江苏省苏州市2020届高三数学期中调研试题

2020学年第一学期高三期中调研试卷

数 学

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 =eI ▲ ． 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 ??=L ▲ ．

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 =ABC △面积的最大值是 ▲ ．

13．已知函数()sin()6f 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．

17．(本题满分15分)

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？

19．(本题满分16分)

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．

2020学年第一学期高三期中调研试卷

数学 (附加) 2020.11 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，0

30

AEC

∠=．

（1）求证：AF FO

=；

（2

）若CF=AD AE

?的值．

B．(矩阵与变换)

（本小题满分10分）

已知矩阵

12

21

??

=??

??

A，

4

2

α

??

=??

??

u r

，求49α

u r

A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

4

2

5

2

5

x 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

ρθ

π

=-≠．

（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．

B

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)

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

231

n +++

+L 与ln(1)n +的大小，并证明你的结论．

2020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,)+∞U 3．充分不必要 4．1 5．13

6．4 7．3π 8．(2,0)(1,2)-U 9．4

5

- 10．(1,2]

11．

1

2018

121 13．2π 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||2

a 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())242f x x π=++

， ∵[0,]4

x π

∈，∴4[,]444x ππ5π+∈，

∴当

444

x π5π+

=，即4

x π=

时，()f x 有最大值

为

1

2

；·················································11分

当

442

x 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2

12b 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

∴

2

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

2131

S S =+得

23

a =符合

21

3a 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12

33n n n λ-?+≤，即2133

n n n

λ--≤对*n ∈N 有

解，··································10分

设2*

1

3()()3n 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 <<<>>>L ， ∴

max 4

[()](4)27

f n f ==

，···························································································14分

∴

4

27

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：

22,01()52(12x x x S 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

9

(1)(1)94()2(224

x x S x x -+

--=-?

=，

当且仅当(1)x -

5

1(1,)2

x =+∈时取“=”， ∴

max 9()4

S x =

，

此时

max 9()24

S x =

>，∴()S x 的最大值为

9

4

，············································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

12

x <<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

1e 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

）

∵

1423

a a a a =，∴

46

a =，又∵

2534

a 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

11

22222n n n a a q q --==，

1121111n n n a a q q ---==，······························

········5分

又

∵

31

2=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

---=??

=++++++=-?≠?-?

L ，

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

1

4

t 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

1

4

t q =??

=?（3t =-舍

去），·······················································································11分

∴

121

222n n n a q --==，

22

212n n a --=，从而对任意

*

n ∈N 有

12n n a -=，···································

·13分

∴0

1

2

1

1222222112

n n n n S --=++++==--L ， 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 =，

B

∵AB 为圆O 的直径，∴90AEB ∠=o ，∴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??

=??-??

u u r

，当23λ=时特征向量为

211

α??

=????

u u r

，···································

··6分

又

∵

12432ααα??==+????

u r

u u

r u u r ，·······························

·······························································8分

∴

5049

4949

1

1225031331αλαλα??

-=+=??+??

u r u u r u u r

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 的距离

为

2

，

即

|

2|

a

a +-=，·······················································8分

解

得

3

a =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 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2

1'()(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 1

n n n +->

+，·····························································································7分

∴1ln 2ln1,21ln 3ln 2,31ln(1)ln ,1n n n ?

->??

?->????

?+->?+?

L L 上述各式相加可得

111ln(1)231

n n +++<++L *()n ∈N ．····················································10分

法二：注意到1ln 22<，11

ln323

+<，……， 故猜想

111

ln(1)231

n n +++<++L *()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 +++<++L ， 在（1）中取1a =，得ln(1)((0,))1

x

x x x +>∈+∞+，

令*1

()1

x k k =∈+N ，有

12

ln()21

k k k +<++，·······································································8分

那么，当1n k =+时，

11111ln(1)ln(1)23122

ln()ln(2)21

k k k k k k k k ++++<++<++++++=++L ，也成立；

由①②可知，

111ln(1)231

n n +++<++L ．·····································································10分