南通市2015届高三第三次调研测试数学试题(word版含答案)

南通市2015届高三第三次调研测试数学试题(word版含答案)
南通市2015届高三第三次调研测试数学试题(word版含答案)

(第5题)

(第4题)

南通市2015届高三第三次调研测试 数学学科参考答案及评分建议

一、填空题:本大题共14小题,每小题5分,共计70分.请把答案填写在答题卡相应位置上......... 1. 设集合A ={3,m },B ={3m ,3},且A =B ,则实数m 的值是 ▲ .

【答案】0

2. 已知复数z =(1i)(12i)+-(i 为虚数单位),则z 的实部为 ▲ .

【答案】3

3. 已知实数x ,y 满足条件||1||1x y ???

≤≤,,则z =2x +y 的最小值是 ▲ .

【答案】-3

4. 为了解学生课外阅读的情况,随机统计了n 名学生的课外阅读时间,所得数据都在[50,

150]中,其频率分布直方图如图所示.已知在[50,

的值为 ▲ . 【答案】1000

5. 在如图所示的算法流程图中,若输出的y 的值为26,则输入的x 的值为 ▲ .

【答案】-4

6. 从集合{1,2,3,4,5,6,7,8,9}中任取一个数记为x ,则log 2x 为整数的概率为 ▲ .

【答案】

4

9

7. 在平面直角坐标系xOy 中,点F 为抛物线

x 2

=8y 的焦点,则F 到双曲线2

2

19

y x -=的渐近

线的距离为 ▲ .

【答案

(第10题)

C

(第11题)

8. 在等差数列{a n }中,若a n +a n +2=4n +6(n ∈N *),则该数列的通项公式a n = ▲ .

【答案】2n +1

9. 给出下列三个命题: ①“a >b ”是“3a >3b ”的充分不必要条件; ②“α>β”是“cos α<cos β”的必要不充分条件;

③“a =0”是“函数f (x ) = x 3+ax 2(x ∈R )为奇函数”的充要条件. 其中正确命题的序号为 ▲ .

【答案】③

10.已知一个空间几何体的所有棱长均为1 cm ,其表面展开图如图所示,则该空间几何体的

体积

V = ▲ cm 3.

【答案

】1+

11. 如图,已知正方形ABCD 的边长为2,点E 为AB 的中点.以A 为圆心,AE 为半径,作

弧交AD 于点F .若P 为劣弧EF 上的动点,则PC PD 的最小值为 ▲ .

【答案

】5-12. 已知函数322301()5 1x x m x f x mx x ?++=?+?

≤≤,,

,>.若函数f (x )的图象与x 轴有且只有两个不同的交

点,则实数m 的取值范围为 ▲ .

【答案】(-5,0)

13.在平面直角坐标系xOy 中,过点P (-5,a )作圆x 2+y 2-2ax +2y -1=0的两条切线,切点分

别为M (x 1,y 1),N (x 2,y 2),且21122112

2

0y y x x x x y y -+-+=-+,则实数a 的值为 ▲ .

【答案】3或-2

14.已知正实数x ,y 满足24

310x y x y

+

++=,则xy 的取值范围为 ▲ . 【答案】[1,8

3

]

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) 如图,在三棱柱ABC -A 1B 1C 1中,B 1C ⊥AB ,侧面BCC 1B 1为菱形. (1)求证:平面ABC 1⊥平面BCC 1B 1;

(2)如果点D ,E 分别为A 1C 1,BB 1的中点,

求证:DE ∥平面ABC 1.

解:(1)因三棱柱ABC -A 1B 1C 1的侧面BCC 1B 1为菱形, 故B 1C ⊥BC 1.……………………………………………………………………… 2分

又B 1C ⊥AB ,且AB ,BC 1为平面ABC 1内的两条相交直线,

故B 1C ⊥平面ABC 1.

5分

因B 1C ?平面BCC 1B 1,

故平面ABC 1⊥平面BCC 1B 1.

7分

(2)如图,取AA 1的中点F ,连DF ,FE . 又D 为A 1C 1的中点,故DF ∥AC 1,EF ∥AB .

因DF ?平面ABC 1,AC 1?平面ABC 1,

故DF ∥面ABC 1. ………………… 10分 同理,EF ∥面ABC 1.

因DF ,EF 为平面DEF 内的两条相交直线,

故平面DEF ∥面ABC 1.……………………………………………………………… 12分 因DE ?平面DEF ,

故DE ∥面ABC 1.…………………………………………………………………… 14分

16.(本小题满分14分)

已知函数()sin()f x A x ω?=+(其中A ,ω,?为常数,

且A >0,ω>0,22

?ππ

-<<)的部分图象如图所示.

(1)求函数f (x )的解析式;

1 (第15题答图)

1

(第15题)

(2)若3()2

f α=,求sin(2)6απ

+的值.

解:(1)由图可知,A =2,…………………………………………………………… 2分 T =2π,故1ω=,所以,f (x ) =2sin()x ?+.…………………………………… 4分

又22(

)2sin()233f ?ππ=+=,且22?ππ-<<,故6

=-. 于是,f (x ) =2sin()6x π

-.…………………………………………………………

7分 (2)由3

()2

f α=

,得3sin()64απ-=.…………………………………………

9分 所以,sin(2)sin 2()cos 2()6626αααππππ???

?+=-+=-???????

?…………………………

12分 =21

12sin ()68

απ--=-.……………………………………

14分

17.(本小题满分14分)

如图,在平面直角坐标系xOy 中,椭圆22

221x y a b

+=(a >b >0)的两焦点分别为F 1

(0),

F 2

0),且经过点

1

2

).

(1)求椭圆的方程及离心率;

(2)设点B ,C ,D 是椭圆上不同于椭圆顶点的三点,点B 与点D 关于原点O 对称.设

直线CD ,CB ,OB ,OC 的斜率分别为k 1,k 2,k 3,k 4,且k 1k 2=k 3k 4. ①求k 1k 2的值; ②求OB 2+OC 2的值.

解:(1)方法一

依题意,c

a 2=

b 2+3,………………………………………………………

2分

(第17题)

由22

13413b b +=+,解得b 2=1(b 2

=34

-,不合,舍去),从而a 2=4. 故所求椭圆方程为:2

214x y +=.

离心率e

5分

方法二

由椭圆的定义知,2a

4,

即a =2.…………………………………………………………………………… 2分

又因c

b 2=1.下略.

(2)①设B (x 1,y 1),C (x 2,y 2),则D (-x 1,-y 1),

于是k 1k 2=21212121y y y y x x x x -+?-+=122

22221y y x x --=22

2122

21(1)(1)44x x x x ----=1

4

-.………………… 8分

②方法一

由①知,k 3k 4=k 1k 2=1

4

-,故x 1x 2=124y y -.

所以,(x 1x 2)2=(-4y 1y 2)2,即(x 1x 2)2

=22

1216(1)(1)44

x x --=2222

1212164()x x x x -++,

所以,22

12x x +=4.…………………………………………………………………… 11分

又2=22221212()()44x x y y +++=22

2212124

x x y y +++,故22

12

1y y +=. 所以,OB 2+OC 2 =2222

1122

x y x y +++=5.………………………………………… 14分

方法二

由①知,k 3k 4=k 1k 2=14

-.

将直线y =k 3x 方程代入椭圆2

214

x y +=中,得2123414x k =+.……………………

9分

同理,2

2

2

4

4

14x k =+.

所以,2212

2234441414x x k k +=+++=2

233

44

11414()4k k +++-=4.…………………… 11分 下同方法一.

18.(本小题满分16分)

为丰富市民的文化生活,市政府计划在一块半径为200 m ,圆心角为120°的扇形地上建造市民广场.规划设计如图:内接梯形ABCD 区域为运动休闲区,其中A ,B 分别在半径OP ,OQ 上,C ,D 在圆弧PQ 上,CD ∥AB ;△OAB 区域为文化展示区,AB

长为;其余空地为绿化区域,且CD 长不得超过....200 m . (1)试确定A ,B 的位置,使△OAB 的周长最大?

(2)当△OAB 的周长最大时,设∠DOC =2θ,试将运动休闲

区ABCD 的面积S 表示为θ的函数,并求出S 的最大值.

解:(1)设(0200]OA m OB n m n ==∈,,,,

, 在△OAB 中,22222cos

3

AB OA OB OA OB π

=+-??,

即222m n mn =++,…………………………………………………… 2分

所以,22

2

2

2()3

()()()44

m n m n mn m n m n +=+-+-=+≥,…………

4分

所以100m n +≤,当且仅当m =n =50时,m n +取得最大值,此时△OAB 周长取得最大值. 答:当OA OB 、都为50 m 时,△OAB 的周长最大. 6分

(2)当△AOB 的周长最大时,梯形ACBD

为等腰梯形. 过O 作OF ⊥CD 交CD 于F ,交AB 于E , 则E F 、分别为AB ,CD 的中点,

所以DOE θ∠=,由CD 200≤,得(0]6 θπ

∈,.

8分

在△ODF 中,200sin 200cos DF OF θθ==,

. 又在△AOE 中,cos

253

OE OA π

==,故200cos 25EF θ=-. 10分

所以,1

400sin )(200cos 25)

2S θθ=-

B C

D

Q

(第18题)

O B

C

D

Q

(第18题答图)

O E

F

=8sin )(8cos 1)θθ-

8sin 64sin cos θθθθ=-+,(0]6

θπ

∈,.…………

12分

(一直没有交代范围扣2分)

令()8sin 64sin cos f θθθθθ=-+,(0]6

θπ

∈,,

()8cos 64cos216sin()64cos26f θθθθθθπ'=--+=-++,(0]6

θπ

∈,,

又y =16sin()6πθ-+及y =cos 2θ在(0]6

θπ

∈,上均为单调递减函数,

故()f θ'在(0]6

θπ

∈,上为单调递减函数.

因1()4)62f π'=-?>0,故()f θ'>0在(0]6

θπ

∈,上恒成立,

于是,()f θ在(0]6

θπ∈,上为单调递增函数.

……… 14分

所以当6

θπ

=时,()f θ有最大值,此时S

有最大值为625(8+. 答:当6

θπ

=

时,梯形ABCD

面积有最大值,且最大值为625(8+ m 2.… 16分

19.(本小题满分16分) 已知数列{a n },{b n }中,a 1=1,2

211

1

(1)n n n n a b a a ++=-?,n ∈N *,数列{b n }的前n 项和为S n .

(1)若12n n a -=,求S n ;

(2)是否存在等比数列{a n },使2n n b S +=对任意n ∈N *恒成立?若存在,求出所有满足条

件的数列{a n }的通项公式;若不存在,说明理由;

(3)若a 1≤a 2≤…≤a n ≤…,求证:0≤S n <2.

解:(1)当a n =12n -时,b n =11(1)42n -?=23

2n +.………………………………………

2分 所以,S n =12

31

133

(1)82

242n n -+++

+

=-

.……………………………………… 4分

(2)满足条件的数列{a n }存在且只有两个,其通项公式为a n =1和a n =1(1)n --.

证明:在2n n b S +=中,令n =1,得b 3=b 1. 设a n =1n q -,则b n =211

(1)n

q q -.………………………………………………… 6分

由b 3=b 1,得2321111(1)(1)q q q q

-

=-. 若q =1±,则b n =0,满足题设条件.此时a n =1和a n =1(1)n --.………………… 8分 若q 1≠±,则

311

q q

=,即q 2 =1,矛盾. 综上,满足条件的数列{a n }存在,且只有两个,一是a n =1,另一是a n =1(1)n --. 10分

(3)因1=a 1≤a 2≤…≤a n ≤…,故0n a >,0<1n n a a +≤1,于是0<2

21n

n a a +≤1.

所以,2

211

1

(1)n n n n a b a a ++=-?≥0,n =1,2,3,….

所以,S n =b 1+b 2+…+b n ≥0.………………………………………………………… 13分

又,2

211

1

(1)n n n n a b a a ++=-?=1111(1)(1)n n n n n a a a a a ++++-?

=11111(1)()n n n n n n a a a a a a ++++

-?≤1

11

2()n n a a +-. 故,S n =b 1+b 2+…+b n ≤1223

1

1111

11

2(

)2()2(

)n n a a a a a a +-+-++- =11112(

)n a a +-=1

12(1)n a +-<2. 所以,0≤S n <2.…………………………………………………………………

16分

20.(本小题满分16分) 已知函数1

()ln f x a x x

=-

-(a ∈R )

. (1)若a =2,求函数()f x 在(1,e 2)上的零点个数(e 为自然对数的底数); (2)若()f x 恰有一个零点,求a 的取值集合;

(3)若()f x 有两零点x 1,x 2(x 1<x 2),求证:2<x 1+x 2<13e a --1.

解:(1)由题设,()f x '=21x

x

-,故()f x 在(1,e 2)上单调递减.…………………… 2分

所以()f x 在(1,e 2)上至多只有一个零点. 又221(1)(e )1()e

f f =?-<0,故函数()f x 在(1,e 2

)上只有一个零点.…………… 4分 (2)()f x '=

2

1x

x -,令()f x '=0,得x =1. 当x >1时,()f x '<0,()f x 在(1 )+∞,

上单调递减; 当0<x <1时,()f x '>0,()f x 在(0,1)上单调递增,

故max [()]f x =f (1)=a -1.……………………………………………………… 6分 ①当max [()]f x =0,即a =1时,因最大值点唯一,故符合题设;…………… 8分

②当max [()]f x <0,即a <1时,f (x )<0恒成立,不合题设; ③当max [()]f x >0,即a >1时,一方面,e a ?>1,1

(e )e a a

f =-

<0; 另一方面,e a -?<1,(e )2e a a f a -=-≤2a -e a <0(易证:e x ≥e x ), 于是,f (x )有两零点,不合题设.

综上,a 的取值集合为{1}.………………………………………………………… 10分 (3)证:先证x 1+x 2>2. 依题设,有a =111ln x x +=221

ln x x +,于是212121

ln x x x x x x -=.

21x x =t ,t >1,则11ln t t tx -=,故11

ln t x t t

-=. 于是,x 1+x 2=x 1(t +1)=21ln t t t

-,x 1+x 2-2=21

2(ln )

2ln t t t t --.

记函数g (x )=21

ln 2x x x

--,x >1.

因2

2

(1)()2x g x x -'=>0,故g (x )在(1 )+∞,上单调递增. 于是,t >1时,g (t )>g (1)=0.

又ln t >0,所以,x 1+x 2>2.…………………………………………………………… 13分

再证x 1+x 2<13e a --1.

因f (x )=0?h (x )=ax -1-x ln x =0,故x 1,x 2也是h (x )的两零点. 由()h x '=a -1-ln x =0,得x =1e a -(记p =1e a -).

仿(1)知,p 是h (x )的唯一最大值点,故有12

()0.h p x p x ???<>,

作函数h (x )=2()ln ln x p x p x p ---+,则2

2

()()()x p h x x x p -'=+≥0,故h (x )单调递增. 故,当x >p 时,h (x )>h (p )=0;当0<x <p 时,h (x )<0. 于是,ax 1-1=x 1ln x 1<

11112()

ln x x p x p x p

-++.

整理,得211(2ln )(2ln 1)p a x p ap p p x p +--+--+>0, 即,21111(3e 1)e a a x x ----+>0.

同理,2

112

2(3e 1)e a a x x ----+<0. 故,2

112

2(3e 1)e a a x x ----+<21111(3e 1)e a a x x ----+, 1212121()()(3e 1)()a x x x x x x -+---<,

于是,1123e 1a x x -+-<.

综上,2<x 1+x 2<13e a --1.………………………………………………………

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.[选修4-1:几何证明选讲](本小题满分10分)

如图,BC为圆O的直径,A为圆O上一点,过点A作圆O的切线交BC的延长线于点P,AH⊥PB于H.

求证:P A·AH=PC·HB.

证:连AC,AB.

因BC为圆O的直径,故AC⊥AB.

又AH⊥PB,故AH2=CH·HB,即AH HB

CH AH

=.………………………………5分

因P A为圆O的切线,故∠P AC=∠B.在Rt△ABC中,∠B+∠ACB=90°.

在Rt△ACH中,∠CAH+∠ACB=90°.所以,∠HAC=∠B.

所以,∠P AC=∠CAH,

所以,PC PA

CH AH

=,即

AH PA

CH PC

=.

所以,PA HB

PC AH

=,即P A·AH=PC·HB.…………………………………………10分

B.[选修4-2:矩阵与变换](本小题满分10分)

在平面直角坐标系xOy中,已知点A(0,0),B(2,0),C(1,2),矩阵

01

1

2

??

??

=

??

-

??M,

点A,B,C在矩阵M对应的变换作用下得到的点分别为A',B',C',求△A B C

'''的面积.

解:因

00

00

????

=

????

????

M,

20

01

????

=

????

-

????

M,

2

1

1

2

2

??

????

=

????

-

??

??

M,

(第21(A)题答图)

(第21(A)题)

1

(00)(01)(2)

2

A B C

'''

--

,,,,,.……………………………………………………6分

1

21

2

S A B''

=??=.………………………………………………………………10分

C .[选修4-4:坐标系与参数方程](本小题满分10分)

在平面直角坐标系xOy 中,曲线C 的参数方程为cos sin x r y r αα=??=?

,(α为参数,r 为常数,r >

0).以原点O 为极点,x 轴的正半轴为极轴建立极坐标系,直线l

的极坐标方程为

cos()204

θπ

++=.若直线l 与曲线C 交于A ,B

两点,且AB =,求r 的值.

cos()204

θπ

++=,得cos sin 20ρθρθ-+=,

即直线l 的方程为20x y -+=.…………………………………………………… 3分

由cos sin x r y r αα=??=?,,

得曲线C 的普通方程为222x y r +=,圆心坐标为(0,0),……… 6分

所以,圆心到直线的距离d

AB =,则2r =.……………… 10分

D .[选修4-5:不等式选讲](本小题满分10分)

已知实数a ,b ,c ,d 满足a >b >c >d ,求证:

14936

a b b c c d a d

++----≥

. 证:因a >b >c >d ,故a -b >0,b -c >0,c -d >0. 故2149[()()()](123)36a b b c c d a b b c c d ??-+-+-++++= ?---??

≥,…………… 6分 所以,

14936

a b b c c d a d

++----≥

.………………………………………………… 10分

【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)

如图,正四棱柱ABCD -A 1B 1C 1D 1中,12AA AB =. (1)求1AD 与面11BB D D 所成角的正弦值;

(2)点E 在侧棱1AA 上,若二面角E -BD -C 1, C

D

A 1

B 1

C 1

D 1

1

AE

AA 的值. 解:(1)以D 为原点,DA ,DC ,DD 1分别为x 轴,y 轴,z 轴, 建立如图所示空间直角坐标系D -xyz . 设1AB =,则D (0,0,0),A (1,0,0), B (1,1,0),C (0,1,0),D 1(0,0,2),

A 1(1,0,2),

B 1(1,1,2),

C 1(0,1,2).

2分

(1)设1AD 与面11BB D D 所成角的大小为θ,

1(102)AD =-,,,

设平面11BB D D 的法向量为n =(x ,y ,z ),

(1,1,0)DB =,1(0,0,2)DD =,则10,0DB DD ?=?=n n ,即0,0x y z +==.

令1x =,则1y =-,所以(110) =-,

,n ,111sin |cos ,|||||||

AD AD AD θ?=<>==

n n n , 所以1AD 与平面11BB D D .………………………… 6分

(2)设E (1,0,λ),0≤λ≤2.

设平面EBD 的法向量为n 1=(x 1,y 1,z 1),平面1BDC 的法向量为n 2=(x 2,y 2,z 2),

(110)(10)DB DE λ==,,,,,,

由1100DB DE ?=?=,n n ,得11110,0x y x z λ+=+=, 令11z =,则11,x y λλ=-=,1(,,1)λλ=-n ,1(0,1,2)DC =,

由22100DB DC ?=?=,n n ,得2222020x y y z +=+=,, 令z 2=1,则x 2=2,y 2=-2,2(2,2,1)=-n ,121212cos ,||||?<>=

=n n n n n n ,

=,得1λ=.所以

1

1

2AE AA =.…………………………… 10分

23.(本小题满分10分)

袋中共有8个球,其中有3个白球,5个黑球,这些球除颜色外完全相同.从袋中随机取出一球,如果取出白球,则把它放回袋中;如果取出黑球,则该黑球不再放回,并且另补一个白球放入袋中.重复上述过程n 次后,袋中白球的个数记为X n . (1)求随机变量X 2的概率分布及数学期望E (X 2);

(2)求随机变量X n 的数学期望E (X n )关于n 的表达式.

解:(1)由题意可知X 2=3,4,5. 当X 2=3时,即二次摸球均摸到白球,其概率是P (X 2=3)=1133

1188C C C C ?=964

当X 2=4时,即二次摸球恰好摸到一白,一黑球,其概率是P (X 2=4)=11113554

11118888C C C C C C C C +=3564;

当X 2=5时,即二次摸球均摸到黑球,其概率是P (X 2=5)=1154

1188C C C C =516

.……

3分

所以随机变量X 2的概率分布如下表:

数学期望E (X 2)=9355267

34564641664

?+?+?=.……………………………… 5分

(2)设P (X n =3+k )=p k ,k =0,1,2,3,4,5.

则p 0+p 1+p 2+p 3+p 4+p 5=1,E (X n )=3p 0+4p 1+5p 2+6p 3+7p 4+8p 5.

P (X n +1=3)=038p ,P (X n +1=4)=58p 0+48p 1,P (X n +1=5)=48p 1+58p 2,P (X n +1=6)=38

p 2+6

8p 3,

P (X n +1=7)=28p 3+78p 4,P (X n +1=8)=18p 4+8

8

p 5,……………………… 7分

所以,E (X n +1)

=3×38p 0+4×(58p 0+48p 1)+5×(48p 1+58p 2)+6×(38p 2+68p 3)+7×(28p 3+78p 4)+8×(18p 4+88

p 5)

=298p 0+368p 1+438p 2+508p 3+578p 4+648p 5

=7

8

(3p 0+4p 1+5p 2+6p 3+7p 4+8p 5)+ p 0+p 1+p 2+p 3+p 4+p 5

=7

8

E(X n)+1.…………………9分

由此可知,E(X n+1)-8=7

8

(E(X n)-8).

又E(X1)-8=

35

8

-,所以E(X n)=1

357

8()

88

n-

-.……………………………10分

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