2016年浦东高三二模

浦东新区2015学年第二学期高三教学质量检测

数学试卷（文理合卷）

一、填空题

1．已知全集U R =,若集合|

01x A x x ??

=>??-??

，则U C A = . 2．已知复数z 满足(1)2z i i ?-=，其中i 为虚数单位，则z = . 3．双曲线2226x y -=的焦距为 .

4．已知6

1ax x ?

?+ ??

?二项展开式中的第五项系数为152，则正实数a .

5．方程22log (97)2log (31)x x +=++的解为 . 6．已知函数311()=

3x f x a x a +??

≠ ?+??

的图像与它的反函数的图像重合，则实数a 的值为 .

7．在ABC ?中，边,,a b c 所对角分别为,,A B C ，若s i n 02c o s a B b A

π??

+ ?

=??，

则ABC ?的形状为 . 8．（理）在极坐标系中，点(2,

)2A π

到直线cos()4

π

ρθ+=的距离为________．

（文）若某几何体的三视图（单位：cm ）如图所示，则此几何体的体积是 3

cm ． 9．（理）离散型随机变量ξ的概率分布列如图，若1E ξ=， 则D ξ的值为________．

（文）设,x y 满足约束条件??

?

??≥≥≥+-≤--0,002063y x y x y x ，

则目标函数2z x y =+的最大值为_____．

10．已知四面体ABCD 中，2==CD AB ，E ，F 分别为BC ，AD 的中点，且异面直线AB 与CD 所成的角为

3

π

，则EF =________． 11．设,m n 分别为连续两次投掷骰子得到的点数，且向量(,)a m n =r ，(1,1)b =-r

，则a r 与b r 的夹角为锐

角的概率是________．

12. （理）已知数列{}n a 的通项公式为(1)2n n n a n =-?+，*

n N ∈，则这个数列的前n 项和n S =_______．

（文）已知数列{}n a 的通项公式为(1)2n n n a n =-?+，*

n N ∈，则这个数列的前2n 项和2n S =________．

13．（理）任意实数,a b ，定义00ab ab a b a ab b

≥??

?=?

的等比数列，且61a =，1239101()()()()()2f a f a f a f a f a a +++++=L ，则1a =_______． （文）已知函数1

()f x x x

=-

，数列{}n a 是公比大于0的等比数列，且61a =，1239101()()()()()f a f a f a f a f a a +++++=- ，则1a =_______．

14．（理）关于x 的方程

11

sin 211

x x π=--在[]2016,2016-上解的个数是 ．

（文）关于x 的方程

11

sin 211

x x π=--在[]6,6-上解的个数是 ．

二、选择题（本大题共有4题，满分20分）； 每小题都给出四个选项，其中有且只有一个选项是正确的，考生应在答题纸相应位置上，选对得 5分，否则一律得零分. 15. “

1

12

x <<”是“不等式11x -<成立”的（ ） （A ）充分非必要条件. （B ）必要非充分条件. （C ）充要条件. （D ）既非充分亦非必要条件. 16.给出下列命题，其中正确的命题为( )

（A ）若直线a 和b 共面，直线b 和c 共面，则a 和c 共面；

（B ）直线a 与平面α不垂直，则a 与平面α内的所有直线都不垂直； （C ）直线a 与平面α不平行，则a 与平面α内的所有直线都不平行； （D ）异面直线a 、b 不垂直，则过a 的任何平面与b 都不垂直.

17．抛物线24y x =的焦点为F ，点(,)Px y 为该抛物线上的动点，又点(1,0)A -，则

PF PA

的最小值是（ ）

（A ）1

2

（B

（C

（D

18.已知平面直角坐标系中两个定点(3,2),(3,2)E F -，如果对于常数λ

，在函数

22

4,[4,4]

y x x x =++--∈-的图像上有且只有6个不同的点P ，使得λ=?PF PE 成立，那么λ的取值范围是( )

（A ）95,5??-- ??

? （B ）9,115??- ?

?? （C ）9,15??-- ??? （D ）()5,11-

三、解答题（本大题共有5题，满分74分）；解答下列各题必须在答题纸的相应位置上，写出必要的步骤． 19．（本题满分12分，第（1）题6分，第（2）题6分）

如图，在圆锥SO 中，AB 为底面圆O 的直径，点C 为?AB 的中点，SO AB =. （1）证明：AB ⊥平面SOC ；

（2）若点D 为母线SC 的中点，求AD 与平面SOC 所成的角.（结果用反三角函数表示）

20. （本题满分14分，第（1）题8分，第（2）题6分）

如图，一智能扫地机器人在A 处发现位于它正西方向的B 处和北偏东?30方向上的C 处分别有需要清扫的垃圾，红外线感应测量发现机器人到B 的距离比到C 的距离少0.4m ，于是选择沿C B A →→路线清扫.已知智能扫地机器人的直线行走速度为0.2m/s,忽略机器人吸入垃圾及在B 处旋转所用时间，10秒钟完成了清扫任务.

（1）B 、C 两处垃圾的距离是多少？（精确到0.1）

（2）智能扫地机器人此次清扫行走路线的夹角B ∠是多少？（用反三角函数表示）

21．（理）（本题满分14分，第（1）题6分，第（2）题8分）

数列{}n a 满足：112,2n n n a a a λ+==+?，且123,1,a a a +成等差数列，其中*n N ∈。 （1）求实数λ的值及数列{}n a 的通项公式； （2）若不等式

216

25n

p p n a +≤

-成立的自然数n 恰有4个，求正整数p 的值． 21．（文）（本题满分14分，第（1）题6分，第（2）题8分）

数列{}n a 满足：112,2n n n a a a λ+==+?，且123,1,a a a +成等差数列，其中*n N ∈。 （1）求实数λ的值及数列{}n a 的通项公式； （2）若不等式16

25n

p n a ≤-成立的自然数n 恰有3个，求正整数p 的值．

22．（理）（满分16分，第1小题4分，第2小题6分，第3小题6分）

教材曾有介绍：圆222r y x =+上的点),(00y x 处的切线方程为200r y y x x =+。我们将其结论推广：椭圆

12

2

22=+b y a x （0>>b a ）上的点),(0

0y x 处的切线方程为12020=+b y y a x x ，在解本题时可以直接应用。已知，直线03=+-y x 与椭圆E ：12

22=+y a

x （1>a ）有且只有一个公共点。

（1）求a 的值；

（2）设O 为坐标原点，过椭圆E 上的两点A 、B 分别作该椭圆的两条切线1l 、2l ，

且1l 与2l 交于点),2(m M 。当m 变化时，求OAB ?面积的最大值；

（3）在（2）的条件下，经过点),2(m M 作直线l 与该椭圆E 交于C 、D 两点，在线段CD 上存在点N ，使|

||

|||||MD MC ND CN =

成立，试问：点N 是否在直线AB 上，请说明理由。

22．（文）（满分16分，第1小题4分，第2小题中的第①小题6分，第②小题6分）

教材曾有介绍：圆222r y x =+上的点),(00y x 处的切线方程为200r y y x x =+。我们将其结论推广：椭圆

12

2

22=+b y a x （0>>b a ）上的点),(0

0y x 处的切线方程为12020=+b y y a x x ，在解本题时可以直接应用。已知，直线03=+-y x 与椭圆E ：12

22=+y a

x （1>a ）有且只有一个公共点。

（1）求a 的值；

（2）设O 为坐标原点，过椭圆E 上的两点A 、B 分别作该椭圆的两条切线1l 、2l ，

且1l 与2l 交于点),2(m M 。 ①设0m ≠，直线AB 、OM 的斜率分别为1k 、2k ，求证：21k k 为定值。 ②设m R ∈，求OAB ?面积的最大值。

011n n a x x x x b -=<<<<=L ，和式11

()()n

i i i f x f x M -=-≤∑恒成立，则称()f x 为[],a b 上的“绝对差

有界函数”。注：

121

n

i

n i a

a a a ==+++∑L 。

（1）证明函数()sin cos f x x x =+在,02π??

-

????

上是“绝对差有界函数”

。 （2）证明函数cos ,01,()20,0

x x f x x

x π?

<≤?

=??=?不是[]0,1上的“绝对差有界函数”。 （3）记集合{[]121212()|0,,,()()}A f x k x x a b f x f x k x x =>∈-≤-存在常数，对任意的有成立证明集合A 中的任意函数()f x 为“绝对差有界函数”，并判断()()2016sin 2016g x x =是否在集合A 中，如果在，请证明并求k 的最小值；如果不在，请说明理由。

011n n a x x x x b -=<<<<=L ，和式11

()()n

i i i f x f x M -=-≤∑恒成立，则称()f x 为[],a b 上的“绝对差

有界函数”。注：

121

n

i

n i a

a a a ==+++∑L 。

（1） 证明函数()sin cos f x x x =+在,02π??

-????

上是“绝对差有界函数”

。 （2） 记集合

{[]121212()|0,,,()()}A f x k x x a b f x f x k x x =>∈-≤-存在常数，对任意的有成立证明集合A

中的任意函数()f x 为“绝对差有界函数”。当[][],1,2a b =时，

判断()g x =A 中，如果在，

请证明并求k 的最小值；如果不在，请说明理由。

（3） 证明函数cos ,01,()20,0

x x f x x

x π?

<≤?

=??=?不是[]0,1上的“绝对差有界函数”。

浦东新区2015学年第二学期高三教学质量检测

数学试卷答案及评分细则（文理合卷）

1． []0,1 2．

． 6 4．

5． {}0,1 6． 3a =- 7．等腰或直角三角形 8．

（理）（文）18 9．（理）0.4 （文）14 10． 1 或3． 11． 512

12. （理）1122,2

52,22

n n n n

n S n n ++?+-??=??--??为奇数为偶数 （文）21222n n S n +=+-

13．（理）4．

14．（理）4031． （文）9． 15. A 16. D 17． B 18. C

19．（1）证明：在圆锥SO 中，SO AB ⊥………………（2分）

∵点C 为?AB 的中点，∴OC AB ⊥…………………（4分）

∴由AB SO AB OC AB SO OC C ⊥?

?

⊥?⊥??=?

I 平面SOC …………（6分）

（2）解：联结OD ，AB ⊥ 平面SOC

ADO ∴∠为AD 与平面SOC 所成的角……………（8分）

设OC a =，则2SO a =

，12OD SC ∴=

== ∴在Rt AOD ?

中，tan OA ADO OD

∠=

==（11分）

ADO ∴∠=12分） 20. 解：（1）设c b a 、、分别是A 、B 、C 所对的边，c b a 、、 均为正数。由题意可知：

?=∠120A ，4.0=-c b ，2102.0=?=+a c ……（3分）

由余弦定理可得： A bc c b a cos 22

2

2

-+=

)2)(4.2()2()4.2(2

22a a a a a --+-+-= 2

得，4.1=a （另一解2.5=a 与题意不符，舍 不舍扣1分）……（4分） 即B 、C 两处垃圾的距离是1.4米。…………………（1分） （2）由题意得：14.2=-=a b ……………（1分） 正弦定理可得：

A

a

B b sin sin = 即，14

3

54.1120sin 1sin =

??=

B …………………（3分） 由题意，B ∠为锐角，得14

3

5arcsin

=∠B …………………（1分） 即，智能扫地机器人此次清扫行走路线的夹角14

3

5arcsin

=∠B ………（1分） （6.0arcsin ≈∠B 、

8.0arccos 14

11

arccos ≈=∠B 均给分，应用余弦定理或建立坐标系等解法参照给分） 21．（理）

解答：（1）由题意：2322,26a a λλ=+=+

∵123,1,a a a +成等差数列，()2(32)226λλ∴+=++，……………………（2分）

解得：1λ=………………………………………………………………………（3分）

∵112,2n n n a a a +==+，

112211()()()2n n n n n n a a a a a a a a ---∴=-+-++-+=L ,……………………（5分）

解得：2n n a =………………………………………………………………………（6分） （2）解：∵ 216

25n

p

p n a +≤

-，18252n p p n -+∴

≤- ∵0p >，1,2n ∴=显然成立……………………………………………………（8分）

当3n ≥时，

125

82

n p n p --≤+， …………………………………………………（9分） 设()1

1252n n b n -??

=-? ?

??

()()1

11112325(72)222n

n n

n n b b n n n -+????

??

∴-=-?--?=- ? ?

???????

………………（11分）

当3n =时，43b b >；当4n ≥时，456b b b >>>L ；

34561357

,,,481632

b b b b ====，有4536b b b b >>>

若

125n p n --≤还需有2解，则35p b b <≤，即15p <≤，………（12分）

解得

840311p <≤，

所以正整数3p =………………………………………（14分）

21．（文）

解答：（1）由题意：2322,26a a λλ=+=+

∵123,1,a a a +成等差数列，()2(32)226λλ∴+=++，……………………（2分）

解得：1λ=………………………………………………………………………（3分）

∵112,2n n n a a a +==+，

112211()()()2n n n n n n a a a a a a a a ---∴=-+-++-+=L ,……………………（5分）

解得：2n n a =………………………………………………………………………（6分）

（2）解：∵

1625n

p n a ≤-，18

252n p n -∴

≤- ∵0p >，1,2n ∴=显然成立……………………………………………………（8分）

当3n ≥时，

125

82

n p n --≤， ……………………………………………………（9分） 设()1

1252n n b n -??

=-? ?

??

()()1

11112325(72)222n

n n

n n b b n n n -+????

??

∴-=-?--?=- ? ?

???????

………………（11分）

当3n =时，43b b >；当4n ≥时，456b b b >>>L ；

345135

,,4816

b b b ===，有453b b b >>

若

12582n p n --≤还需有1解，则548p b b <≤，即53

1688

p <≤，………（12分） 解得

5

32p <≤，

所以正整数p 的值为3………………………………（14分） 22．（理）

解：（1）联立???

??=++=13

2

22y a

x x y 整理得0232)11(22=+++x x a 依题意0=?即202)11

(

4)32(2

2

=?=?+?-a a …………………………（4分） （2）设),(11y x A 、),(22y x B 于是直线1l 、2l 的方程分别为

1211=+y y x x 、12

22=+y y x

x 将),2(m M 代入l 、l 的方程得01=-+my x 且01=-+my x

所以直线AB 的方程为01=-+my x ……………………（6分）

联立????

??=+=-+12

012

2y x m y x 012)2(22=--+my y m 显然0>?，由21,y y 是该方程的两个实根， 有22221+=

+m m y y ，2

1

22

1+-=m y y ………………（8分） OAB ?面积112212x y S x y =

的绝对值，即121

||2

S y y =-

即2

121

1)1(2)2()1(2]4)[(4122222212

212

≤++++=++=-+=m m m m y y y y S

当0=m 时，S 取得最大值

2

2

………………（10分） （3）点N 在直线AB 上………………（11分） 因为

|

||

|||||MD MC ND CN =

设),(C C y x C 、),(D D y x D 、),(00y x N ，且→

-→

-=ND CN λ（1.0≠>λλ） 于是→

-→

--=MD CM λ 即

01x x x D C =++λλ、01y y y D C =++λλ、21=--λλD C x x 、m y y D

C =--λ

λ1

又1222

=+C C y x ，122

2

=+D D y x …………（13分）

?2

2

2

C C y x +222

21)2(λλ-=+-D D y x

?+--?++?λλλλ1121D C D C

x x x x 111=--?++λλλλD C D C y y y y ，…………（15分） ?=+???

122

1

00m y x 0100=-+my x ，即N 在直线AB 上。…………（16分） 22．（文）

解：（1）联立??

???=++=13

2

22y a x x y 整理得0232)11(22=+++x x a 依题意0=?即202)11

(

4)32(22

=?=?+?-a a

…………………………（4分）

（2）①设),(11y x A 、),(22y x B 于是直线1l 、2l 的方程分别为

1211=+y y x x 、12

22=+y y x

x 将),2(m M 代入1l 、2l 的方程得0111=-+my x 且0122=-+my x 所以直线AB 的方程为01=-+my x ……………………（7分）

m k 11-

=，22m k =，所以2

1

21-=k k 为定值………………（10分） ②依题意联立????

??=+=-+12

12

2y x m y x 012)2(22=--+my y m 显然0>?，由21,y y 是该方程的两个实根， 有22221+=

+m m y y ，2

12

21+-=m y y ………………（12分） OAB ?面积112212x y S x y =

的绝对值，即121

||2

S y y =-……（14分）

即2

121

1)1(2)2()1(2]4)[(4122222212

212

≤++++=++=-+=m m m m y y y y S

当0=m 时，S 取得最大值2

2

………………（16分）

23.（理）

解：（1）因为(

)+

4f x x π?

?

??

?

在区间,02π??

-

????

上为单调递增函数，…………（2分） 所以当1,0,1,2,1i i x x i n +<=-L 时，有()()1,0,1,2,1i i f x f x i n +<=-L ， 所以

()11

()()022n

i i i f x f x f f π-=??

-=--= ???

∑

。

从而对区间,02π

??

-????

的任意划分：01102n n x x x x π

--=<<<<=L ，存在2M =，

11

()()2n

i i i f x f x -=-≤∑

成立。

综上，函数()sin cos f x x x =+在,02π??

-????上是“绝对差有界函数”

。…………（4分） （2）取区间[]0,1的一个划分：111012212

n n <<<<<-L ，……………………（6分） 则有：

211

148121(21)12()()|

cos 0||cos cos |2221222

1211111111

|cos cos |1++++222244881616111=1++++222

n

i i i n

i n n n f x f x n n n i ππππ

π-==--=-+-+-+-=>+++++++∑

∑L L L L

14442444314444244443

L L

个

个

所以对任意常数0M >，只要n 足够大，就有区间[]0,1的一个划分：

111012212n n <<<<<-L 满足11

()()n i i i f x f x M -=->∑。……………………（10分）

（3）证明：任取()f x A ∈，存在常数[]120,,,k x x a b >∈，对任意的有1212()()f x f x k x x -≤-成立。从

而

对

区

间

[]

,a b 的任意划分：

011n n a x x x x b

-=<<<<=L ，和式

()111

1

()()n

n i i i i i i f x f x k x x k b a --==-≤-=-∑

∑成立。

取()M k b a =-，所以集合A 中的任意函数()f x 为“绝对差有界函数”。…………（14分） 因为|sin |||x x ≤，所以对任意的[]12,,,x x a b ∈有

()()()()()()()1212121212212

|||2016sin 20162016sin 2016|

20162sin 10081008cos 1008100820162sin 100810082016g x g x x x x x x x x x x x -=-=-+≤-≤-， 所以k 的最小值为2

2016。………………………………………………………………（18分） 23.（文）

解：（1）因为(

)+

4f x x π?

?

??

?

在区间,02π??

-

????

上为单调递增函数，…………（2分） 所以当1,0,1,2,1i i x x i n +<=-L 时，有()()1,0,1,2,1i i f x f x i n +<=-L ， 所以

()11

()()022n

i i i f x f x f f π-=??

-=--= ???

∑

。

从而对区间,02π

??

-????

的任意划分：01102n n x x x x π

--=<<<<=L ，存在2M =，

11

()()2n

i i i f x f x -=-≤∑

成立。

综上，函数()sin cos f x x x =+在,02π??

-

????

上是“绝对差有界函数”

。…………（4分） （2）证明：任取()f x A ∈，存在常数[]120,,,k x x a b >∈，对任意的有1212()()f x f x k x x -≤-成立。从

而

对

区

间

[]

,a b 的任意划分：

a x x x x b

=<<<<=L ，和式

()111

1

()()n

n

i i i i i i f x f x k x x k b a --==-≤-=-∑

∑成立。

取()M k b a =-，所以集合A 中的任意函数()f x 为“绝对差有界函数”。……（8分） 对于任意的[]

12,1,2x x ∈，

()(

)12121

2

g x g x x x -=

=

≤-。

所以k 的最小值为

1

2

。……………………………………………………………………（12分） （3）取区间[]0,1的一个划分：111012212

n n <<<<<-L ，……………………（14分） 则有：

211

148121(21)12()()|

cos 0||cos cos |2221222

1211111111

|cos cos |1++++222244881616111=1++++222

n

i i i n

i n n n f x f x n n n i πππ

π

π-==--=-+-+-+-=>+++++++∑

∑L L L L

14442444314444244443

L L

个

个

所以对任意常数0M >，只要n 足够大，就有区间[]0,1的一个划分：

111012212n n <<<<<-L 满足11

()()n i i i f x f x M -=->∑。

所以函数cos ,01,

()20,0

x x f x x

x π?

<≤?=??=?不是[]0,1的“绝对差有界函数”。……………（18分）

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