江苏省苏州市2017-2018学年第一学期高三期中调研试卷数学(理)

2017—2018学年第一学期高三期中调研试卷 数
学 2017.11
注意事项：
1．本试卷共4页．满分160分，考试时间120分钟．
2．请将填空题的答案和解答题的解题过程写在答题卷上，在本试卷上答题无效． 3．答题前，务必将自己的姓名、学校、准考证号写在答题纸的密封线内．
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 ϕϕπ
=+<<图象的一条对称轴是12
x π
=
，则ϕ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) 15．(本题满分14分)
已知函数1
())(0,0)242
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240a 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．
已知数列{}n a 的前n 项和是n S ，且满足11a =，*131()n n S S n +=+∈N ． （1）求数列{}n a 的通项公式； （2）在数列{}n b 中，13b =，*1
1()n n n n
a b b n a ++-=
∈N ，若不等式2n 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
(02
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) 2017.11
注意事项：
1．本试卷共2页．满分40分，考试时间30分钟． 2．请在答题卡上的指定位置作答，在本试卷上作答无效．
3．答题前，请务必将自己的姓名、学校、考试证号填写在答题卡的规定位置．
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
，求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 +
+-+≥．
【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)
B
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
111
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 121 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
())242
f x x b π=+++， 又∵()f x 的图象与x 轴相切，
∴1||022
b b +=>，·
······················································6分
∴1
2
b =-；
··········································································································8分 （2）由（1
）可得())4f x x π=+， ∵[0,]4x π∈，∴4[,]444
x ππ5π
+∈，
∴当444x π5π+
=
，即4
x π
=时，()f x
有最大值为12；·················································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122b 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
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
∴2n 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)()333
n 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(1)x -=
51(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，111222n 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，∴221323
24()()(),()()(),S t S t S t S t S t S t ⎧+=++⎪⎨+=++⎪⎩ 即22(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），·······················································································11分
∴121222n n n a q --==，22212n n a --=，从而对任意*n ∈N 有12n n a -=，····································13分 ∴01211222222112n 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
B
（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 ∴50494949
11225031331αλαλα⎡⎤-=+=⎢⎥+⎣⎦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分
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|a a +-，·······················································8分 解得3a =或13a =-
．·······························································································10分
D ．(不等式选讲，本小题满分10分)
证：∵0,0,0x y x y >>->， ∴222
11222()2()x y x y x xy y x y +-=-+-+-
21()()3()x y x y x y =-+-+
=-≥，。