2019届江苏省苏州市高三上学期期初调研考试数学(理)试题(word版)

EDCBA2019届江苏省苏州市高三上学期期初调研考试数学(理)试题（正卷）2018．9注 意 事 项考生在答题前请认真阅读本注意事项及各题答题要求 1．本试卷共4页，包括填空题（第1题～第14题）、解答题（第15题～第20题）两部分．本试卷满分160分，考试时间120分钟． 2．答题前，请您务必将自己的姓名、考试号用0.5毫米黑色墨水的签字笔填写在答题卡的指定位置． 3．答题时，必须用0.5毫米黑色墨水的签字笔填写在答题卡的指定位置，在其它位置作答一律无效． 4．如有作图需要，可用2B 铅笔作答，并请加黑加粗，描写清楚．5. 请保持答题卡卡面清洁，不要折叠、破损．一律不准使用胶带纸、修正液、可擦洗的圆珠笔． 方差公式：2222121[()()()]n s x x x x x x n =-+-++-，其中121()n x x x x n=+++.锥体体积公式：1=3V Sh 锥体(S 为锥体底面面积，h 为锥体的高).一、填空题：本大题共14小题，每小题5分，共70分．不需要写出解答过程，请把答案直接填在答题卡相应位置上．．．．．．．．． 1．已知集合{1,0,1}A =-，集合{|0}B x x =>，则AB = ▲ ．2．若复数12i z =+，22i z a =-(i 为虚数单位)，且12z z 为实数，则实数a = ▲ ．3．一组数据1，2，3，4，a 的平均数为2，则该组数据的方差等于 ▲ ． 4．如图是某一算法的伪代码，则输出值n 等于 ▲ ．5．一只口袋中装有5个大小相同的球，其中3个黑球，2个白球，从中一次 摸出2只球，则摸出1个黑球和1个白球的概率等于 ▲ ．6．已知函数222(0)()(0)x x x f x x ax x ⎧-⎪=⎨-+<⎪⎩≥为奇函数，则实数a 的值等于 ▲ ．7．已知函数()sin(2)f x x ϕ=+(0ϕπ<≤)的一条对称轴是512x π=-，则ϕ= ▲ ．8．已知等比数列{}n a 的前 项和为n S ，若264,,S S S 成等差数列，则246a a a +的值为 ▲ ．9．已知△ABC 的三边上高的长度分别为2，3，4，则△ABC 最大内角的余弦值等于 ▲ ． 101(cm)的圆形纸片按如图所示的实线裁剪，并按虚线折叠为各棱长均相等的四棱锥，则折叠所成的四棱锥的体积为 ▲ cm 3．11．如图，已知AC 与BD 交于点E ，AB ∥CD ，AC =26AB CD ==，则当tan 3A =时，BE CD ⋅= ▲ ．(第4题)(第11题)12．已知函数f (x )＝|x 2－6|，若0a b >>，且f (a )＝f (b )，则a 2b 的最大值是 ▲ ． 13．在斜三角形ABC 中，已知11tan 0tan tan C A B++=，则tan C 的最大值等于 ▲ ． 14．已知⊙C 的方程为：222(3)(2)(0)x y r r -+-=>，若直线33x y +=上存在一点P ，在⊙C 总存在不同的两点M ，N ，使得点M 是线段PN 的中点，则⊙C 的半径r 的取值范围是 ▲ ．二、解答题：本大题共6小题，共计90分．请在答题卡指定区域．．．．．．．内作答，解答时应写出必要的文字说明、证明过程或演算步骤． 15．（本题满分14分）已知πcos (0,)2αα=∈． （1）求πsin()4α+的值；（2）若()11πcos ,(0,)142αββ+=∈，求β的值．16．（本题满分14分）如图，已知矩形CDEF 和直角梯形ABCD ，AB ∥CD ，90ADC ∠=︒，DE =DA ， M 为AE 的中点．(1)求证：AC ∥平面DMF ； (2)求证：BE ⊥DM ．MFE DCBA (第16题)17．（本题满分14分）如图，有一块半圆形的空地，政府计划在空地上建一个矩形的市民活动广场ABCD 及矩形的停车场EFGH ，剩余的地方进行绿化．其中半圆的圆心为O ，半径为r ，矩形的一边AB 在直径上，点C ，D ，G ，H 在圆周上，E ，F 在边CD 上，且∠BOG =60︒，设BOC θ∠=．(1)记市民活动广场及停车场的占地总面积为()f θ，求()f θ的表达式； (2)当cos θ为何值时，可使市民活动广场及停车场的占地总面积最大．18．（本题满分16分）已知椭圆C ：22221(0)x y a b a b +=>>的左、右顶点分别为A ，B ，离心率为12，点P (1,32)为椭圆上一点．(1)求椭圆C 的标准方程；(2)如图，过点(0,1)C 且斜率大于1的直线l 与椭圆交于M ，N 两点，记直线AM 的斜率为1k ，直线BN 的斜率为2k ，若122k k =，求直线l 斜率的值．FEGH C DOBA(第17题)19．（本小题满分16分）已知数列{}n a 的奇数项是首项为1的等差数列，偶数项是首项为2的等比数列，数列{}n a 前n 项和为n S ，且满足34S a =，523a a a =+．(1)求数列{}n a 的通项公式；(2)若12m m m a a a ++=，求正整数m 的值； (3)是否存在正整数m ，使得221mm S S -恰好为数列{}n a 中的一项？若存在，求出所有满足条件的m 值，若不存在，说明理由.20．（本小题满分16分）若对任意的实数k ，b ，函数()y f x kx b =++与直线y kx b =+总相切，则称函数()f x 为“恒切函数”．(1)判断函数2()f x x =是否为“恒切函数”；(2)若函数()ln f x m x nx =+(0m ≠)是“恒切函数”，求实数m ，n 满足的关系式； (3)若函数()(e 1)e x x f x x m =--+是“恒切函数”，求证：104m -<≤．2018~2019学年第一学期期初教学质量调研卷高三数学（附加卷）2018．921．【选做题】在A，B，C，D 四小题中只能选做两题．．．．．．，每小题10分，共计20分．请在答题卡．．．指定区域．．．．内作答，解答时应写出文字说明、证明过程或演算步骤．A．选修4—1：几何证明选讲如图，⊙O为△ABC的外接圆，∠BAC的平分线交⊙O于点D，过点D作⊙O的切线分别与AB，AC交于点E，F．求证：BC∥EF．B．选修4—2：矩阵与变换已知矩阵21mn⎡⎤=⎢⎥⎣⎦M的两个特征向量为11⎡⎤=⎢⎥⎣⎦α，21⎡⎤=⎢⎥⎣⎦α，若21⎡⎤=⎢⎥⎣⎦β，求5Mβ．C．选修4—4：坐标系与参数方程已知曲线C 的极坐标方程为4cos ρθ=．以极点为平面直角坐标系的的原点，极轴为x 轴的非负半轴，建立平面直角坐标系．直线l 的参数方程是1cos sin x t y t αα=+⎧⎨=⎩(t 为参数) ．(1)将曲线C 的极坐标方程化为直角坐标方程；(2)若直线l 与曲线C 相交于A ，B两点，且||AB l 的倾斜角α的值．D ．选修4—5：不等式选讲已知正实数x ，y ，z 满足x y z xyz ++=，求证：111x y y z x z +++++．【必做题】第22题、第23题，每题10分，共计20分．请在答题卡指定区域．．．．．．．内作答， 解答时应写出文字说明、证明过程或演算步骤． 22．（本小题满分10分）已知直三棱柱111ABC A B C -，AB ⊥AC ，3AB =，4AC =，B 1C ⊥AC 1．现以A 为原点，分别以AB ，AC ，AA 1所在直线为x ，y ，z 轴建立空间直角坐标系如图所示． (1)求1AA 的长度；(2)若1BP =，求二面角1P A C A --的正弦值．23．（本小题满分10分）设()()n f n a b =+(2n ≥，N*n ∈)，若在()f n 的展开式中，存在连续的三项的二项式系数依次成等差数列，则称()f n 具有性质P ． (1)求证：(7)f 具有性质P ；(2)若存在2018n ≤，使得()f n 具有性质P ，求n 的最大值．2018~2019学年第一学期期初教学质量调研卷高三数学（正卷）参考解答与评分标准一、填空题：（每题5分，满分70分） 1．{1} 2．4 3．2 4．4 5．356．−2 7．3π 8．2 9．1124-1011．1212．1613．-14．)+∞ 二、解答题（共6小题，满分90分） 15．（本题满分14分） 解：(1)由πcos (0,)2αα=∈，得1sin 7α=，············································· 2分 所以πππsin()sin cos cos sin 444ααα+=+ ·············································· 4分17=+=． ································ 6分 (2)因为π,(0,)2αβ∈，所以(0,π)αβ+∈．又()11cos 14αβ+=，则()sin αβ+== 8分所以()()()sin sin sin cos cos sin βαβααβααβα=+-=+-+ ·············· 10分11111472=-⨯=． ············································· 12分 因为π(0,)2β∈，所以π6β=． ······················································ 14分16．（本题满分14分）证明：(1)连接EC 交DE 于N ，连接MN ．∵矩形CDEF ，∴EC ，DF 相互平分，∴N 为EC 中点． ·· 2分 又∵M 为EA 中点，∴MN ∥AC ． ··································· 4分 又∵AC ⊄平面DMF ，且MN ⊂平面DMF ．∴AC ∥平面DMF ． ·················································· 7分 (2)∵矩形CDEF ，∴CD ⊥DE ．又∵AB ∥CD ，∴AB ⊥DE ． ·························································· 8分 又∵直角梯形ABCD ，AB ∥CD 且90ADC ∠=︒，∴AB ⊥AD ．∵DE AD =D ，∴AB ⊥平面ADE ． ··············································· 10分又∵DM ⊂平面ADE ，∴AB ⊥DM ．∵AD DE =，M 为AE 的中点，∴AE ⊥DM ． ·································· 11分 又∵AB AE A =，∴MD ⊥平面A BE ． ·········································· 13分 ∵BE ⊂平面ABE ，∴BE ⊥MD ． ··················································· 14分 17．（本题满分14分）解：(1)∵半圆的半径为r ，BOC θ∠=，∠OBC =90°．∴在直角三角形OBC 中，cos OB r θ=，sin BC r θ=，∴2cos AB r θ=．∴22sin cos ABCD S AB BC r θθ=⋅=矩形． ················································ 2分 又∵∠BOG =60︒，由半圆的对称性可知，∠HOA =60︒，∴∠HOG =60︒． ∴△HOG 为等边三角形，∴HG =r ，HE sin r θ-=sin )r θ． ∴2sin )EFGH S EF EH r θ=⋅=-矩形． ·············································· 4分 ∴()ABCD EFGH f S S θ=+=矩形矩形2(2sin cos sin r θθθ-+，其中(0,)3πθ∈． ································································································· 7分(2) ∵222()(2cos 2sin cos )f r θθθθ'=--=22(4cos cos 2)r θθ--． ··········· 9分 令()0f θ'=，即24cos cos 20θθ--=， 解得：cos θ=cos θ=(舍去)． ·································· 11分 令0cos θ=0(0,)3πθ∈． 1︒当0(0,)θθ∈时，()0f θ'>，()f θ单调递增；2︒当0(,)3πθθ∈时，()0f θ'<，()f θ单调递减．∴当0θθ=时，()f θ取得最大值． ···················································· 13分 答：当cos θ= · 14分18．（本题满分16分）解：(1)∵椭圆的离心率为12，∴2a c =．ABODC H GEF又∵222a b c =+，∴b =．∴椭圆的标准方程为：2222143x y c c+=． ··············· 3分又∵点P (1,32)为椭圆上一点，∴22914143c c +=，解得：1c =． ··············· 5分∴椭圆的标准方程为：22143x y +=． ················································· 6分 (2)由椭圆的对称性可知直线l 的斜率一定存在，设其方程为1y kx =+． 设1122(,),(,)M x y N x y ．联列方程组：221431x y y kx ⎧+=⎪⎨⎪=+⎩，消去y 可得：22(34)880k x kx ++-=． ∴由韦达定理可知：122834k x x k +=-+，122834x x k =-+． ····················· 8分 ∵1112y k x =+，2212yk x =-，且122k k =，∴1212222y y x x =+-． ·················· 10分即221222124(2)(2)y y x x =+-．①又∵1122(,),(,)M x y N x y 在椭圆上， ∴22113(4)4y x =-，22223(4)4y x =-．②将②代入①可得：121224(2)22x x x x -+=+-，即1212310()120x x x x +++=． ······· 12分 ∴22883()10()1203434k k k-+-+=++，即2122030k k -+=． ················· 14分 解得：16k =或32k =．又∵k >1，∴32k =． ······································ 16分 19．（本小题满分16分）解：(1)设奇数项的等差数列公差为d ，偶数项的等比数列公比为q ． ∴数列{}n a 的前5项依次为：1，2，1+d ，2q ，1+2d ．∵34523S a a a a =⎧⎨=+⎩，∴42123d q d d +=⎧⎨+=+⎩，解得：23d q =⎧⎨=⎩． ···························· 2分∴12()23()nn n n a n -⎧⎪=⎨⎪⋅⎩为奇数为偶数． ······························································ 4分 (2) ∵12m m m a a a ++=．1︒若2m k =(N*k ∈)则22122k k k a a a ++=，∴123(21)23k k k -⋅⨯+=⋅，即213k +=，∴1k =，即2m =． ································································································· 6分2︒若21m k =-(N*k ∈)则21221k k k a a a -+=，∴1(21)2321k k k --⨯⋅=+，∴12122312121k k k k -+⋅==+--． ∵123k -⋅为整数，∴221k -必为整数，∴211k -=，∴1k =，此时0233⋅≠． 不合题意． ················································································· 8分 综上可知：m =2． ········································································ 9分 (3)∵21321242()()m m m S a a a a a a -=++⋅⋅⋅++++⋅⋅⋅+=(121)2m m +-+2(13)13m --=231m m +-． ································· 10分21122122312331m m m m m m S S a m m ---=-=+--⋅=+-． ··························· 11分∴221m m S S -=2123131m m m m -+-+-=2122(1)3331m m m ---+-≤． ······································ 12分 若221mm S S -为数列{}n a 中的项，则只能为123,,a a a ． 1︒2211m m S S -=，则2122(1)3131m m m ---=+-，∴130m -=，m 无解． ··················· 13分 2︒2212m m S S -=，则2122(1)3231m m m ---=+-，∴12310m m -+-=． 当1m =时，等式不成立； 当2m =时，等式成立；当3m ≥时，令1221()31313x x f x x x -=+-=⋅+-．∴ln3()323xf x x '=⋅-，2ln 3()323x f x ''=⋅-． 当3x ≥时，()0f x ''>，∴()f x '在[3,)+∞上单调递增． 又∵(3)9ln 360f '=->，∴()0f x '>在[3,)+∞上恒成立， ∴()f x 在[3,)+∞上单调递增．∵(3)10f =>，∴当3m ≥时，方程12310m m -+-=无解． ···················· 14分 3︒2213m m S S -=，则2122(1)3331m m m ---=+-，∴210m -=，即1m =． ·············· 15分 综上可知：1m =或2m =． ···························································· 16分20．（本小题满分16分）解：(1)函数()f x 为“恒切函数”，设切点为00(,)x y ．则0000()()f x kx b kx b f x k k ++=+⎧⎨'+=⎩，∴00()0()0f x f x =⎧⎨'=⎩． ······································· 2分对于函数2()f x x =，()2f x x '=．设切点为00(,)x y ，∴20020x x ⎧=⎪⎨=⎪⎩， ······················································· 3分解得：00x =．∴2()f x x =是“恒切函数”． ······································· 4分 (2)若函数()ln f x m x nx =+(0m ≠)是“恒切函数”，设切点为00(,)x y ．∵()mf x n x '=+，∴000ln 00m x nx m n x +=⎧⎪⎨+=⎪⎩， ·············································· 5分解得：0ln 1x =，即0x e =． ···························································· 7分 ∴实数m ，n 满足的关系式为：0m ne +=． ······································· 8分 (3) 函数()(1)x x f x e x e m =--+是“恒切函数”，设切点为00(,)x y ． ∵()(22)xxf x e x e '=--，∴000000(1)0(22)0x x x x e x e m e x e ⎧--+=⎪⎨--=⎪⎩， ∴0000(1)22x x x m e x e e x ⎧=---⎪⎨=+⎪⎩． ······························································· 10分考查方程22x e x =+的解，设()22x g x e x =--． ∵()21x g x e '=-，令()0g x '=，解得：ln 2x =-． ∴当(,ln 2)x ∈-∞-时，()0g x '<，()g x 单调递减； 当(ln 2,)x ∈-+∞时，()0g x '>，()g x 单调递增．∴min ()(ln 2)ln 210g x g =-=-<． ··················································· 12分1︒当(,ln 2)x ∈-∞-时∵24(2)0g e -=>，2(1)10g e-=-<． ∴()22x g x e x =--在(,ln 2)-∞-上有唯一零点0(2,1)x ∈--．又∵00(1)x x m e x e =---=001(2)4x x +，∴1(,0)4m ∈-． ························ 14分2︒当(ln 2,)x ∈-+∞时∵(0)0g =，∴()22x g x e x =--在(ln 2,)-+∞上有唯一零点0，∴0m =． ································································································ 15分 综上可知：104m -<≤． ································································ 16分2018~2019学年第一学期期初教学质量调研卷高三数学附加卷参考解答21A ．选修4—1：几何证明选讲(本题满分10分) 证明：连接BD ．∵EF 为⊙O 的切线，∴∠BDE =∠BAD ． ············· 2分 ∵AD 为∠BAC 的平分线，∴∠BAD =∠DAF ，则∠BDE =∠DAF ． ·········································· 4分 又∵∠CBD =∠DAF (同弧所对的圆周角相等) ． ····· 6分 ∴∠CBD =∠BDE ． ·········································· 8分 ∴BC ∥EF ． ················································· 10分 21B ．选修4—2：矩阵与变换(本题满分10分)解:设矩阵M 的两个特征向量1α，2α相对应的特征值分别为1λ，2λ．∴111222M M αλααλα=⎧⎨=⎩，解得：0m n ==，12λ=，21λ=． ······························ 4分又∵1221022101βαα⎡⎤⎡⎤⎡⎤==+=+⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦． ··················································· 6分∴55551211221064M M (2)264011βααλαλα⎡⎤⎡⎤⎡⎤=+=+=+=⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦． ··················· 10分21C ．选修4—4：坐标系与参数方程(本题满分10分) 解：(1)∵4cos ρθ=，∴24cos ρρθ=．又∵222cos x y xρρθ⎧=+⎨=⎩，∴224x y x +=． ·················································· 4分∴曲线C 的直角坐标方程为：2240x y x +-=． (2)设A ，B 两点对应的参数值为1t ，2t将1cos sin x t y t αα=+⎧⎨=⎩(t 为参数)代入方程2240x y x +-=整理可得： 22cos 30t t α--=．由韦达定理可知：12122cos 3t t t t α+=⎧⎨=-⎩．∴12||||AB t t =-== ························· 6分 解得：1cos 2α=±． ········································································· 8分又∵[0,)απ∈，∴3πα=或23πα=． ················································· 10分 21D ．选修4—5：不等式选讲(本题满分10分) 证明：∵x ，y ，z 为正实数，由基本不等式可得：111x y y z x z ++≤+++． (当且仅当x y z ==时，等号成立) ······················································ 5分1(1112=≤．(当且仅当x y z ==时，等号成立) ····················································· 10分 22．（本小题满分10分） 解：(1)设AA 1=t ，则(0,0,0)A ，1(0,4,)C t ，1(3,0,)B t ，(0,4,0)C ． ∴1(0,4,)AC t =，1(3,4,)B C t =--．∵11B C AC ⊥，∴110B C AC ⋅=，即2160t -=，解得：4t =．∴14AA =． ··············································· 3分 (2)∵1BP =，∴(3,0,1)P ，又∵1(3,0,)B t ，(0,4,0)C ． ∴1(0,4,4)AC =-，1(3,0,3)A P =-． 设(,,)n x y z =是平面1PAC 的法向量． 则1100n AC n A P ⎧⋅=⎪⎨⋅=⎪⎩，∴440330y z x z -=⎧⎨-=⎩．令1z =，则1x =，1y =，∴(1,1,1)n =是平面1PA C 的一个法向量． ·········· 5分 ∵直三棱柱111ABC A B C -，∴1AA ⊥平面ABC ，AB ⊂平面ABC ． ∴1AB AA ⊥，又∵AB ⊥AC ，且1AA AC A =．∴AB ⊥平面AA 1C ，∴(3,0,0)AB =是平面AA 1C 的一个法向量． ·············· 7分∴3cos ,||||AB n AB n AB n ⋅〈〉==··························································· 9分 ∴26sin ,1cos ,AB n AB n 〈〉=-〈〉=．∴二面角1P AC A -- ··············································· 10分 23．（本小题满分10分）解：(1)∵7(7)()f a b =+的展开式中第2,3,4项的二项式系数分别为：177C =，2721C =，3735C =，∴17C ，27C ，37C 成等差数列． ∴(7)f 具有性质P ． ········································································ 3分 (2)假设()f n 具有性质P ，则一定存在N*k ∈，11k n -≤≤，使得1k n C -，k n C ，1k n C +成等差数列，∴112k k k n n n C C C -+=+．∴!!!2!()!(1)!(1)!(1)!(1)!n n n k n k k n k k n k ⨯=+---++--．化简可得：224420k nk n n -+--=． ················································· 5分 ∴2(2)2k n n -=+．∵,N*k n ∈，∴2n +是完全平方数． ·················································· 8分 ∵2018n ≤，2244202045<<，∴n 的最大值为：2442-=1934．此时989k =或945k =． ·································································· 10分。