k k +-===-=++， ··········· 11分 所以直线OT 的斜率3'4Q Q y k x k==-，所以直线OT 的方程为：34y x k =-， ········································ 12分 从而3(4,)T k-，此时TF 的斜率30141TF k k k--==--，··············································································· 13分 所以11TF MN k k k k⋅=-⋅=-，所以TF ⊥MN . ································································································· 14分21．（本小题满分14分） 【解析】(1)依题意，2ln '()()x ax x f x x a +-=+， ·················································································································· 1分 所以211'(1)(1)1a f a a+==++，又由切线方程可得'(1)1f =，即111a =+，解得0a =， 此时ln ()x f x x =，21ln '()xf x x -=， ············································································································· 3分令'()0f x >，所以1ln 0x ->，解得0x e <<；令'()0f x <，所以1ln 0x -<，解得x e >， 所以()f x 的增区间为：(0,)e ，减区间为：(,)e +∞. ···················································································· 5分(2)解法一：由(1)知，函数()f x 在(,)e +∞上单调递减，所以(2014)(2015)f f >，即2015201420152014ln 2014ln 2015201420152015ln 20142014ln 2015ln 2014ln 201520142015>⇔>⇔>⇔> ··············································································································································································· 9分 解法二：201420142015201520151201420142014⎛⎫=⨯⎪⎝⎭，因为 201420142233201420142014201420142015112014201411111()()()20142014201411122!3!2014!1112122320132014111112(1)()()223201320141320143C C C ⎛⎫⎛⎫=+ ⎪ ⎪⎝⎭⎝⎭=+++++<++++<++++⨯⨯⨯=+-+-++-=-< 所以2014201520153120142014<<，所以2014201520152014<. ·················································································· 9分 (3)若()2kx f x >+对任意0x >恒成立，则2ln 2x k x x >+，记2ln 2()x g x x x=+，只需max ()k g x >.又32312ln 2122ln '()x x xg x x x x ---=-=， ································································································· 10分 记()122ln h x x x =--，则2'()20h x x=--<，所以()h x在(0,)+∞上单调递减.又(1)10h =-<，3()12ln 1ln 21ln 2ln 0222h ==>-+=>，所以存在唯一0(2x ∈，使得0()0h x =，即00122ln 0x x --=， ·················································· 11分 当0x >时，(),'(),()h x g x g x 的变化情况如下：····································································································· 12分所以00max 022ln ()()x x g xg x x +==，又因为00122ln 0x x --=，所以0022ln 1x x +=， 所以200000220000(22ln )212111()()222x x x x g x x x x x +++===⋅+， 因为0(2x ∈，所以01x ∈，所以03()12g x <<， ·························································· 13分 又max ()(1)2g x g ≥=，所以02()1g x ≤<，因为max ()k g x >，即0()k g x >，且k ∈Z ，故k 的最小整数值为3.所以存在最小整数3k =，使得()2kx f x >+对任意0x >恒成立. ··························································· 14分。