江苏省苏州市2018届第一学期期末高三调研测试数学试卷及参考答案
苏州市2018届高三调研测试
一、填空题:本大题共14小题,每小题5分,共计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i
为虚数单位,复数3
i 2
z 的模为 . 2. 已知集合{1,2}a A =,{1,1,4}B =-,且A B ?,则正整数a = . 3. 在平面直角坐标系xOy 中,抛物线28y x =-的焦点坐标为 . 4. 苏州轨道交通1号线每5分钟一班,其中,列车在车站停留0.5
分钟,假设乘客到达站台的时刻是随机的,则该乘客到达站台
立即能乘上车的概率为 .
5. 已知42a =,log 2a x a =,则正实数x = .
6. 秦九韶是我国南宋时期的数学家,他在所著的《数书九章》中
提出的多项式求值的秦九韶算法,至今仍是比较先进的算法. 右边的流程图是秦九韶算法的一个实例.若输入n ,x 的值分别 为3,3,则输出v 的值为 .
7. 已知变量x ,y 满足03,0,30,x x y x y ??
+??-+?
≤≤≥≤则23z x y =-的最大值为 .
8. 已知等比数列{}n a 的前n 项和为n S ,且63198S S =-,4215
8
a a =--,则3a 的值为 .
9. 鲁班锁是中国传统的智力玩具,起源于中国古代建筑中首创的
榫卯结构,它的外观是如图所示的十字立方体,其上下、左右、 前后完全对称,六根等长的正四棱柱体分成三组,经90°榫卯 起来.若正四棱柱的高为5,底面正方形的边长为1,现将该鲁
班锁放进一个球形容器内,则该球形容器的表面积至少为 .(容器壁的厚度忽略不计,结果保留π)
10.如图,两座建筑物AB ,CD 的高度分别是9m 和15m ,从建筑物AB
的顶部A 看建筑物CD 的张角45CAD ∠=?,则这两座建筑物AB 和CD 的底部之间的距离BD = m .
11.在平面直角坐标系xOy 中,已知过点(2,1)A -的圆C 和直线 x + y = 1相切,且圆心在直线 y = -2x
上,则圆C 的标准方程为 .
12.已知正实数a ,b ,c 满足11
1a b
+=,
111a b c +=+,则c 的取值范围是 .
13.如图,△ABC 为等腰三角形,120BAC ∠=?,4AB AC ==,以A 为圆心,1为半径的圆分别
交AB ,AC 与点E ,F ,点P 是劣弧EF 上的一点,则PB PC ?的取值范围是 .
14.已知直线y =a 分别与直线22y x =-,曲线2e x y x =+交于点A ,B ,则线段AB 长度的最小值
为 . D
C
B
A
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,解答时应写出文字说明、证明过程或演算步骤. 15.(本小题满分14分)
已知函数2()sin )2f x x x x =+-.
(1)求函数()f x 的最小值,并写出()f x 取得最小值时自变量x 的取值集合;
(2)若,22x ππ??
∈-????
,求函数()f x 的单调增区间.
16.(本小题满分14分)
如图,在正方体1111ABCD A B C D -中,已知E ,F ,G ,H 分别是A 1D 1,B 1C 1,D 1D ,C 1C 的中点.
(1)求证:EF ∥平面ABHG ; (2)求证:平面ABHG ⊥平面CFED .
A 1
B 1
C 1
D 1
A
B
C
D
E
F G H
如图,B ,C 分别是海岸线上的两个城市,两城市间由笔直的海滨公路相连,B ,C 之间的距离为100km ,海岛A 在城市B 的正东方50km 处.从海岛A 到城市C ,先乘船按北偏西θ角(π
2
αθ<≤,其中锐角α的正切值为12
)航行到海岸公路P 处登陆,再换乘汽车到城市C .已知船速为25km/h ,车速为75km/h .
(1)试建立由A 经P 到C 所用时间与θ的函数解析式; (2)试确定登陆点P 的位置,使所用时间最少,并说明理由.
18.(本小题满分16分)
在平面直角坐标系xOy 中,椭圆22
22:1(0)x y C a b a b
+=>>
P 到一
个焦点的距离的最小值为1). (1)求椭圆C 的标准方程;
(2)已知过点(0,1)M -的动直线l 与椭圆C 交于 A ,B 两点,试判断以AB 为直径的圆是否恒过定点,并说明理由.
A
已知各项是正数的数列{}n a 的前n 项和为n S .
(1)若212
3
n n n a S S -++=(n ∈N *,n ≥2),且12a =.
① 求数列{}n a 的通项公式;
② 若12n n S λ+?≤对任意*n ∈N 恒成立,求实数λ的取值范围;
(2)数列{}n a 是公比为q (q >0, q ≠1)的等比数列,且{a n }的前n 项积.为10n T .若存在正整数k ,对任意n ∈N *,使得(1)k n kn
T T +为定值,求首项1a 的值.
已知函数32,0,
()e ,0.
x x x x f x ax x ?-+=?-??≥
(1)当2a =时,求函数()f x 的单调区间;
(2)若方程()()e 3x f x f x -+=-在区间(0,+∞)上有实数解,求实数a 的取值范围; (3)若存在实数,[0,2]m n ∈,且||1m n -≥,使得()()f m f n =,求证:1e e 1
a
-≤≤.
苏州市2018届高三调研测试数学试卷参考答案
一、填空题(共70分) 1
2.2
3.(2,0)-
4.
110
5.12
6.48 7.9- 8.
94 9.30π
10.18 11.22(1)(2)2x y -++= 12.4
(1,]3
13.[11,9]--
14.
3ln 2
2
+ 二、解答题(共90分)
15. 解(1
)2()sin )2f x x x x =+-
223cos cos sin 2x x x x x =++-
3(1cos2)1cos2222
x x
x +-=+ ·
··················································· 2分
cos 222x x =-+2cos(2)23
x π
=++. ··········································· 4分
当223
x k π+=π+π,即()3x k k π
=π+∈Z 时,()f x 取得最小值0.
此时,()f x 取得最小值时自变量x 的取值集合为,3x x k k π??
=π+∈????
Z .
····································································································· 7分
(注:结果不写集合形式扣1分)
(2)因为()2cos(2)23
f x x π
=++,
令2222()3
k x k k π
π+π+π+π∈Z ≤≤, ··············································· 8分
解得()36k x k k π5π+π+π∈Z ≤≤, ·
···················································· 10分 又[,]22
x ππ∈-,令1k =-,,26x ππ??∈--????,令0k =,,32x ππ??
∈????,
所以函数在[,]22
ππ-的单调增区间是,26ππ??--????和,32ππ??
????. ························ 14分
(注:如果写成两区间的并集,扣1分,其中写对一个区间给2分) 16. 证明:(1)因为E ,F 是A 1D 1,B 1C 1的中点,所以11EF A B ∥, 在正方体1111ABCD A B C D -中,A 1B 1∥AB , (注:缺少A 1B 1∥AB 扣1分)
所以EF AB ∥. ········································ 3分 又EF ?平面ABHG ,AB ?平面ABHG , (注:缺少AB ?平面ABHG 不扣分)
所以EF ∥平面ABHG . ······························· 6分 (2)在正方体ABCD ?A 1B 1C 1D 1中,CD ⊥平面BB 1C 1C ,
又BH ?平面11BB C C ,所以BH CD ⊥.① ············································ 8分 设BH
CF P =,△BCH ≌△1CC F ,所以1HBC FCC ∠=∠,
因为∠HBC +∠PHC =90?,所以1FCC ∠+∠PHC =90?.
所以90HPC ∠=?,即BH CF ⊥.② ···················································· 11分 由①②,又DC
CF C =,DC ,CF ?平面CFED ,
所以BH ⊥平面CFED .
A 1
B 1
C 1
D 1 A B C D
E F
G H P
又BH ?平面ABHG ,
所以平面ABHG ⊥平面CFED . ··························································· 14分 (注:缺少BH ?平面ABHG ,此三分段不给分)
17. 解(1)由题意,轮船航行的方位角为θ,所以90BAP θ∠=?-,50AB =,
则5050cos(90)sin AP θθ
==?-,50sin(90)50cos 50tan(90)cos(90)sin BP θθ
θθθ?-=?-==
?-. 50cos 100100sin PC BP θ
θ
=-=-
. ························································· 4分 (注:AP ,BP 写对一个给2分)
由A 到P 所用的时间为12
25sin AP t θ
==
, 由P 到C 所用的时间为250cos 10042cos sin 7533sin t θ
θθθ
-
=
=-, ·························· 6分 所以由A 经P 到C 所用时间与θ的函数关系为
12242cos 62cos 4
()sin 33sin 3sin 3
t f t θθθθθθ-==
+=++-. ·
································ 8分 函数()f θ的定义域为(,]2
απ,其中锐角α的正切值为1
2.
(2)由(1),62c o s 4()3sin 3
f θθθ-=+,(,]2θαπ
∈,
2(13cos )()9si 6n f θθθ
-'=,令()0f θ'=,解得1
cos 3θ=
, ······························· 10分 设θ0∈(0,)
π,使01
cos θ=
················
·······················································
···········
··················· 12分
所以,当0θθ=时函数f (θ)取得最小值,此时BP =
0050cos sin θθ≈17.68 km ,
答:在BC 上选择距离B 为17.68 km 处为登陆点,所用时间最少.·
···········
14分
(注:结果保留根号,不扣分)
18. 解(1)由题意
c a =,故a =, ··················································· 1分 又椭圆上动点P 到一个焦点的距离的最小值为1),所以3a c -=, ······
······························································································· 2分 解得3c =,a =2229b a c =-=, ········································· 4分
所以椭圆C 的标准方程为22
1189
x y +=. ·
·················································· 6分 (2)当直线l 的斜率为0时,令1y =-,则4x =±,
此时以AB 为直径的圆的方程为2(1)16x y ++=. ···································· 7分
联立222
(1)16,9,x y x y ?++=??+=??
解得0,3x y ==,即两圆过点(0,3)T . 猜想以AB 为直径的圆恒过定点(0,3)T . ··············································· 9分 对一般情况证明如下:
设过点(0,1)M -的直线l 的方程为1y kx =-与椭圆C 交于1122(,),(,)A x y B x y ,
则22
1,218,
y kx x y =-??+=?整理得22(12)4160k x kx +--=, 所以1212
22
416
,1212k x x x x k k +==-++. ················································· 12分 (注:如果不猜想,直接写出上面的联立方程、韦达定理,正确的给3分) 因为1122121212(,3)(,3)3()9TA TB x y x y x x y y y y ?=-?-=+-++
121212(1)(1)3(11)9x x kx kx kx kx =+----+-+21212(1)4()16k x x k x x =+-++
222222
16(1)1616(12)16160121212k k k k k k
-+-+=-+=+=+++, 所以TA TB ⊥.
所以存在以AB 为直径的圆恒过定点T ,且定点T 的坐标为(0,3). ·············· 16分
19. 解(1)①当2n ≥时,由2
12
,3
n n n a S S -++= ①
则2112
,3
n n n a S S ++++= ②
②-①得22
111()3
n n n n a a a a ++-=-,即13n n a a +-=,2n ≥
···························· 2分 当2n =时,由①知2212123
a a a a +++=,即2
2
23100a a --=, 解得25a =或22a =-(舍),
所以213a a -=,即数列{}n a 为等差数列,且首项13a =,
所以数列{}n a 的通项公式为31n a n =-. ················································· 5分 (注:不验证213a a -=扣1分)
②由①知,31n a n =-,所以2(312)322n n n n n S -++==
, 由题意可得212322n n n S n n
λ+++=≥对一切*n ∈N 恒成立,
记2232
n n n n
c ++=,则211
3(1)(1)2n n n n c -+-+-=,2n ≥, 所以212
3114
2n n n n n c c -+-+--=,2n ≥, ················································ 8分
当4n >时,1n n c c -<,当4n =时,41316c =
,且31516c =,27
8
c =,112c =,
所以当3n =时,2232n n n n c ++=取得最大值15
16
,
所以实数λ的取值范围为15
[,)16
+∞. ······················································· 11分
(2)由题意,设11n n a a q -=(0,1q q >≠),1210n T n a a a ???=,两边取常用对数,
12lg lg lg n n T a a a ++
+=.
令1lg lg lg lg n n b a n q a q ==+-,
则数列{}n b 是以1lg a 为首项,lg q 为公差的等差数列, ····························· 13分
若
(1)k n kn
T T +为定值,令
(1)k n kn
T T μ+=,则
11(1)[(1)1]
(1)lg lg 2(1)
lg lg 2
k n k n k n a q
kn kn kn a q
μ++-++
=-+, 即2
2
2
1{[(1)]lg }[(1)](lg )lg 0a k k q n k k q q
μμ+-++-=对*n ∈N 恒成立,
因为0,1q q >≠,问题等价于22
2
1(1)0,
(1)0.
k k k k a q μμ?+-=??+-==??或
将
1
k k
+=(1)0k k μ+-=,解得01μμ==或. 因为*k ∈N ,所以0,1μμ>≠,
所以21a q =,又0,n a >
故1a =. ························································ 16分
20. 解(1)当2a =-时,32
,0,
()e +2,0,
x x x x f x x x ?-+=???≥
当0x <时,32()f x x x =-+,则2()32(32)f x x x x x '=-+=--,
令()0f x '=,解得0x =或2
3
x =
(舍),所以0x <时,()0f x '<, 所以函数()f x 在区间(,0)-∞上为减函数. ··············································· 2分 当0x ≥时,()e 2x f x x =-,()e 2x f x '=-,
令()0f x '=,解得ln2x =,当0ln2x <<时,()0f x '<,当ln2x >时,()0f x '>, 所以函数()f x 在区间(0,ln 2)上为减函数,在区间(ln 2,)+∞上为增函数, 且(0)10f =>. ················································································· 4分 综上,函数()f x 的单调减区间为(,0)-∞和(0,ln 2),单调增区间为(ln 2,)+∞.
····································································································· 5分 (注:将单调减区间为(,0)-∞和(0,ln 2)写出(,ln 2)-∞的扣1分) (2)设0x >,则0x -<,所以32()()e x f x f x x x ax -+=++-, 由题意,32e e 3x x x x ax ++-=-在区间(0,)+∞上有解, 等价于23
a x x x
=++在区间(0,)+∞上有解. ············································· 6分 记23
()(0)g x x x x x
=++
>,
则322222
323(1)(233)
()21x x x x x g x x x x x +--++'=+-==
, ························ 7分 令()0g x '=,因为0x >,所以22330x x ++>,故解得1x =, 当(0,1)x ∈时,()0g x '<,当(1,)x ∈+∞时,()0g x '>,
所以函数()g x 在区间(0,1)上单调递减,在区间(1,)+∞上单调递增,
故函数()g x 在1x =处取得最小值(1)5g =. ············································· 9分 要使方程()a g x =在区间(0,)+∞上有解,当且仅当min ()(1)5a g x g ==≥, 综上,满足题意的实数a 的取值范围为[5,)+∞. ······································· 10分 (3)由题意,()e x f x a '=-,
当0a ≤时,()0f x '>,此时函数()f x 在[0,)+∞上单调递增,
由()()f m f n =,可得m n =,与条件||1m n -≥矛盾,所以0a >. ·············· 11分 令()0f x '=,解得ln x a =,
当(0,ln )x a ∈时,()0f x '<,当(ln ,)x a ∈+∞时,()0f x '>, 所以函数()f x 在(0,ln )a 上单调递减,在(ln ,)a +∞上单调递增.
若存在,[0,2]m n ∈,()()f m f n =,则ln a 介于m ,n 之间, ······················ 12分 不妨设0ln 2m a n <<≤≤,
因为()f x 在(,ln )m a 上单调递减,在(ln ,)a n 上单调递增,且()()f m f n =, 所以当m x n ≤≤时,()()()f x f m f n =≤,
由02m n <≤≤,||1m n -≥,可得1[,]m n ∈,故(1)()()f f m f n =≤, 又()f x 在(,ln )m a 上单调递减,且0ln m a <≤,所以()(0)f m f ≤.
所以(1)(0)f f ≤,同理(1)(2)f f ≤. ··················································· 14分
即2
e 1,
e e 2,
a a a -??--?≤≤解得2e 1e e a --≤≤, 所以1e e 1
a
-≤≤.·
············································································· 16分
2018届高三调研测试数学附加题参考答案
21B 选修4-2 矩阵与变换
解 矩阵M 的特征多项式为212
()2321
f λλλλλ--=
=----, ··················· 2分
令()0f λ=,解得123,1λλ==-,解得
属于λ1的一个特征向量为111??
=????
α,属于λ2的一个特征向量为211??=??-??α. ······· 5分
令12m n =+βαα,即111711m n ??????=+??????-??????,所以1,
7,m n m n +=??-=?
解得4,3m n ==-.
····································································································· 7分 所以44441212(43)4()3()=-=-M M M M βαααα
44441122113214()3()433(1)11327λλ??????
=-=?-?-=??????-??????
αα. ············· 10分
解 由曲线C 的极坐标方程是2
2cos =
sin θ
ρθ
,得ρ2sin 2θ=2ρcos θ. 所以曲线C 的直角坐标方程是y 2=2x . ··················································· 2分
由直线l 的参数方程1,
3x t y t =+??=-? (t 为参数),得40x y --=,
所以直线l 的普通方程为40x y --=. ················································· 4分 将直线l 的参数方程代入曲线C 的普通方程y 2=2x ,得2870t t -+=,
设A ,B 两点对应的参数分别为t 1,t 2,
所以12|AB t t -=== ············· 7分 因为原点到直线40x y --=
的距离d ==,
所以△AOB
的面积是11
1222
S AB d =
??=??=. ·
···················· 10分 21D 选修4-5 不等式选讲
解 因为a ,b ,c ∈R ,2221a b c ++=,
由柯西不等式得2222()()(111)3a b c a b c -+++++=≤, ·························· 4分
因为2|1||1|()x x a b c -++-+≥对一切实数a ,b ,c 恒成立, 所以|1||1|3x x -++≥. 当1x <-时,23x -≥,即3
2
x -
≤; 当11x -≤≤时,23≥不成立; 当1x >时,23x ≥,即32
x ≥;
综上,实数x 的取值范围为3
3(,][,)22
-∞-+∞. ··················
22. 解(1)因为平面ABCD ⊥平面ABEP ,平面ABCD ∩平面ABEP =AB ,BP ⊥AB ,
所以BP ⊥平面ABCD ,又AB ⊥BC ,所以直线BA ,BP ,BC 两两垂直,
以B 为原点,分别以BA ,BP ,BC 为x 轴,y 轴,z 轴建立如图所示的空间直角坐标系,则P (0,2,0),B (0,0,0),D (2,0,1),E (2,1,0),C (0,0,1),
因为BC ⊥平面ABPE ,所以(0,0,1)BC =为平面ABPE 的一个法向量, 2分
(2,2,1),(2,0,0)PD CD =-=,设平面PCD 的一个法向量为(,,)x y z =n , 则0,0,
CD PD ??=???=??n n 即20,220,x x y z =??-+=?令1y =,则2z =,故(0,1,2)=n ,
4分
设平面PCD 与平面ABPE 所成的二面角为θ,则cos ||||1BC BC θ?===??n n ,
显然π02θ<<
,所以平面PCD 与平面ABPE ·
···· 6分 (2)设线段PD 上存在一点N ,使得直线BN 与平面PCD 所成角α的正弦值等于25
. 设(2,2,)(01)PN PD λλλλλ==-≤≤,(2,22,)BN BP PN λλλ=+=-. ·
·· 7分 由(1)知,平面PCD 的一个法向量为(0,1,2)=n , 所以2cos ,5
BN BN BN ?<>=
=
=
?n n |||n |
, 即29810λλ--=,解得1λ=或1
λ=-(舍去). ·
································· 9分
当点N 与点D 重合时,直线BN 与平面PCD 所成角的正弦值为2
5
. ··········· 10分 23. 解(1)因为()[(1)1]2[2(1)]f n f n f n ++=-+,整理得4()
(1)()2
f n f n f n -+=+,
由(1)2f =,代入得421(2)222f -==+,1472(3)15
22
f -
==+,所以719(3)(2)5210f f -=-=. 2分 (2)由(1)2f =,1(2)2f =,可得41
,55
a b =-=. ·
································ 3分 以下用数学归纳法证明
存在实数,41,5
5a b =-=
,使1
()1431()525
n f n =+---成立.
① 当1n =时,显然成立. ·
································································ 4分 ② 当n k =时,假设存在41
,55a b =-=,使得1()1431()525
k f k =+---成立,
····································································································· 5分
那么,当1n k =+时,141431()()4()525(1)1()212431()()525
k k f k f k f k ??
-+??
---??-??+=
=+++--- 11238()11525111232631431()()()525525525
k k k k +-+
==+=+-------,
即当1n k =+时,存在41
,55a b =-=,使得11(1)1431()525k f k ++=+---成立.9分
由①,②可知,存在实数,41
,55a b =-=,使1()13()2
n f n a b =+--对任意正整
数n 恒成立. ·
·················································································· 10分