苏州市2018届高三第一学期期中调研试卷
数 学
一、填空题(本大题共14小题，每小题5分，共70分，请把答案直接填写在答卷纸．．．相应的位置) 1．已知集合{1,2,3,4,5},{1,3},{2,3}U A B ===，则()U A B = ▲ ． 2．函数1
ln(1)
y x =
-的定义域为 ▲ ．
3．设命题:4p x >；命题2:540q x x -+≥，那么p 是q 的 ▲ 条件（选填“充分不必要”、“必要不充分”、“充要”、“既不充分也不必要”）．
4．已知幂函数2
2*()m m y x m -=∈N 在(0,)+∞是增函数，则实数m 的值是 ▲ ． 5．已知曲线3()ln f x ax x =+在(1,(1))f 处的切线的斜率为2，则实数a 的值是 ▲ ． 6．已知等比数列{}n a 中，32a =，4616a a =，则
79
35
a a a a -=- ▲ ．
7．函数sin(2)(0)2
y x ??π
=+<<图象的一条对称轴是12
x π
=
，则?的值是 ▲ ． 8．已知奇函数()f x 在(,0)-∞上单调递减，且(2)0f =，则不等式()
01
f x x >-的解集为 ▲ ．
9．已知tan()24
απ
-=，则cos2α的值是 ▲ ．
10．若函数8,2
()log 5,2a
x x f x x x -+?=?
+>?≤(01)a a >≠且的值域为[6,)+∞，则实数a 的取值范围是 ▲ ． 11．已知数列{},{}n n a b 满足1111
,1,(*)21
n n n n a a b b n a +=+==∈+N ，则122017b b b ??
= ▲ ．
12．设ABC △的内角,,A B C 的对边分别是,,a b c ，D 为AB 的中点，若cos sin b a C c A =+
且CD =，
则ABC △面积的最大值是 ▲ ．
13．已知函数()sin()6
f x x π=-，若对任意的实数5[,]62
αππ
∈-
-，都存在唯一的实数[0,]m β∈，使()()0f f αβ+=，则实数m 的最小值是 ▲ ．
14．已知函数ln ,0
()21,0x x f x x x >?=?+?
≤，若直线y ax =与()y f x =交于三个不同的点(,()),(,()),A m f m B n f n
(,())C t f t （其中m n t <<），则1
2n m
+
+的取值范围是 ▲ ． 二、解答题(本大题共6个小题，共90分，请在答题卷区域内作答，解答时应写出文字说明、证明过程或
演算步骤)
已知函数
21 ()sin(2)(
0,0)
42
f x ax b a b
π
=-+++>>的图象与x轴相切，且图象上相邻两个最高点之
间的距离为
2
π
．
（1）求,a b的值；
（2）求()
f x在[0,]
4
π
上的最大值和最小值．
16．(本题满分14分)
在ABC
△中，角A,B,C所对的边分别是a,b,c，已知sin sin sin()
B C m A m
+=∈R，且240
a bc
-=．（1）当
5
2,
4
a m
==时，求,b c的值；
（2）若角A为锐角，求m的取值范围．
17．(本题满分15分)
已知数列{}
n
a的前n项和是
n
S，且满足
1
1
a=，*
1
31()
n n
S S n
+
=+∈N．
（1）求数列{}
n
a的通项公式；
（2）在数列{}
n
b中，
1
3
b=，*
1
1
()
n
n n
n
a
b b n
a
+
+
-=∈N，若不等式2
n n
a b n
λ+≤对*
n∈N有解，求实数λ的取值范围．
18．(本题满分15分)
如图所示的自动通风设施．该设施的下部ABCD是等腰梯形，其中AB为2米，梯形的高为1米，CD 为3米，上部CmD是个半圆，固定点E为CD的中点．MN是由电脑控制可以上下滑动的伸缩横杆（横杆面积可忽略不计），且滑动过程中始终保持和CD平行．当MN位于CD下方和上方时，通风窗的形状均为矩形MNGH（阴影部分均不通风）．
（1）设MN与AB之间的距离为
5
(0
2
x x<
≤且1)
x≠米，试将通风窗的通风面积S（平方米）表示成关于x的函数()
y S x
=；
（2）当MN与AB之间的距离为多少米时，通风窗的通风面积S取得最大值？
已知函数2()ln ,()f x x g x x x m ==--． （1）求过点(0,1)P -的()f x 的切线方程；
（2）当0=m 时，求函数()()()F x f x g x =-在],0(a 的最大值；
（3）证明：当3m ≥-时，不等式2()()(2)e x f x g x x x +<--对任意1[,1]2
x ∈均成立（其中e 为自然对数的底数,e 2.718...=）．
20．(本题满分16分)
已知数列{}n a 各项均为正数，11a =,22a =，且312n n n n a a a a +++=对任意*n ∈N 恒成立，记{}n a 的前n 项和为n S ．
（1）若33a =，求5a 的值；
（2）证明：对任意正实数p ，221{}n n a pa -+成等比数列；
（3）是否存在正实数t ，使得数列{}n S t +为等比数列．若存在，求出此时n a 和n S 的表达式；若不存在，说明理由．
2017—2018学年第一学期高三期中调研试卷
数学(附加题部分)
21．【选做题】本题包括A 、B 、C 、D 四小题，请选定其中两题，并在相应的答题区域内作答．．．．．．．．．．．．．．．．．．．．若多做，则按作答的前两题评分．解答时应写出文字说明、证明过程或演算步骤． A ．(几何证明选讲)（本小题满分10分）
如图，AB 为圆O 的直径，C 在圆O 上，CF AB ⊥于F ，点D 为线段CF 上任意一点，延长AD 交圆O 于
E ，030AEC ∠=．
（1）求证：AF FO =；
（2
）若CF =，求AD AE ?的值．
B ．(矩阵与变换)（本小题满分10分）
已知矩阵1221??=????A ，42α??
=??
??
，求49αA 的值．
C ．(极坐标与参数方程)（本小题满分10分）
在平面直角坐标系中，直线l 的参数方程为425
25x t y t ?
=-+????=??
(t 为参数)，以原点O 为极点，x 轴正半轴为
极轴建立极坐标系，圆C
的极坐标方程为cos()(0)4
a ρθπ
-≠． （1）求直线l 和圆C 的直角坐标方程；
（2）若圆C 任意一条直径的两个端点到直线l
a 的值．
D ．(不等式选讲)（本小题满分10分）
设,x y 均为正数，且x y >，求证：22
1
2232x y x xy y
++-+≥．
B
【必做题】第22、23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． 22．(本小题满分10分)
在小明的婚礼上，为了活跃气氛，主持人邀请10位客人做一个游戏．第一轮游戏中，主持人将标有数字1,2,…,10的十张相同的卡片放入一个不透明箱子中，让客人依次去摸，摸到数字6,7,…,10的客人留下，其余的淘汰，第二轮放入1,2,…,5五张卡片，让留下的客人依次去摸，摸到数字3,4,5的客人留下，第三轮放入1,2,3三张卡片，让留下的客人依次去摸，摸到数字2,3的客人留下，同样第四轮淘汰一位，最后留下的客人获得小明准备的礼物．已知客人甲参加了该游戏． （1）求甲拿到礼物的概率；
（2）设ξ表示甲参加游戏的轮数．．，求ξ的概率分布和数学期望()E ξ．
23．(本小题满分10分)
（1）若不等式(1)ln(1)x x ax ++≥对任意[0,)x ∈+∞恒成立，求实数a 的取值范围； （2）设*n ∈N ，试比较11
1
23
1
n +++
+与ln(1)n +的大小，并证明你的结论．
2017—2018学年第一学期高三期中调研试卷
数 学 参 考 答 案
一、填空题(本大题共14小题，每小题5分，共70分)
1．{1} 2．(1,2)
(2,)+∞ 3．充分不必要 4．1 5．1
3
6．4 7．3
π
8．(2,0)(1,2)- 9．45- 10．(1,2]
11．12018 12
1 13．2
π
14．1(1,e )e +
二、解答题(本大题共6个小题，共90分) 15．(本题满分14分)
解：（1）∵()f x 图象上相邻两个最高点之间的距离为
2
π
， ∴()f x 的周期为
2
π，∴202||2a a ππ=>且，·
·····································································2分 ∴2a =，··················································································································4分
此时1
())42
f x x b π=+++， 又∵()f x 的图象与x 轴相切，
∴1||02b b +=>，·
······················································6分
∴1
22
b =-；
··········································································································8分 （2）由（1
）可得())242
f x x π=-++
， ∵[0,]4x π∈，∴4[,]444
x ππ5π
+∈，
∴当444x π5π+
=
，即4
x π
=时，()f x
；·················································11分 当442x ππ+=，即16
x π
=时，()f x 有最小值为0．························································14分
16．(本题满分14分)
解：由题意得b c ma +=，240a bc -=．···············································································2分
（1）当52,4a m ==
时，5
,12
b c bc +==， 解得212b c =???=??或122
b c ?
=??
?=?；································································································6分
（2）2
2
222222
22
()()22cos 23222
a ma a
b
c a b c bc a A m a bc bc
--+-+--====-，····························8分 ∵A 为锐角，∴2cos 23(0,1)A m =-∈，∴
23
22
m <<，·
···················································11分 又由b c ma +=可得0m >，·························································································13分
∴
6
22
m <<·
····································································································14分 17．(本题满分15分)
解：（1）∵*131()n n S S n +=+∈N ，∴*131(,2)n n S S n n -=+∈N ≥，
∴*13(,2)n n a a n n +=∈N ≥，·························································································2分 又当1n =时，由2131S S =+得23a =符合213a a =，∴*13()n n a a n +=∈N ，······························3分 ∴数列{}n a 是以1为首项，3为公比的等比数列，通项公式为1*3()n n a n -=∈N ；·····················5分 （2）∵*1
13()n n n n
a b b n a ++-=
=∈N ，∴{}n b 是以3为首项，3为公差的等差数列，·
···················7分 ∴*33(1)3()n b n n n =+-=∈N ，·····················································································9分 ∴2
n n a b n λ+≤，即1
2
3
3n n n λ-?+≤，即2133
n n n
λ--≤对*n ∈N 有解，·
·································10分 设2*
1
3()()3
n n n f n n --=∈N ， ∵2221(1)3(1)32(41)
(1)()333n n n
n n n n n n f n f n -+-+---++-=-=
， ∴当4n ≥时，(1)()f n f n +<，当4n <时，(1)()f n f n +>， ∴(1)(2)(3)(4)(5)(6)f f f f f f <<<>>>，
∴max 4
[()](4)27
f n f ==，···························································································14分 ∴427
λ≤
．·············································································································15分 18．(本题满分15分)
解：（1）当01x <≤时，过A 作AK CD ⊥于K （如上图），
则1AK =，1
22
CD AB DK -==，1HM x =-， 由2AK MH DK DH ==，得122
HM x DH -==， ∴322HG DH x =-=+，
∴2()(1)(2)2S x HM HG x x x x =?=-+=--+；·······························································4分
当5
12
x <<
时，过E 作ET MN ⊥于T ，连结EN （如下图）， 则1ET x =-，2
2239(1)(1)224MN TN x x ??
==---- ???
∴29
2
(1)4
MN x =-- ∴29
()2
(1)(1)4
S x MN ET x x =?=---，
······································································8分 综上：22
2,01()952(1)(1)142x x x S x x x x ?--+
=?---<
≤；·································································9分 （2）当01x <≤时，221
9
()2()24
S x x x x =--+=-++
在[0,1)上递减， ∴max ()(0)2S x S ==；································································································11分
2?当5
12
x <<时，22
29
(1)(1)994()2(1)(1)2424
x x S x x x -+
--=---?
=≤，
当且仅当29(1)(1)4x x -=
--325
1(1,)42
x =+∈时取“=”， ∴max 9()4S x =
，此时max 9
()24
S x =>，∴()S x 的最大值为94，············································14分 答：当MN 与AB 之间的距离为32
14
+米时，通风窗的通风面积S 取得最大值．·
···················15分 19．(本题满分16分)
解：（1）设切点坐标为00(,ln )x x ，则切线方程为000
1
ln ()y x x x x -=
-， 将(0,1)P -代入上式，得0ln 0x =，01x =，
∴切线方程为1y x =-；·······························································································2分 （2）当0m =时，2()ln ,(0,)F x x x x x =-+∈+∞， ∴(21)(1)
(),(0,)x x F x x x
+-'=-
∈+∞，
············································································3分 当01x <<时，()0F x '>，当1x >时，()0F x '<，
∴()F x 在(0,1)递增，在(1,)+∞递减，·············································································5分 ∴当01a <≤时，()F x 的最大值为2()ln F a a a a =-+；
当1a >时，()F x 的最大值为(1)0F =；········································································7分 （3）2()()(2)e x f x g x x x +<--可化为(2)e ln x m x x x >-+-，
设1
()(2)e ln ,[,1]2x h x x x x x =-+-∈，要证3m ≥-时()m h x >对任意1[,1]2
x ∈均成立， 只要证max ()3h x <-，下证此结论成立．
∵1()(1)(e )x h x x x
'=--，∴当1
12x <<时，10x -<，·
······················································8分 设1()e x u x x =-，则21()e 0x u x x '=+>，∴()u x 在1
(,1)2
递增，
又∵()u x 在区间1[,1]2
上的图象是一条不间断的曲线，且1
()202
u <，(1)e 10u =->，
∴01(,1)2
x ?∈使得0()0u x =，即00
1
e x x =
，00ln x x =-，····················································11分 当01(,)2
x x ∈时，()0u x <，()0h x '>；当0(,1)x x ∈时，()0u x >，()0h x '<； ∴函数()h x 在01[,]2
x 递增，在0[,1]x 递减， ∴0max 000000000
12
()()(2)e ln (2)212x h x h x x x x x x x x x ==-+-=-?
-=--，
····························14分 ∵212y x x =-
-在1
(,1)2
x ∈递增，∴0002()121223h x x x =--<--=-，即max ()3h x <-，
∴当3m ≥-时，不等式2()()(2)e x f x g x x x +<--对任意1
[,1]2
x ∈均成立．··························16分 20．(本题满分16分)
解：（1）∵1423a a a a =，∴46a =，又∵2534a a a a =，∴543
92
a a ==；·
······································2分
（2）由312
1423
n n n n n n n n a a a a a a a a +++++++=??=?，两式相乘得2134123n n n n n n n a a a a a a a ++++++=，
∵0n a >，∴2*42()n n n a a a n ++=∈N ，
从而{}n a 的奇数项和偶数项均构成等比数列，···································································4分 设公比分别为12,q q ，则1122222n n n a a q q --==，1121111n n n a a q q ---==，······································5分 又∵
312=
n n n n a a a a +++，∴422311
22a a q
a a q ===，即12q q =，···························································6分 设12q q q ==，则2212223()n n n n a pa q a pa ---+=+，且2210n n a pa -+>恒成立，
数列221{}n n a pa -+是首项为2p +，公比为q 的等比数列，问题得证；····································8分 （3）法一：在（2）中令1p =，则数列221{}n n a a -+是首项为3，公比为q 的等比数列，
∴22212223213 ,1
()()()3(1),11k k k k k k k q S a a a a a a q q q
---=??
=++++
++=-?≠?-?
， 121221
32 ,13(1)2,11k k k k k k k q q S S a q q q q ---?-=?
=-=?--≠?-?
，·····································································10分 且12341,3,3,33S S S q S q ===+=+，
∵数列{}n S t +为等比数列，∴2
2132
324()()(),
()()(),S t S t S t S t S t S t ?+=++??+=++?? 即2
2(3)(1)(3),(3)(3)(33),
t t q t q t t q t ?+=+++??++=+++??，即26(1),3,t q t t q +=+??=-? 解得14t q =??=?
（3t =-舍去）， (13)
分
∴224121k k k S =-=-，212121k k S --=-， 从而对任意*n ∈N 有21n n S =-， 此时2n n S t +=，
12n n S t
S t
-+=+为常数，满足{}n S t +成等比数列，
当2n ≥时，1
11222n n n n n n a S S ---=-=-=，又11a =，∴1*2()n n a n -=∈N ，
综上，存在1t =使数列{}n S t +为等比数列，此时1*2,21()n n n n a S n -==-∈N ．······················16分 法二：由（2）知，则122n n a q -=，121n n a q --=，且12341,3,3,33S S S q S q ===+=+，
∵数列{}n S t +为等比数列，∴2
2132
324()()(),
()()(),S t S t S t S t S t S t ?+=++??+=++?? 即2
2
(3)(1)(3),(3)(3)(33),
t t q t q t t q t ?+=+++??++=+++??，即26(1),3,t q t t q +=+??=-? 解得1
4t q =??=?
（3t =-舍去）， (11)
∴121222n n n a q --==，22212n n a --=，从而对任意*n ∈N 有12n n a -=，····································13分 ∴0
1
2
1
122222
2112
n
n n n S --=++++==--， 此时2n n S t +=，
12n n S t
S t
-+=+为常数，满足{}n S t +成等比数列，
综上，存在1t =使数列{}n S t +为等比数列，此时1*2,21()n n n n a S n -==-∈N ．······················16分 21．【选做题】本题包括A 、B 、C 、D 四小题，请选定其中两题，并在相应的答题区域内作答．．．．．．．．．．．．．．．．．．．．若多做，则按作答的前两题评分．解答时应写出文字说明、证明过程或演算步骤． A ．(几何证明选讲，本小题满分10分)
解：（1）证明 ：连接,OC AC ，∵030AEC ∠=，∴0260AOC AEC ∠=∠=，
又OA OC =，∴AOC ?为等边三角形，
∵CF AB ⊥，∴CF 为AOC ?中AO 边上的中线，
∴AF FO =；······································································5分 （2）解：连接BE ，
∵CF ，AOC ?是等边三角形， ∴可求得1AF =，4AB =，
∵AB 为圆O 的直径，∴90AEB ∠=，∴AEB AFD ∠=∠， 又∵BAE DFA ∠=∠，∴AEB ?∽AFD ?，∴
AD AF
AB AE
=
， 即414AD AE AB AF ?=?=?=．··················································································10分 B ．(矩阵与变换，本小题满分10分) 解：矩阵A 的特征多项式为21
2
()2321
f λλλλλ--=
=----， 令()0f λ=，解得矩阵A 的特征值121,3λλ=-=，····························································2分
当11λ=-时特征向量为111α??=??-??，当23λ=时特征向量为211α??=????，·····································6分
又∵12432ααα??
==+????
，······························································································8分
∴5049
49
49
1
1225031331αλαλα??
-=+=??+??
A ．···········································································10分
C ．(极坐标与参数方程，本小题满分10分)
解：（1）直线l 的普通方程为220x y +-=； (3)
B
圆C 的直角坐标方程为2
22()()222
a a a x y -+-=；·······························································6分
（2）∵圆C 任意一条直径的两个端点到直线l
∴圆心C 到直线l
|
2|
2a
a +-=，·······················································8分 解得3a =或1
3
a =-．·······························································································10分
D ．(不等式选讲，本小题满分10分) 证：∵0,0,0x y x y >>->，
∴222
11
222()2()
x y x y x xy y x y +
-=-+-+-
21()()3()x y x y x y =-+-+
=-≥， ∴22
1
2232x y x xy y +
+-+≥．
····················································································10分 22．(本题满分10分)
解：（1）甲拿到礼物的事件为A ，
在每一轮游戏中，甲留下的概率和他摸卡片的顺序无关，
则13211()253210
P A =
???=， 答：甲拿到礼物的概率为1
10
；·······················································································3分 （2）随机变量ξ的所有可能取值是1,2,3,4．·
····································································4分 ()1
12
P ξ==，
()121
2255P ξ==?=，
()1311
325310P ξ==??=，
()1321
42535
P ξ==??=，
随机变量ξ的概率分布列为：
ξ
1 2 3 4
P
12 15 110 15 所以1111
()1234225105
E ξ=?+?+?+?=．
····································································10分 ·············································8分
23．(本题满分10分)
解：（1）原问题等价于ln(1)01
ax
x x +-
+≥对任意[0,)x ∈+∞恒成立， 令()ln(1)1
ax
g x x x =+-+，则21'()(1)x a g x x +-=+，
当1a ≤时，2
1'()0(1)
x a
g x x +-=
+≥恒成立，即()g x 在[0,)+∞上单调递增， ∴()(0)0g x g =≥恒成立；
当1a >时，令'()0g x =，则10x a =->，
∴()g x 在(0,1)a -上单调递减，在(1,)a -+∞上单调递增， ∴(1)(0)0g a g -<=，即存在0x >使得()0g x <，不合题意；
综上所述，a 的取值范围是(,1]-∞．················································································4分 （2）法一：在（1）中取1a =，得ln(1)((0,))1
x
x x x +>∈+∞+， 令*1()x n n =
∈N ，上式即为11
ln()1
n n n +>
+， 即1
ln(1)ln 1n n n +->+，·····························································································7分
∴1ln 2ln1,21ln 3ln 2,31ln(1)ln ,1n n n ?
->??
?->????
?+->?+?
上述各式相加可得
111
ln(1)231n n +++<++*()n ∈N ．
····················································10分 法二：注意到1ln 22<，11
ln323+<，……，
故猜想111
ln(1)231
n n +++<++*()n ∈N ，·
···································································5分 下面用数学归纳法证明该猜想成立．
证明：①当1n =时，1
ln 22
<，成立；·············································································6分
②假设当n k =时结论成立，即111ln(1)231k k +++<++， 在（1）中取1a =，得ln(1)((0,))1x
x x x +>∈+∞+，
令*1
()1
x k k =∈+N ，有
12ln()21k k k +<++，·······································································8分 那么，当1n k =+时，
11
111ln(1)ln(1)23
122ln()ln(2)21
k k k k k k k k +++
+<++<++++++=++，也成立；
由①②可知，111
ln(1)23
1
n n +++
<++．·
····································································10分