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2017届高三徐州二模-解析版

2017届高三徐州二模-解析版
2017届高三徐州二模-解析版

i ←1

While i < 6 i ←i +2 S ←2i +3 End While Print S

(第3题)

扬州、南通、泰州、淮安、宿迁、徐州六市2017届高三第二次调研测

一、填空题:本大题共14小题,每小题5分,共计70分.

1. 已知集合{} 03 4 A =,,,{} 102 3 B =-,,,,则A B = ▲ . 【答案】{}03,

2. 已知复数3i

1i

z -=+,其中i 为虚数单位,则复数z 的模是 ▲ .

3. 根据如图所示的伪代码,可知输出的结果S 是 ▲ .

【答案】17

4. 现有1 000根某品种的棉花纤维,从中随机抽取50根,纤维长度(单位:mm )的数据分

组及各组的频数见右上表,据此估计这1 000根中纤维长度不小于37.5 mm 的根数是 ▲ . 【答案】180

5. 100张卡片上分别写有1,2,3,…,100.从中任取1张,则这张卡片上的数是6的倍

数的概率是 ▲ . 【答案】425

(或0.16)

6. 在平面直角坐标系xOy 中,已知抛物线24y x =上一点P 到焦点的距离为3,则点P 的横 坐标是 ▲ .

【答案】2

7. 现有一个底面半径为3 cm ,母线长为5 cm 的圆锥状实心铁器,将其高温融化后铸成一个 实心铁球(不计损耗),则该铁球的半径是 ▲ cm .

(第4题)

8. 函数

()f x =的定义域是 ▲ . 【答案】[]22-,

9. 已知{}n a 是公差不为0的等差数列,n S 是其前n 项和.若2345a a a a =,927S =,则1a 的

值是 ▲ . 【答案】5-

10.在平面直角坐标系xOy 中,已知圆1C :()()2

2

481x y -+-=,圆2C :()()2

2

669x y -++=.

若圆心在x 轴上的圆C 同时平分圆1C 和圆2C 的圆周,则圆C 的方程是 ▲ . 【答案】2281x y +=

11.如图,在平面四边形ABCD 中,O 为BD 的中点,且3OA =,5OC =.若AB →·AD →

=-7, 则BC →·DC →

的值是 ▲ .

【答案】9

12.在△ABC 中,已知2AB =,226AC BC -=,则tan C 的最大值是 ▲ .

13.已知函数20()1 0x m x f x x x -+

≥,,

,,其中0m >.若函数()()1y f f x =-有3个不同的零点,

则m 的取值范围是 ▲ . 【答案】(01),

14.已知对任意的x ∈R ,()()3sin cos 2sin 2 3 a x x b x a b ++∈R ≤,

恒成立,则当a b +取得最 小值时,a 的值是 ▲ . 【答案】45

- 二、解答题:本大题共6小题,共计90分. 15.(本小题满分14分)

(第11题)

已知()

πsin 4α+=,()

ππ2α∈,.

求:(1)cos α的值; (2)()

πsin 24

α-的值.

解:(1)法一:因为()ππ2α∈,,所以()

π3π5π444

α+∈,,

又()

πsin 4α+=,

所以()πcos 4α+=. …… 3分

所以()

ππcos cos 44αα??=+-????

()()

ππππ

cos cos sin sin 4444αα=+++

=

3

5=-. …… 6分

法二:由()

πsin 4α+=得,ππsin cos cos sin 44αα+=, 即1sin cos 5

αα+=. ① …… 3分

又22sin cos 1αα+=. ②

由①②解得3cos 5α=-或cos α=45

因为()

ππ2α∈,

,所以3cos 5

α=-. …… 6分 (2)因为()

ππ2α∈,

,3cos 5

α=-,

所以4sin 5

α==. …… 8分 所以()4324sin 22sin cos 25525

ααα==??-=-,

()2

2

37cos22cos 12525

αα=-=?-=-. …… 12分

所以()

πππsin 2sin 2cos cos2sin 444

ααα-=-

()()

2472525=--

= …… 14分

16.(本小题满分14分)

如图,在直三棱柱111ABC A BC -中,AC BC ⊥,A 1B 与AB 1交于点D ,A 1C 与AC 1交于点E . 求证:(1)DE ∥平面B 1BCC 1; (2)平面1A BC ⊥平面11A ACC . 证明:(1)在直三棱柱111ABC A BC -中,

四边形A 1ACC 1为平行四边形. 又E 为A 1C 与AC 1的交点,

所以E 为A 1C 的中点. …… 2分

同理,D 为A 1B 的中点,

所以DE ∥BC . …… 4分 又BC ?平面B 1BCC 1,DE ?平面B 1BCC 1,

所以DE ∥平面B 1BCC 1. …… 7分

(2)在直三棱柱111ABC A BC -中,

1AA ⊥平面ABC ,

又BC ?平面ABC ,

所以1AA BC ⊥. …… 9分 又AC BC ⊥,1AC

AA A =,1AC AA ?,平面11A ACC ,

所以BC ⊥平面11A ACC . …… 12分 因为BC ?平面1A BC ,

所以平面1A BC ⊥平面11A ACC . …… 14分

17.(本小题满分14分)

如图,在平面直角坐标系xOy 中,已知椭圆22

22 1 (0)y x a b a b

+=>>的离心率为23,C 为椭

B

C 1

A

C

A 1

B 1 D

(第16题)

E

圆上位于第一象限内的一点.

(1)若点C 的坐标为()

523

,,求a ,b 的值;

(2)设A 为椭圆的左顶点,B 为椭圆上一点,且AB →=12OC →

,求直线AB 的斜率.

解:(1)因为椭圆的离心率为23

23=,即2

259b a

=.①

又因为点C ()

523

,在椭圆上,

所以22

425

19a b +=. ② …… 3分 由①②解得2295a b ==,

. 因为0a b >>,

所以3a b ==,

…… 5分 (2)法一:由①知,2259b a =,所以椭圆方程为2222915y x a a

+=,即222595x y a +=.

设直线OC 的方程为x my =()0m >,11()B x y ,,22()C x y ,.

由222

595x my x y a =??+=?

,得2222

595m y y a +=, 所以222559a y m =

+.因为20y >

,所以2

y =

. …… 8分 因为AB →=12OC →

,所以//AB OC .可设直线AB 的方程为x my a =-.

由222

595x my a x y a

=-??+=?,得22(59)100m y amy +-=, 所以0y =或21059am y m =+,得121059am y m =+. …… 11分

因为AB →=12OC →

,所以()()

11221122x a y x y +=,,,于是212y y =,

22059

am m =+()0m >

,所以m =.

所以直线AB

的斜率为1m . …… 14分

(第17题)

法二:由(1)可知,椭圆方程为222595x y a +=,则(0)A a -,.

设11()B x y ,,22()C x y ,.

由AB →=12OC →

,得()()

11221122

x a y x y +=,,,

所以1212x x a =-,1212y y =. …… 8分 因为点B ,点C 都在椭圆222595x y a +=上, 所以()()

22222222225951

595.22x y a y x a a ?+=??-+=??

, 解得24a x =

,2y = …… 12分

所以直线AB

的斜率22y k x =

=

…… 14分

18.(本小题满分16分)

一缉私艇巡航至距领海边界线l (一条南北方向的直线)3.8海里的A 处,发现在其北偏 东30°方向相距4海里的B 处有一走私船正欲逃跑,缉私艇立即追击.已知缉私艇的最 大航速是走私船最大航速的3倍.假设缉私艇和走私船均按直线方向以最大航速航行. (1)若走私船沿正东方向逃离,试确定缉私艇的追击方向,使得用最短时间在领海内拦截

成功;(参考数据:sin17

°

5.7446) (2)问:无论走私船沿何方向逃跑,缉私艇是否总能在领海内成功拦截?并说明理由. 解:(1)设缉私艇在C 处与走私船相遇(如图甲),

依题意,3AC BC =. …… 2分 在△ABC 中,由正弦定理得,

sin sin BC BAC ABC AC ∠=∠sin1203

==.

因为sin17°,所以17BAC ∠=°.

从而缉私艇应向北偏东47方向追击. …… 5分

(第18题)

在△ABC 中,由余弦定理得,

2224cos1208BC AC BC

+-=,

解得BC = 1.68615≈.

又B 到边界线l 的距离为3.84sin30 1.8-=.

因为1.68615 1.8<,所以能在领海上成功拦截走私船. …… 8分 (2)如图乙,以A 为原点,正北方向所在的直线为y 轴建立平面直角坐标系xOy . 则(2B ,,设缉私艇在()P x y ,处(缉私艇恰好截住走私船的位置)与走私 船相遇,则3PA PB

=3 整理得,()(

22

9944x y -+=, …… 12 所以点()P x y ,的轨迹是以点(9432

为半径的圆. 因为圆心(94到领海边界线l : 3.8x =的距离为1.55,大于圆半径32,

所以缉私艇能在领海内截住走私船. …… 14分 答:(1)缉私艇应向北偏东47方向追击;

(2)缉私艇总能在领海内成功拦截走私船. …… 16分

19.(本小题满分16分)

已知函数1()e

x f x =,()ln g x x =,其中e 为自然对数的底数. (1)求函数()()y f x g x =在x =1处的切线方程;

(2)若存在12x x ,()12x x ≠,使得[]1221()()()()g x g x f x f x λ-=-成立,其中λ为常数,

求证:e λ>;

(3)若对任意的(]01x ∈,

,不等式()()(1)f x g x a x -≤恒成立,求实数a 的取值范围. A B

C

图甲 60

解:(1)因为ln ()()e x x

y f x g x ==,所以()

211e ln e ln e e x x x x x x

x x y ?-?-'==,故11e x y ='=. 所以函数()()y f x g x =在x =1处的切线方程为1(1)e

y x =-,

即e 10x y --=. …… 2分

(2)由已知等式[]1221()()()()g x g x f x f x λ-=-得1122()()()()g x f x g x f x λλ+=+.

记()()()ln e

x p x g x f x x λλ=+=+,则e ()e x

x x p x x λ-'=. …… 4分 假设e λ≤.

① 若λ≤0,则()0p x '>,所以()p x 在()0+∞,上为单调增函数.

又12()()p x p x =,所以12x x =,与12x x ≠矛盾. …… 6分 ② 若0e λ<≤,记()e x r x x λ=-,则()e x r x λ'=-.

令()0r x '=,解得0ln x λ=.

当0x x >时,()0r x '>,()r x 在()0x +∞,

上为单调增函数; 当00x x <<时,()0r x '<,()r x 在()00x ,上为单调减函数. 所以0()()=1ln )0r x r x λλ-≥(≥,所以()0p x '≥, 所以()p x 在()0+∞,

上为单调增函数. 又12()()p x p x =,所以12x x =,与12x x ≠矛盾.

综合①②,假设不成立,所以e λ>. …… 9分 (3)由()()(1)f x g x a x -≤得ln e (1)x x a x --≤0. 记ln e (1)x F x x a x --()=,0x <≤1, 则()

2

11e e e x x x

F x ax x a x x '-=-()=. ① 当1e a ≤时,因为211e

e x x ≥,e 0x x >,所以0F x '()≥, 所以F x ()在(]0+∞,

上为单调增函数,所以(1)F x F ()≤=0,

故原不等式恒成立. …… 12分 ② 法一:

当1e

a >时,由(2)知e e x x ≥,3

211e e a x F x a x x x -'-=()≤,

当()

1

3

e 1a x -<<时,0F x '<(),()F x 为单调减函数,

所以(1)F x F >()=0,不合题意. 法二:

当1e

a >时,一方面1=1e 0F a '-<(). 另一方面,111e x a ?=<,()

()111121111e e e e 10F x a x x a x a a x x '-=-=->()≥.

所以01(1)x x ?∈,,使0=0F x '(),又F x '()在(0)+∞,上为单调减函数, 所以当01x x <<时,0F x '<(),故F x ()在0(1)x ,上为单调减函数, 所以(1)F x F >()=0,不合题意.

综上,1e

a ≤. …… 16分

20.(本小题满分16分)

设数列{}n a 的前n 项和为S n ()*n ∈N ,且满足:

①12 a a ≠;②()()()22112n n r n p S n n a n n a +-=++--,其中r p ∈R ,,且0r ≠. (1)求p 的值;

(2)数列{}n a 能否是等比数列?请说明理由; (3)求证:当r =2时,数列{}n a 是等差数列. 解:(1)n =1时,211(1)220r p S a a -=-=, 因为12a a ≠,所以20S ≠,

又0r ≠,所以p =1. …… 2分 (2){}n a 不是等比数列.理由如下:

假设{}n a 是等比数列,公比为q ,

当n =2时,326rS a =,即211(1)6ra q q a q ++=,

所以2(1)6r q q q ++=, (i ) …… 4分 当n =3时,431212+4rS a a =,即2321112(1)124ra q q q a q a +++=+,

所以232(1)62r q q q q +++=+, (ii ) …… 6分

由(i )(ii )得q =1,与12a a ≠矛盾,所以假设不成立.

故{}n a 不是等比数列. …… 8分

(3)当r =2时,易知3122a a a +=.

由22112(1)()(2)n n n S n n a n n a +-=++--,得

2n ≥时,1

1(1)(1)(2)211n n n n a n n a S n n +++-=

+

--, ① 11

2(1)(2)(1)(2)2n n n n a n n a S n n

++++-+=+

,② ②-①得,211

2(1)(2)(1)(2)21(1)

n n n n n a n n a n n a a n n n n +++++-+=-+

--, …… 11分 即11121(1)(2)()(1)()

2()1

n n n n n a a n n a a a a n n ++++-+--=

-

-, 211112()(2)()()

11

n n n a a n a a n a a n n n ++-+--=-

+-, 即

()

2111111

121

n n n n a a a a n a a a a n n n n +++-----=-

+- ()

111

(1)2212

n n n n a a a a n n ----=

-

?-- =…… ()

3121

(1)3202223121

n n a a a a -?????--=-=??????--,

所以

11121121n n a a a a a a

n n ----==???=--,

令21a a -=d ,则

1

1

n a a d n -=-(2)n ≥. …… 14分

所以1(1)(2)n a a n d n =+-≥. 又1n =时,也适合上式, 所以*1(1)()n a a n d n =+-∈N . 所以*1()n n a a d n +-=∈N .

所以当r =2时,数列{}n a 是等差数列. …… 16分

数学Ⅱ(附加题)

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,已知△ABC 内接于⊙O ,连结AO 并延长交⊙O 于点D ,ACB ADC ∠=∠. 求证:2AD BC AC CD ?=?. 证明:连结OC .

因为ACB ADC ∠=∠,ABC ADC ∠=∠,

所以ACB ABC ∠=∠.

3分 因为OC =OD ,所以OCD ADC ∠=∠. 所以ACB OCD ∠=∠.

所以△ABC ∽△ODC . …… 8分 所以AC BC OC CD

=,即AC CD OC BC ?=?.

因为12OC AD =,所以2AD BC AC CD ?=?. …… 10分

B .[选修4-2:矩阵与变换](本小题满分10分)

设矩阵A 满足:A 1206??=??

??1203--??

????,求矩阵A 的逆矩阵1-A . 解:法一:设矩阵a b c d ??=????

A ,则1206a b c d ????=????????1203--??

????, 所以1a =-,262a b +=-,0c =,263c d +=. …… 4分 解得0b =,12d =,所以10102-??

??=??

??A . …… 6分 根据逆矩阵公式得,矩阵11002--??

=????

A . …… 10分 (第21—A 题)

法二:在A 1206??=??

??1203--??

????

两边同时左乘逆矩阵1-A 得, 1206??=??

??

1-A 1203--??????. …… 4分 设1-=A a b c d ??????,则1206??=????a b c d ??????1203--??

???

?, 所以1a -=,232a b -+=,0c -=,236c d -+=. …… 6分 解得1a =-,0b =,0c =,2d =,从而11002--??=????

A . …… 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,

已知直线32x y ?=-????,(l 为参数)与曲线218x t y t

?=???=?,

(t 为参数)

相交于A ,B 两点,求线段AB 的长.

解:法一:将曲线218x t y t

?=???=?,

(t 为参数)化为普通方程为28y x =. …… 3分

将直线32x y ?=-+???=?,(l 为参数)代入28y x =得,

2240l -+=, …… 6分

解得1l =

2l =

则12l l -=,

所以线段AB

的长为 …… 10分 法二:将曲线218x t y t

?=???=?,

(t 为参数)化为普通方程为28y x =, …… 3分

将直线3

2x y ?=-+???=?,

(l 为参数)化为普通方程为302

x y -+=, …… 6分

由28302

y x x y ?=?

?-+=??,得,122x y ?=???=?,或926.x y ?=???=?, 所以AB

= …… 10分

D .[选修4-5:不等式选讲](本小题满分10分)

设x y z ,,均为正实数,且1xyz =,求证:33

3111xy yz zx x y y z z x

++++≥. 证明:因为x y z ,,均为正实数,且1xyz =,

所以3122xy yz x x y +=≥,3122yz xz y y z +=≥,3

122xz xy z z x +=≥. …… 8分 所以33

3111xy yz zx x y y z z x

++++≥. …… 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)

某乐队参加一户外音乐节,准备从3首原创新曲和5首经典歌曲中随机选择4首进行演唱. (1)求该乐队至少演唱1首原创新曲的概率;

(2)假定演唱一首原创新曲观众与乐队的互动指数为a (a 为常数),演唱一首经典歌曲观 众与乐队的互动指数为2a .求观众与乐队的互动指数之和X 的概率分布及数学期望. 解:(1)设“至少演唱1首原创新曲”为事件A ,

则事件A 的对立事件A 为:“没有1首原创新曲被演唱”.

所以()

4

5

48C 13()1114

C P A P A =-=-=.

答:该乐队至少演唱1首原创新曲的概率为1314. …… 4分

(2)设随机变量x 表示被演唱的原创新曲的首数,则x 的所有可能值为0,1,2,3.

依题意,()24X ax a x =+-,故X 的所有可能值依次为8a ,7a ,6a ,5a .

则45

48C 1(8)(0)14C P X a P x =====,

1335

48C C 3(7)(1)7C P X a P x =====,

22

3548C C

3(6)(2)7C P X a P x =====,

31

35

48

C C 1(5)(3)14C P X a P x =====.

从而X 的概率分布为:

…… 8分

所以X 的数学期望()133191876514771414E X a a a a a =?+?+?+?=.…… 10分

23.(本小题满分10分)

设*2n n ∈N ≥,.有序数组()12n a a a ???,,,经m 次变换后得到数组()

12m m m n b b b ???,,,,,,,

其中11i i i b a a +=+,,111m i m i m i b b b --+=+,,,(i =1,2,???,n ),11n a a +=,1111m n m b b -+-=,,(2)m ≥. 例如:有序数组()123,

,经1次变换后得到数组()122331+++,,,即()354,,;经第 2次变换后得到数组()897,

,. (1)若 (12)i a i i n ==???,,,,求35b ,的值;

(2)求证:0C m

j

m i i j m j b a +==∑,,其中i =1,2,???,n .

(注:当i j kn t +=+时,*k ∈N ,t =1,2,???,n ,则i j t a a +=.) 解:(1)依题意,()12345678n ???,

,,,,,,,, 经1次变换为:()35791113151n ???+,,

,,,,,,,

经2次变换为:()812162024284n ???+,

,,,,,,, 经3次变换为:()202836445212n ???+,

,,,,,, 所以3552b =,. …… 3分

(2)下面用数学归纳法证明对*

m ∈N ,0

C m

j

m i i j m j b a +==∑,,其中12i n =???,,,.

(i )当1m =时,1

1110

C j i i i i j j b a a a ++==+=∑,,其中12i n =???,,,,结论成立;

(ii )假设*

()m k k =∈N 时,k i b =

,0

C k

j i j

k j a

+=∑,其中12i n =???,,,. …… 5分

则1m k =+时,11k i k i k i b b b ++=+,,,

10

C C k

k

j

j i j k

i j k j j a a +++===+∑∑

110

1

C C k

k j j i j k

i j k j j a a +-++===+∑∑

()0

111

C C C C k

j j k

i k

i j k k i k k j a a a -+++==+++∑

01

1

1111

C C C k

j k i k i j k i k k j a a a +++++++==++∑ 1

10

C k j i j k j a +++==∑,

所以结论对1m k =+时也成立.

由(i )(ii )知,*

m ∈N ,0

C m

j

m i i j m j b a +==∑,,其中12i n =???,,,. …… 10分

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2017·全国卷Ⅱ(物理)
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15.D4[2017·全国卷Ⅱ] 一静止的铀核放出一个α 粒子衰变成钍核,衰变方程为23982U→23940 Th+42He.下列说法正确的是( )
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图1

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