上海市2018届高考二模数学试题含答案-

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2017-2018学年第二学期高三数学质量调研
考生注意: 1. 本试卷共4页,21道试题,满分150. 考试时间120分钟. 2. 本考试分试卷和答题纸. 试卷包括试题与答题要求. 作答必须涂(选择题)或写(非选择题)在答题纸上,在试卷上作答一律不得分. 3. 答卷前,务必用钢笔或圆珠笔在答题纸正面清楚地填写姓名、准考证号,并将核对后的条码贴在指定位置上,在答题纸反面清楚地填写姓名. 一、填空题(本大题共有12题,满分54分)考生应在答题纸相应编号的空格内直接填写结果,每个空格填对前6题得4分、后6题得5分,否则一律得零分. 1. 抛物线x212y的准线方程为_______. 2. 若函数f(x3. 若函数f(x1是奇函数,则实数m________.
x2m12x3的反函数为g(x,则函数g(x的零点为________.
4. 书架上有上、中、下三册的《白话史记》和上、下两册的《古诗文鉴赏辞典》,现将这五本书从左到右摆放在一起,则中间位置摆放中册《白话史记》的不同摆放种数为_______(结果用数值表示).
5. 在锐角三角形ABC中,角ABC的对边分别为abc,若(b2c2a2tanAbc,则角A的大小为________.
36. (x1n的展开式中含有非零常数项,则正整数n的最小值为_________. 2x1207. 某单位年初有两辆车参加某种事故保险,对在当年内发生此种事故的每辆车,单位均可.设这两辆车在一年内发生此种事故的概率分别为获赔(假设每辆车最多只获一次赔偿)1,且各车是否发生事故相互独立,则一年内该单位在此种保险中获赔的概率为_________21(结果用最简分数表示). x8.在平面直角坐标系xOy中,直线l的参数方程为y2t22t为参数),椭圆C2t4
xcos参数方程为为参数),则直线l与椭圆C的公共点坐标为__________.
1ysin29.设函数f(xlogmxm0m1),若m是等比数列annN*)的公比,且
222f(a2a4a6a20187,则f(a12f(a2f(a3f(a2018的值为_________.
xy02xy210. 设变量xy满足条件,若该条件表示的平面区域是三角形,则实数my0xym取值范围是__________. 11.x1My|y,xR21Ny|y1x1m1x2,1x2,若NM,则实数m的取值m1范围是. x212. F1F2分别是椭圆C:y21的左、右两焦点,点N为椭圆C的上顶点,若动点22M满足:MN2MF1MF2,则MF12MF2的最大值为__________.
二、选择题(本大题共有4题,满分20分)每题有且只有一个正确答案,考生应在答题纸的相应编号上,将代表答案的小方格涂黑,选对得5分,否则一律得零分. 13. 已知i为虚数单位,若复数(ai2i为正实数,则实数a的值为„„„„„„(
(A2B1C0D1
14.如图所示的几何体,其表面积为(55,下部圆柱的底面直径与该圆柱的高相等,上部圆锥的母线长为5,则该几何体的主视图的面积为 „„„„„„„„„„(
(A4B6C8D10
Sn存在”是 15. Sn是无穷等差数列an的前n项和(nN*),则limn“该数列公差d0”的 „„„„„„„„„„„„„„„„„„(
14题图
(A充分非必要条件 B必要非充分条件
C充要条件 D既非充分也非必要条件
16.已知kN*x,y,zR,若k(xyyzzx5(x2y2z2,则对此不等式描叙正 确的是„„„„„„„„„„„„„„„„„„„„„„„„„„„„„(
(Ak5,则至少一个以x,y,z为边长的等边三角形 ..Bk6,则对任意满足不等式的x,y,z存在x,y,z为边长的三角形 ..Ck7,则对任意满足不等式的x,y,z存在x,y,z为边长的三角形 ..Dk8,则对满足不等式的x,y,z存在x,y,z为边长的直角三角形 ..三、解答题(本大题共有5题,满分76分)解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤
17. (本题满分14分)本题共有2个小题,第1小题满分6分,第2小题满分8
如图所示的正四棱柱ABCDA1B1C1D1的底面边长为1,侧棱AA12,E在棱CC1上,
D1 A1 B1 E C1 CE=CC1(0. 11)当=时,求三棱锥D1EBC的体积;
2
2DCarccos2)当异面直线BE1所成角的大小为时,求的值. 3

18.(本题满分14分)本题共有2个小题,第1小题满分8分,第2小题满分6 已知函数f(x=sinxcosxsin2xxR. 1)若函数f(x在区间[a,D A B 17题图
C
16]上递增,求实数a的取值范围;
2)若函数f(x的图像关于点Q(x1,y1对称,且x1[,],求点Q的坐标. 44


19.(本题满分14分)本题共有2个小题,第1小题满分8分,第2小题满分6 某市为改善市民出行,大力发展轨道交通建设.规划中的轨道交通s号线线路示意图如图所.已知M,N是东西方向主干道边两个景点,P,Q是南北方向主干道边两个景点,四个景点距离城市中心O均为52km线路AB段上的任意一点到景点N的距离比到景点M的距离都多10km线路BC段上的任意一点到O的距离都相等,线路CD段上的任意一点到景点Q的距离比到景点P的距离都多10km,以O为原点建立平面直角坐标系xOy. 1)求轨道交通s号线线路示意图所在曲线的方程;
2规划中的线路AB段上需建一站点G到景点Q的距离最近,问如何设置站点G的位置?

20. (本题满分16分)本题共有3小题,第1小题4分,第2小题6分,第3小题6. 定义在R上的函数f(x满足:对任意的实数x,存在非零常t,都有f(xttf(x成立. 1)若函数f(xkx3,求实数kt的值;
2)当t2时,若x[0,2]f(xx(2x,求函数f(x在闭区间[2,6]上的值域;
3)设函数f(x的值域为[a,a],证明:函数f(x为周期函数.

21.(本题满分18分)本题共有3小题,第1小题4分,第2小题6分,第3小题8.
若数列an同时满足条件:①存在互异的p,qN*使得apaqcc为常数); ②当npnq时,对任意nN*都有anc,则称数列an为双底数列. 1)判断以下数列an是否为双底数列(只需写出结论不必证明); ann6n ansin ann3n5
2n1012n,1n50a2nn50若数列an是双底数列,求实数m的值以及数列an2m,n50n项和Sn
93)设ankn3,是否存在整数k,使得数列an为双底数列?若存在,求出所10有的k的值;若不存在,请说明理由.

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2017学年第二学期高三数学质量调研评分标准(参考)
一、填空题
1 y3
2 1
23 4 24
5 6 5
x3
9 611 7 2
21

二、选择题
8 (22,2410 4(0,1][,
3
12 1990 (1,0
610

13 D
三、解答题
14 B
15 A
16 B
1 17.1CE=CC1CE1 又正四棱柱ABCDAD1C1平面EBC1B1C1D121zVD1EBCSRtECBD1C1„„„„„„„„„„„ 4
3

111CEBC.„„„„„„„„„„ 6 3262)以D为原点,射线DADCDD1x轴、y轴、z轴的正半轴,建立空间直角坐标系(如图),„„„„„„ 2 B(1,1,0E(0,1,2D1(0,0,2C(0,1,0
x
yDC(0,1,2BE(1,0,2„„„„„„„„„„„„„„„„„„„ 4
1又异面直线BED1C所成角的大小为arccos0(110(2242D1CBE,„„„„„„„„„ 6
223D1CBE514520化简整理得1625,又0,即2
3
5. „„„„„„„„„„„„„„„ 8
4
=sinxcosxsin2x18.1f(x1cos2x1sin2x,„„„„„„„„„„2 22x21sin(2x,„„„„„„„„„„4 2423 8216时,则2x42164又函数f(x[a,16]上递增,则2a42,即a3,„„„„„„„„„7
8则实数a的取值范围为a[3,. „„„„„„„„„„„„„„„„„„„8 8162)若函数f(x的图像关于点Q(x1,y1对称,则sin(2x1 2x1
kZk0,则点Q的坐标为(40 „„„„„„2
4kkZ),则x1k[,],„„„„„„„„„„„„4 28441,. „„„„„„„„„„„„„„„„6 8219.1)因为线路AB段上的任意一点到景点N的距离比到景点M的距离都多10km,所以线路AB段所在曲线是以定点MN为左、右焦点的双曲线的左支,
则其方程为x2y225(x0,y0 „„„„„„„„„„„„„„„„„„„3 因为线路BC段上的任意一点到O的距离都相等,所以线路BC段所在曲线是以O为圆心、OB长为半径的圆,由线路AB段所在曲线方程可求得B(5,0
则其方程为x2y225(x0,y0 „„„„„„„„„„„„„„„„„„„5 因为线路CD段上的任意一点到景点Q的距离比到景点P的距离都多10km,所以线路CD段所在曲线是以定点QP为上、下焦点的双曲线下支,
则其方程为x2y225(x0,y0 „„„„„„„„„„„„„„„„„„„7 故线路示意图所在曲线的方程为xxyy25. „„„„„„„„„„„„„„8 2)设G(x0,y0,又Q(0,52,则GQ由(1)得x0y025,即GQ
2
22x0(y0522
222y0102y075,„„„„„„„„„„„„3

GQ502(y052522,即当时,GQmin52 y022则站点G的坐标为

556,2,可使G到景点Q的距离最近.„„„„„„„„6 2220.(1)由f(xttf(x得,k(xt3t(kx3xR恒成立,
k(t10(kktx(k3t30xR恒成立,则(k3t30,„„„„„„„„2
t0k0. „„„„„„„„„„„„„„„„„„„„„„„„„„„„„4
t12)当x[0,2]时,f(xx(2x1(x12[0,1],„„„„„„„„„„„2 x[2,0]时,即x2[0,2] f(x22f(xf(x11f(x2,则f(x[,0],„„„„„„„„3 22x[2,4]时,即x2[0,2]
f(x22f(xf(x2f(x2,则f(x[2,0] „„„„„„„„4 x[4,6]时,即x2[2,4]
f(x2f(x2f(x[0,4] „„„„„„„„„„„„„„„„„„„5 综上得函数f(x在闭区间[0,6]上的值域为[2,4]. „„„„„„„„„„„„„„6 3)(证法一)由函数f(x的值域为[a,a]得,f(xt的取值集合也为[a,a]
t0时,f(xttf(x[ta,ta],则taa,即t1.„„„„„„„„2
taaf(x1f(xf(x2f(x1f(x
则函数f(x是以2为周期的函数. „„„„„„„„„„„„„„„„„„„„„„3

taaf(xttf(x[ta,ta]t0时,,则,即t1.„„„„„„„„5
taaf(x1f(x,则函数f(x是以1为周期的函数. 故满足条件的函数f(x为周期函数. „„„„„„„„„„„„„„„„„„„„„6 (证法二)由函数f(x的值域为[a,a]得,必存在x0R,使得f(x0a t1时,对t1,有f(x0ttf(x0taa
t1,有f(x0ttf(x0taa,则t1不可能;
0t1时,即111f(x0f(x0t
ttf(x的值域为[a,a]得,必存在x0R,使得f(x0ta 仿上证法同样得0t1也不可能,则必有t1 ,以下同证法一.
21. 1)①③是双底数列,②不是双底数列;„„„„„„„„„„„„„„„„„4 2)数列an1n50时递减,当n50时递增,
由双底数列定义可知a50a51,解得m1,„„„„„„„„„„„„„„„„„2 1n50时,数列成等差,Snn991012n100nn2
222n501 n50时,Sn1005050212122n49n2548 „„„„„„„„„„„„„„„5
100nn2,1n50综上,Snn49.„„„„„„„„„„„„„„„„„„„„6
2n2548,n5093an1anknk310nn19kn3
10n99knk3kn3
1010
199k3kn „„„„„„„„„„„„„„2 1010若数列an是双底数列,则9k3kn有解(否则不是双底数列), 9n3n,„„„„„„„„„„„„„„„„„„„„„„„„„„„3
kk1k3k1k3
n8n12n10n639故当k1时,an1an6n
10101n5时,an1an;当n6时,an1an;当n7时,an1an 从而 a1a2a3a4a5a6a7a8 ,数列an不是双底数列; 同理可得:
k3时,a1a2a8a9a10a11 ,数列an不是双底数列; k1时,a1a2a12a13a14a15 ,数列an是双底数列; k3时,a1a2a10a11a12a13 ,数列an是双底数列; „„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„7 综上,存在整数k1k3,使得数列an为双底数列.„„„„„„„„„„8

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