静安区2012学年高三年级第一学期期末教学质量检测
数学试卷(文理科合并)
一、填空题(本大题满分56分)本大题共有14题,考生应在答题纸相应编号的空格内直接填写结果,每个空格填对得4分,否则一律得零分.
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9.(文)已知0 >++-x a ax 的解集是 . 10.(文)已知α、β为锐角,且2)2 tan 1)(2 tan 1(=++β α ,则βαtan tan = . 11.(文)数列{}n a 的前n 项和为2 2n S n =(*N n ∈),对任意正整数n ,数列{}n b 的项都满足 等式022 121=+-++n n n n n a b a a a ,则n b = . 第12题 12.(文)机器人“海宝”在某圆形区域表演“按指令行走”.如图所示,“海宝”从圆心O 出发,先沿北偏西13 12arcsin 方向行走13米至点A 处,再沿正南方向行走14米至点B 处,最后沿正东 方向行走至点C 处,点B 、C 都在圆O 上.则在以圆心O 为坐标原点,正东方向为x 轴正方向, 正北方向为y 轴正方向的直角坐标系中圆O 的方程为 . 13(文)设P 是函数x x y 2+ =(0>x )的图像上任意一点, 过点P 分别向直线x y =和y 轴作垂线,垂足分别为A 、B ,则PA 14.(文)设复数i a a z )sin 2()cos (θθ-++=(i 为虚数单位),若对任意实数θ,2≤z ,则实数a 的取值范围为 . 二、选择题(本大题满分20分)本大题共有4题,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,填上正确的答案,选对得5分,否则一律得零分. 15.(文)某地区的绿化面积每年平均比上一年增长10.4%,经过x 年,绿化面积与原绿化面积之 比为y ,则y=f(x)的图像大致为 ( ) (文)16.若复数021≠z z ,则2121z z z z =是12z z =成立的( ) (A) 充要条件 (B) 既不充分又不必要条件 (C) 充分不必要条件 (D) 必要不充分条件 17.(文)函数])3,1[(4 2)(2 ∈+-= x x x x x f 的值域为 ( ) (A) ]3,2[ (B) ]5,2[ (C) ]3,3 7 [ (D) ]4,3 7 [ 18.(文)已知向量a 和b 满足条件:a b ≠且0≠?b a .若对于任意实数t ,恒 有-≥-,则在a 、b 、b a +、b a -这四个向量中,一定具有垂直关系的两个向 量是( ) (A) a 与b a - (B) b 与b a - (C) a 与b a + (D)b 与b a + 三、解答题(本大题满分74分)本大题共5题,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 . 19.(文)已知数列}{n a 的递推公式为?? ?=∈≥+-=-. 2),2(,3231*1a N n n n a a n n (1)令n a b n n -=,求证:数列}{n b 为等比数列;(2)求数列}{n a 的前 n 项和. 20. (文)已知c b a ,,分别为△ABC 三个内角A 、B 、 C 所对的边长,且c A b B a 53cos cos =-. (1)求: B A tan tan 的值; (2)若0 60=A ,5=c ,求a 、b . 21. 某仓库为了保持库内的湿度和温度,四周墙上均装有如图所示的自动通风设施.该设施的下部ABCD 是正方形,其中AB =2米;上部CDG 是等边三角形,固定点E 为AB 的中点.△EMN 是由电脑控制其形状变化的三角通风窗(阴影部分均不通风),MN 是可以沿设施边框上下滑动且始终保持和AB 平行的伸缩横杆. (1)设MN 与AB 之间的距离为x 米,试将△EMN 的面积S (平方米)表示成关于x 的函数; (2)求△EMN 的面积S (平方米)的最大值. 22. 已知椭圆 12 22 2=+ b y a x 的两个焦点为)0,(1c F -、)0,(2c F ,2c 是2a 与2 b 的等差中项,其 中a 、b 、c 都是正数,过点),0(b A -和)0,(a B 的直线与原点的距离为2 3. (1)求椭圆的方程; (2)(文)过点A 作直线交椭圆于另一点M ,求AM 长度的最大值; (3)已知定点)0,1(-E ,直线t kx y +=与椭圆交于C 、D 相异两点.证明:对任意的0>t ,都存在实数k ,使得以线段CD 为直径的圆过E 点. A B C (文21题) 23已知x x f 2 1log )(=,当点),(y x M 在)(x f y =的图像上运动时,点),2(ny x N -在函数 )(x g y n =的图像上运动(*N n ∈). (1)求)(x g y n =的表达式; (2)若方程)2()(21a x g x g +-=有实根,求实数a 的取值范围; (3)设) (2 )(x g n n x H =,函数)()()(11x g x H x F +=(b x a ≤≤<0)的值域为 ]2 2 log ,2 2 [log 4 2 5 2 ++a b ,求实数a ,b 的值. 高三年级 文理科数学试卷答案及评分标准 1.4 1= a ; 2.64; 3.(理)65 33arccos ;(文)14-n 4.(理) 3 16;(文)同理3 5.(理)13;(文)2 6.(理)12-?n n ;(文))3,2( 7.(理)(2)(3)(4);(文)3 16 8.(理)4;(文)同理5 9.(理) 9 1093;(文))2,2( a 10.(文理)1; 11.(理)2252 2=+y x ;(文)1 41422 -+=n n b n ; 12.(理)x y 162=; (文)同理11 13.(理)),3[+∞-;(文)-1 14.(理) 6 π ;(文)]5 5, 5 5[- . 15.(文理)D ; 16.(理)C ;(文)D ; 17.(文理)A ;18.(文理)B 19(理)解:(1) ①如图1所示,当MN 在矩形区域滑动, 即0<x ≤1时, △EMN 的面积S =x ??221 =x ; ··························· 1分 ②如图2所示,当MN 在三角形区域滑动, 即1<x <31+时, 如图,连接EG ,交CD 于点F ,交MN 于点H , ∵ E 为AB 中点, ∴ F 为CD 中点,GF ⊥CD ,且FG =3. 又∵ MN ∥CD , ∴ △MNG ∽△DCG . ∴ GF GH DC MN = ,即M N = . ························ 4分 故△EMN 的面积S =12 x ? =x x )3 31(3 32+ +-; ············································· 6分 综合可得: C 图2 N 图1 ( )( 2011113 3x x S x x x ??=?? ?-+++ ?? ????,<≤.<< ···························································7分 (2)①当MN 在矩形区域滑动时,x S =,所以有10≤ ·······································8分 ②当MN 在三角形区域滑动时,S =x x )3 31(3 32 + +- . 因而,当2 3 1+=x (米)时,S 得到最大值,最大值S =33 2 1 + (平方米). ∵ 13 321>+ , ∴ S 有最大值,最大值为3 32 1+ 平方米. ··································································· 12分 (文)解:(1)11113))1((3333323----=--=+-=-+-=-=n n n n n n b n a n a n n a n a b , 2≥n 又1111=-=a b ,所以0≠n b (*N n ∈), )2(31 ≥=-n b b n n 所以,数列}{n b 是以1为首项3为公比的等比数列. ··················································6分 (2)1 3-=n n b ,n b a n n +=······················································································8分 所以数列}{n a 的前 n 项和)21()(21n b b b S n n +++++++= = 2 1 32 -++n n n . ·································································································································· 14分 20(理)解:(1)∵a 、b 、c 成等比数列,∴b 2 =ac ························································1分 则cos B = ac b c a 22 2 2 -+= ac ac c a 22 2 -+ ·················································································3分 而a 2 +c 2 ≥2ac ∴cos B = ac ac c a 22 2 -+≥2 12=ac ac ,等号当且仅当a =c 时取得,即 2 1≤cos B <1,得到 3 0π ≤ (2)cos2x -4sin( 2 4x π+)sin( 24 x π- )=cos2x -4sin(2 4 x π+ )cos(2 4 x π+ ) =2cos x 2 -2cos x -1=2(cos x - 2 1)2-2 3 ·················································································· 11分 ∵x =B ∴ 2 1≤cos x <1 ∴2(cos x - 2 1)2- 2 3≥- 23 则由题意有:-m <-2 3即m > 2 3··················································································· 14分 (说明:这样分离变量1cos 2cos 22cos cos 22++-=->x x x x m 参照评分) (文)解:(1)由正弦定理 C c B b A a sin sin sin = = 得C A B B A sin 5 3cos sin cos sin = -,2分 又B A B A B A C sin cos cos sin )sin(sin +=+=,所以A B B A cos sin 5 8cos sin 5 2= ,5分 可得 4cos sin cos sin tan tan == A B B A B A . · ··················································································7分 (2)若060=A ,则2 3s i n = A ,2 1cos = A ,3tan =A ,得4 3t a n = B ,可得 19 194cos = B ,19 193sin ?= B . ··········································································· 10分 38 1935sin cos cos sin )sin(sin ?= +=+=B A B A B A C , 由正弦定理 C c B b A a sin sin sin ==得 19sin sin =?= A C c a ,2sin sin =?= B C c b ························································ 14分 21(理)解:(1)当1=n 时,3 12 1a a =,由01≠a 得11=a .····································1分 当2=n 时,3 2221)1(a a +=+,由02≠a 得22=a 或12-=a .当3=n 时,3 33 22 321)1(a a a a ++=++,若22=a 得33=a 或23-=a ;若12-=a 得13=a ; 5分 综上讨论,满足条件的数列有三个: 1,2,3或1,2,-2或1,-1,1. ·············································································6分 (2)令n n a a a S +++= 21,则3 3 23 12 n n a a a S +++= (*N n ∈). 从而3 13 3 23 12 1)(++++++=+n n n n a a a a a S . ·······················································7分 两式相减,结合01≠+n a ,得12 12++-=n n n a a S . ······················································8分 当1=n 时,由(1)知11=a ;当2≥n 时,)(221--=n n n S S a =)()(2 12 1n n n n a a a a ---++,即0)1)((11=--+++n n n n a a a a ,所以n n a a -=+1或11+=+n n a a . ·························· 12分 又11=a ,20122013-=a ,所以无穷数列{}n a 的前2012项组成首项和公差均为1的等差数列,从第2013项开始组成首项为-2012,公比为-1的等比数列.故 ???>-?≤≤=)2012() 1(2012)20121(n n n a n n . ··········································································· 14分 (说明:本题用余弦定理,或者正弦定理余弦定理共同使用也可解得,请参照评分) (文)解:(1) ①如图1所示,当MN 在正方形区域滑动, 即0<x ≤2时, △EMN 的面积S =x ??221 =x ; ··························· 2分 ②如图2所示,当MN 在三角形区域滑动, 即2<x <32+ 时, 如图,连接EG ,交CD 于点F ,交MN 于点H , ∵ E 为AB 中点, ∴ F 为CD 中点,GF ⊥CD ,且FG =3. 又∵ MN ∥CD , ∴ △MNG ∽△DCG . ∴ GF GH DC MN =,即3 ) 23(2x MN -+= . ················· 5分 故△EMN 的面积S = x x ?-+?3 ) 23(221 =x x )3 321(3 32++-; ··········································· 7分 综合可得: ?? ? ??+<<++-≤<=322,)33 21(332 0,2x x x x x S ····························································8分 说明:讨论的分段点x=2写在下半段也可. (2)①当MN 在正方形区域滑动时,x S =,所以有20≤ ②当MN 在三角形区域滑动时,S =x x )3 321(3 32 + +-. 因而,当22 31<+ =x (米),S 在)32,2(+上递减,无最大值,20< 所以当2=x 时,S 有最大值,最大值为2平方米. ···················································· 14分 22.解:(1)在椭圆中,由已知得2 2 22 2 2 b a b a c += -= ···········································1分 E N 图1 E B C 图2 过点),0(b A -和)0,(a B 的直线方程为 1=-+ b y a x ,即0=--ab ay bx ,该直线与原点的距离 为2 3,由点到直线的距离公式得: 2 32 2 = +b a a b ···················································3分 解得:1,32 2==b a ;所以椭圆方程为 11 3 2 2 =+ y x · ···················································4分 (2)(理))0,2(1-F ,直线11A F 的方程为22+=x y ,611= A F ,当椭圆上的点P 到直 线11A F 距离最大时,△11PA F 面积取得最大值·······························································6分 设与直线11A F 平行的直线方程为d x y += 2,将其代入椭圆方程 11 3 2 2 =+ y x 得: 0122372 2 =-++d x d x ,0=?,即03 283 2882 2 =+ - d d ,解得72 =d , 当7-=d 时,椭圆上的点P 到直线11A F 距离最大为 3 7 2+ ,此时△11PA F 面积为 2 14 223 7 26 2 1+ = + ··························································································9分 (文)设),(y x M ,则)1(322y x -=,422)1(2 222 ++-=++=y y y x AM ,其中1 1≤≤-y ·····································································································································6分 当2 1= y 时,2 AM 取得最大值 2 9,所以AM 长度的最大值为 2 23····························9分 (3)将t kx y +=代入椭圆方程,得0336)31(2 2 2 =-+++t ktx x k ,由直线与椭圆有两个交 点,所以0)1)(31(12)6(222>-+-=?t k kt ,解得3 12 2 ->t k ································ 11分 设),(11y x C 、),(22y x D ,则2 21316k kt x x +- =+,2 2 2131)1(3k t x x +-= ?,因为以CD 为直径的圆 过E 点,所以0=?ED EC ,即0)1)(1(2121=+++y y x x , ····································· 13分 而))((2121t kx t kx y y ++==221212)(t x x tk x x k +++,所以 01316) 1(31)1(3) 1(2 2 2 2 2 =++++-+-+t k kt tk k t k ,解得t t k 3122 -= ······························ 14分 如果3 12 2 -> t k 对任意的0>t 都成立,则存在k ,使得以线段CD 为直径的圆过E 点. 09)1(3 1)312( 2 2 222 2 2 >+-= -- -t t t t t t ,即3 12 2 -> t k .所以,对任意的0>t ,都存在k , 使得以线段CD 为直径的圆过E 点. ··········································································· 16分 23(理)解:(1)x y = ,3 x y =是对等函数; · ························································4分 (2)研究对数函数x y a log =,其定义域为),0(+∞,所以x x a a log log =,又0log ≥x a , 所以当且仅当0log ≥x a 时)()(x f x f =成立.所以对数函数x y a log =在其定义域),0(+∞内 不是对等函数. ·············································································································6分 当10< ,此时x y a log =是对等函数; 当1>a 时,若),1[+∞∈x ,则0log ≥x a ,此时x y a log =是对等函数;