当前位置:文档之家› 厦门市2018-2019学年度第一学期高三质检理科数学(含参考答案解析)

厦门市2018-2019学年度第一学期高三质检理科数学(含参考答案解析)

厦门市2018-2019学年度第一学期高三年级质量检测理科数学注意事项:1. 答卷前,考生务必将自己的姓名、准考证号填写在答题卡上.2.回答选择题时,选出每小题答案后,用铅笔把答题卡上对应题目的答案标号涂黑.如需改动,用橡皮擦干净后,再选涂其它答案标号.回答非选择题时,将答案写在答题卡上,写在本试卷上无效. 3. 考试结束后,将答题卡交回.一、选择题:本题共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2{|60},{|0}M x x x N x x =+-≤=>,则MN =A .(0,2]B .[3,2]-C .(]0,3D .[)3,-+∞2.设a ∈R ,则“1-=a ”是“直线01=-+y ax 与直线05=++ay x 平行”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 3.实数,x y 满足x y >,则下列不等式成立的是A .1yx<B .22x y --<C .lg()0x y ->D .22x y > 4.设,x y 满足约束条件0,,290,x y x x y ≥⎧⎪≤⎨⎪+-≤⎩则3z x y =+的最大值为A .0B .92C .12D .27 5.已知角α的顶点为坐标原点,始边与x 轴的非负半轴重合,终边上有一点(sin 47,cos 47)P ︒︒, 则sin(13)α-︒=A .12B .2 C . 12-D .2-6.已知函数()3,0,1,02x x x f x x ≤⎧⎪=⎨⎛⎫-> ⎪⎪⎝⎭⎩,则()()2log 3f f =A .9-B .1-C .13-D .127-7.长江某地南北两岸平行,一艘游船从南岸码头A 出发航行到北岸,假设游船在静水中的航行速度v 1的 大小为|v 1|=10 km/h ,水流的速度v 2的大小为|v 2|=4 km/h .设v 1和v 2的夹角为()0180θθ︒<<︒,北岸 的点A ′在A 的正北方向,游船正好到达A ′处时,cos θ= A .521 B .521-C .52 D .52-8.已知函数()21sin 2f x x =-,若将其图象沿x 轴向右平移()0ϕϕ>个单位,所得图象关于原点对称, 则实数ϕ的最小值为A .πB .43π C .2πD .4π9.函数()[]()cos ln 12π,2πy x x x =++∈-的图象大致为A BC D10.直线l 与双曲线2222:1(0,0)x y E a b a b-=>>的一条渐近线平行,l 过抛物线2:4C y x =的焦点,交C于,A B 两点,若5AB =,则E 的离心率为A .2B .CD.211.已知圆锥的顶点为P ,母线长为2,底面半径为r ,点,,,A B C D 在底面圆周上,当四棱锥P ABCD -体积最大时,r =AB .83 C .3D .27 12.在平面四边形ABCD 中,A C D ∆面积是ABC ∆面积的2倍,数列{}n a 满足13a =,且()()132n n CA a CB a CD +=-+-,则5a =A .31B .33C . 63D .65二、填空题:本题共4小题,每小题5分,共20分.13.已知复数z 满足()12i i z +=,其中i 为虚数单位,则z = ▲ .14.《张丘建算经》卷上第22题有如下内容:今有女善织,日益功疾,初日织五尺,今一月日织九匹三丈.其意思为:现有一善于织布的女子,从第2天开始,每天比前一天多织相同量的布,第1天织布5尺,现在一个月(按30天计算)共织布390尺.那么,该女子本月中旬(第11天到第20天)共织布 ▲ 尺.15.某三棱柱的三视图如图所示,则该三棱柱外接球的表面积为 ▲ .16. 已知偶函数()f x 满足:当x ≥0时,()()()log 111x a f x a x a =-+-> 若()f x 恰有三个零点,则a 的取值范围是 ▲ .俯视图侧视图正视图第15题图三、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、23题为选考题,考生根据要求作答. (一)必考题:共60分. 17.( 12分)在ABC ∆中,内角,,A B C 的对边分别为,,a b c ,ABC ∆的面积为S ,已知222a b c +-=. (1)求角C ;(2)若2c =a -的取值范围.18.( 12分)已知数列{}n a 的前n 项和为n S ,且22n n S a n =--. (1)求证:{}1+n a 是等比数列; (2)数列{}n b 满足221log (1)log (1)n n n a b a ++=+,数列{}n c 满足nn n b b c 1+=,求数列{}n c 的前n 项和n T .19.(12分)如图,在四棱锥P ABCD -中,PB ⊥平面PAC ,四边形ABCD 为平行四边形,且4,135A D A B B A D ==∠=︒.(1)证明:AC ^平面PAB ;(2)当直线PC 与平面PAB所成角的正切值为时,求二面角A PC D --的余弦值.第19题图B C20.(12分)已知圆22:(16E x y ++=,点F ,动点P 在E 上,线段PF 的垂直平分线与直线PE 相交于点Q ,Q 的轨迹是曲线C . (1)求C 的方程;(2)已知过点(2,1)-的直线l 与C 交于,A B 两点,M 是C 与y 轴正半轴的交点,设直线,MA MB 的斜率分别为12,k k ,证明:12k k +为定值.21.(12分)已知函数()()()e e11xxf x a a x a -=--+<,若()f x 存在极大值点1x 和极小值点2x .(1)求实数a 的取值范围;(2)若()()12f x kf x >,求实数k 的取值范围.(二)选考题:共10分.请考生在第22、23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4-4:坐标系与参数方程](10分)在同一直角坐标系中,经过伸缩变换1,2x x y y ⎧'=⎪⎨⎪'=⎩后,曲线C 变为曲线221x y ''+=.以坐标原点为极点,x 轴正半轴为极轴建立极坐标系,直线l的极坐标方程为sin()3πρθ-=(1)求C 和l 的直角坐标方程;(2)过点()1,0P 作l 的垂线交C 于A ,B 两点,点A 在x 轴上方,求11PA PB -.23.[选修4-5:不等式选讲](10分)函数()2f x ax =+,不等式()f x a ≤的解集为{}20x x -≤≤. (1)求a 的值;(2)求证:对任意x R ∈,存在1m >,使得不等式1(2)(2)1f x f x m m -+≥+-成立.厦门市2018—2019学年度第一学期高三年级质量检测数学(理科)参考答案一、选择题:本题共12小题,每小题5分,共60分.1—5:AABCA 6—10:BDDAC 11-12:CB11. 解析:设圆锥的高为h ,,AC BD 相交于点M ,AMB,则0,2h,224r h ,22111sin 22222ABCDS AC BD AC BD r r ,当且仅当,22ACBD r 时,ABCD S 取到最大值, 则2231222443333PABCDABCD V S h r h h hh h ,令34f h h h ,则234f hh ,令0f h,解得233h, 所以f h 在230,3上单调递减,在23,23上单调递增, 所以min2316339h h,则四棱锥P ABCD . 所以当四棱锥PABCD 体积最大时,22643rh . 12. 解析:设AC 和BD 交于点E ,ACD 和ABC 的高分别为1h ,2h ,∵ACD 的面积是ABC 面积的2倍,∴212h h 2DE EB ,∴2DE EB ,即2CE CD CB CE ,∴2133CECB CD , 又132n n CAa CB a CD ,由A 、C 、E 三点共线,设CA CE2133CB CD , 由平面向量基本定理得:123,3123n na a , ∴1322n n a a ,即1121n n a a , ∴数列1n a 是以112a 为首项,以2为公比的等比数列,∴11222nn n a ,即21n n a ,所以,532133a .二、填空题:本题共4小题,每小题5分,共20分.13.514. 130 15. 283 16. 1e a16.因为当x ≥0时,1log 11x a f xa x a ,所以00f ,又因为f x 为偶函数,所以f x 恰有三个零点等价于f x 在0+,恰有一个零点,令0f x,得1=log 1x a a x ,所以1x g xa 与函数log 1a h xx 的图象恰有一个交点,因为函数y g x 与函数y h x 的图象关于yx 对称,解法一:由于1a ,当1x g xa 的图象与直线yx 相切时,设切点为00,x y ,则A CDBE2h 1hln 1x a a 且001xa x ,所以0ln ln x a ,011ln ln 1ln x a a,设ln t a ,则ln 1t t t ,设ln xx x x ,则'ln xx ,所以x 在0,1单调递增,在1,单调递减,又因为11,所以ln 1ta ,e a ,由图可知,a 的取值范围为1e a .解法二:如图,由于1a ,函数1x g x a 的图象与直线y x 有一个公共点为0,0,当函数1x g xa 的图象与直线y x 切于原点时,ln 1a ,e a ,由图可知,a 的取值范围为1e a .三、解答题:本题共6小题,共70分.17. 本题考查正弦定理、余弦定理、三角恒等变换等基础知识;考查推理论证能力、运算求解能力等;考查函数与方程思想、化归与转化思想等.满分12分. 解:(1)∵1sin 2Sab C ,22243a b c S ,∴22223sin a b c ab C , ······························································ 1分在ABC 中,由余弦定理,得2222cos a b c ab C ,∴cos 3sin C C ,········································································· 4分 ∴3tan 3C , ∵0,C,∴6C. ····································································· 6分(2)由正弦定理,得4sin sin sin ab c A B C, ············································· 7分所以3b a43sin 4sin B A 543sin()4sin 6A A23cos 2sin A A 4sin()3A, ···························································· 10分因为5(0,)6A ,所以7(,)336A , ·············································· 11分 所以3(2,4]b a ,即3b a 的取值范围为(2,4]. ···························· 12分18. 本题考查等比数列的定义、递推数列、数列求和等基础知识;考查推理论证能力、运算求解能力、创新意识等;考查分类与整合思想、化归与转化思想等.满分12分.解:(1)1n 当时,111212S a a ,13a , ················································· 1分 当2n时,22nnS a n , ①112(1)2n n S a n , ② ············································ 2分 由①-②得,122(1)nn n a a a n n ,121n n a a ,···················································································· 4分112(1)n n a a 2n ,1140a ,1121n n a a ,1n a 是首项为4,公比为2的等比数列. ················································· 6分(2)由(1)得,111422nn n a , ····························································· 7分1222212log (1)log 21log (1)log 22n n nn n a n b a n , ························································ 8分 11221n nnn n c b b n n 11111122112n n n n , ················ 10分111111222233412nT n n1111112233412nn n2114922222224nn nnnn n n .············································· 12分19. 本题考查线面垂直、直线与平面所成角、二面角等基础知识;考查空间想象能力、运算求解能力、推理论证能力;考查数形结合思想、化归与转化思想等. 满分12分. (1)证明:由已知,得454ABCBC ,,在△ABC 中,222cos 22ACBA BC BA BC ABC, ····················· 2分∴222BC AB AC ,即AC AB , ························································ 3分∵PB 平面PAC ,AC 平面PAC , ∴PB AC , ······················································································ 4分 又∵PB AB B ,AB 平面PAB ,PB 平面PAB , ∴AC 平面PAB . ···················································································· 5分(2)解:∵AC 平面PAB ,∴CPA 为直线PC 与平面PAB 所成角, ······················································· 6分∴tan 2ACCPA AP,∴2PA ,在Rt PAB △中,222PB AB PA ,取AB 的中点H ,连结PH ,则PH AB , ∵AC 平面PAB ,PH 平面PAB , ∴AC PH ,又∵AC AB A ,AC 平面ABCD ,AB 平面ABCD ∴PH 平面ABCD , ··············································································· 8分 以A 点为坐标原点,建立如图空间直角坐标系A xyz , 则2,2,0,2,2,0,0,4,0,1,1,2B C D P , ∴1,3,2,1,5,2PCPD ,设平面PCD 的法向量为,,n x y z ,则320,520n PC x y z n PDx yz ,取1x ,解得1,1,22n, ···························· 9分又平面PAC 的法向量为1,1,2PB , ··················································· 10分∴10cos ,5PB n PB nPB n, ··································································· 11分 ∴二面角A PCD 的余弦值为5. ·························································· 12分 20.本题考查椭圆的定义、直线与圆锥曲线的位置关系等基础知识;考查运算求解能力、推理论证能力等,考查数形结合、化归与转化等数学思想. 满分12分.解:(1)依题意,QF QP ,则4QE QF QE QP EP EF , ·············· 3分所以Q 的轨迹为以,E F 为焦点,4为长轴长的椭圆, 所以2,3,1ac b ,所以点Q 的轨迹方程为2214x y . ························································ 5分 (2)依题意得直线AB 的斜率存在,设直线:1(2)AB y k x ,即21y kx k ,设1122(,),(,)A x y B x y ,联立2221,14y kx k x y,消去y 并整理得222(14)8(21)4(21)40k x k k x k ···························· 6分所以0,1228(21)14k k x x k ,21224(21)414k x x k , ······························· 8分因为M 是C 与y 轴正半轴的交点,所以(0,1)M ,所以12121211y y k k x x12211211y x y x x x1221122222kx k x kx k x x x121212221kx x k x x x x ··················································· 10分16121216(1)k k k k k k221kk 1所以12k k 为定值,且定值为1. ··························································· 12分21. 本题考查函数的极值、导数及其应用、不等式等基础知识;考查推理论证能力、运算求解能力、创新意识等;考查函数与方程思想、化归与转化思想、分类与整合思想等.满分12分. 解法一:(1)由e e1x xf xa a x 得,'e e 1x xf xa a , ························· 1分即'ee 1e 1xx x f x a ,①当0a时,当,0x时,'0f x;当0,x时,'0f x;所以f x 在,0单调递增,在0,单调递减,f x 不存在极小值点,不合题意;····················································································································· 3分②当01a 时,令'0f x 得,120,ln x x a ,因为01a ,所以12x x , 当,0x 时,'0f x ;当0,ln xa 时,'0f x ; 当ln ,xa 时,'0f x;所以f x 在,0单调递增,在0,ln a 单调递减,在ln ,a 单调递增,所以f x 存在极大值点10x 和极小值点2ln x a ,符合题意; 综上,实数a 的取值范围为01a . ··································································· 5分 (2)由(1)知01a ,且f x 的极大值点为10x ,极小值点为2ln x a ,此时11f x a ,211ln f x a a a , ······················································· 6分依题意,得111ln a k a a a 对任意01a 恒成立,由于此时210f x f x ,所以0k ; ······························································ 7分所以11ln 11a aa k,即11ln 11a ak a ,设11ln 11x g x x k x ,则222122111'11x x k k g xxx x , ··························· 8分 令2210x x k(*), ①当1k 时,2440k ,所以'0g x ,g x 在0,1单调递增, 所以10g a g ,即11ln 11a a k a ,符合题意; ········································· 10分②当01k 时,2440k ,设(*)的两根为34,x x ,且34x x ,则343420,1x x x x k,因此3401x x , ······················································· 11分 则当31x x 时,'0g x ,g x 在3,1x 单调递增,所以当31x a 时,10g a g ,即11ln 11a a k a ,所以12f x kf x ,矛盾,不合题意;综上,k 的取值范围是1k . ············································································ 12分 解法二:(1)同解法一;(2)由(1)知01a ,且f x 的极大值点为10x ,极小值点为2ln x a ,此时11f x a ,211ln f x a a a , ······················································· 6分依题意,得111ln a k a a a 对任意01a 恒成立, 设1ln 11g x k x xkx ,则1'ln 1ln 1x kg xk x k k xxx, 22111''k x g xk xx x , ········································································ 7分①当0k时,当0,1x时,''0g x ,所以'g x 在0,1单调递增,所以''110g x g k ,所以'g x 在0,1单调递减,所以10g a g ,即111ln a k a a a ,不符合题意; ······················································· 8分 ②当1k时, 当0,1x时,''0g x,所以'g x 在0,1单调递减,所以''110g x g k ,所以'g x 在0,1单调递增,所以10g a g , 即111ln a k a a a ,符合题意; ···························································· 10分③当01k 时,''0g x,所以'g x 在0,1单调递减,又因为'110g k ,11'ee2kkg k ,设e 2x h xx,则当1x 时,21e '0xx h xx , 所以h x 在1,单调递增,所以11e 20hh k,即1'e0kg ,所以1'e '10kg g ,即'g x 在0,1恰有一个零点0x , ································· 11分且当0,1x x 时,'0g x,'g x 在0,1x 单调递减, 所以当01x a 时,10g a g ,即111ln a k aa a ,不合题意;综上,k 的取值范围是1k . ········································································· 12分 解法三:(1)同解法一.(2)由(1)知01a ,且f x 的极大值点为10x ,极小值点为2ln x a , 此时11f x a ,211ln f x a a a , ······················································· 6分依题意,得111ln a k a a a 对任意01a 恒成立,由于此时210f x f x ,所以0k , ······························································ 7分①当1k 时,122f x f x kf x ,不等式成立,符合题意; ······························ 9分②当01k 时,11ln 11a aa k,即11ln 11a a k a ,设11ln 11x g x x k x ,则222122111'11x x k k g xxx x , ·························· 10分 令2210x x k(*),则2440k ,设(*)的两根为34,x x ,且34x x , 则343420,1x x x x k,因此3401x x , ······················································· 11分 则当31x x 时,'0g x ,g x 在3,1x 单调递增,所以当31x a 时,10g a g ,即11ln 11a a k a ,所以12f x kf x ,矛盾,不合题意;综上,k 的取值范围是1k . ············································································ 12分 22.本题考查曲线的伸缩变换的概念、极坐标方程、直线的参数方程等基础知识;考查运算求解能力;考查数形结合、函数与方程思想.满分10分.解:(1)将1',2'x x y y,代入22''1x y 得:曲线C 的方程为2214x y , ·························· 2分 由sin()33得:sin coscos sin333, ··································· 3分因为cos ,sinxy ,代入上式并整理得: 直线l 的直角坐标方程为3230xy . ······················································ 5分 (2)因为直线l 的倾斜角为3,所以其垂线的倾斜角为56, 过点1,0P 作l 的垂线的参数方程为51cos,650sin 6x t y t (t 为参数),即31,212x t y t(t 为参数), ··································· 6分 代入曲线C 的方程整理得:2743120t t , ·················································· 7分 设A 、B 两点对应的参数为1t 、2t ,由题意知10t ,20t则121243,712,7t t t t 且2(43)47120, ····················································· 8分 所以121212111133t t PA PBt t t t . ··························································· 10分。

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