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广东省韶关市2016届高三数学上学期1月调研测试试题 理(含解析)

广东省韶关市2016届高三数学上学期1月调研测试试题 理(含解析)
广东省韶关市2016届高三数学上学期1月调研测试试题 理(含解析)

韶关市2016届高三调研测试

数学(理科)试题

第Ⅰ卷

一、本大题共12小题,每小题5分,满分60分,在每小题给出的四个选项中,只有一项是符合题目要求的.

(1) 若复数z 满足(1)i z i -=,则复数z 模为( )

A .

12

B C D .2

(2)22cos 165sin 15-= ( )

A .

12 B (3)已知命题:p 对任意x R ∈,总有20x >;:q "1"x >是"2"x >的充分不必要条件,则下列命题为真命题的是( )

A .p q ∧

B .p q ??∧

C . p q ?∧

D . p q ?∧

(4) 曲线

221(6)106x y m m m +=<--与曲线22

1(59)59x y n n n

+=<<--的( ) A .焦距相等 B . 离心率相等 C .焦点相同 D .顶点相同

(5)如图,圆柱内有一个直三棱柱,三棱柱的底面在圆柱底面内,且底面是正三角形. 如果三棱柱的体积为 A .π12 B .π14 C .π16 D .π18

(6)已知随机变量ξ服从正态分布(1,1)N ,若(3)0.977P ξ<=,则(13)P ξ-<<=( ) A. 0.683 B. 0.853 C. 0.954 D. 0.977

(7)如图给出的是计算111

1352015

++++L L 的值的程序框图,其中判断框内应填入的是( ) A .2012i ≤ B .2014i ≤ C .2016i ≤

D .2018i ≤

(8) 某校开设10门课程供学生选修,其中A B C 、、三门由于上 课时间相同,至多选一门,学校规定:每位同学选修三门,则每位 同学不同的选修方案种数是( )

A .70 B. 98 C . 108 D .120

(9)在△ABC 中,∠C =90°,且BC =3,点M 满足BM 2MA =

, 则CM CB ?

等于( )

A .2

B .3

C .4

D .6

(10)已知函数()sin()(0,0)f x x ω?ωπ?=+>-<<的最小正周期是π, 将函数()f x 图象向左平移

3

π

个单位长度后所得的函数图象过点(0,1)P ,则函数()sin()f x x ω?=+ ( )

A.在区间[,]63ππ-

上单调递减 B.在区间[,]63ππ-上单调递增 C.在区间[,]36ππ

-

上单调递减 D.在区间[,]36

ππ

-上单调递增 (11)某几何体的三视图如图所示,正视图为直角三角形,侧视图为等边 三角形,俯视图为等腰直角三角形,则其外接球的表面积为( )

A .π5

B .

π320 C .π8 D .π3

28

(12)已知定义在R 上的函数)(x f y =满足:函数(1)y f x =-的图象关于直线

1x =对称,且当(,0),()'()0x f x xf x ∈-∞+<('()f x 是函数()f x 的导函数)成立, 若11(sin )(sin )22a f =,(2)(2)b ln f ln =,121

2()4

c f log =,则,,a b c 的大小关系是( )

A . a b c >>

B .b a c >>

C .c a b >>

D .a c b >>

第Ⅱ卷

本卷必考题与选考题两部分,第(13)至(21)题是必考题,每个试题考生必须做答,第(22)至(24)是选考题,考生根据要求做答。

二.填空题(本大题共4小题,每小题5分,满分20分).

(13)已知函数()f x 的图像在点(1,(1))A f 处的切线方程是2310x y -+=,'

()f x 是函数()f x 的

导函数,则(1)'(1)f f += .

(14)抛物线2

4y x =的焦点为F ,倾斜角等于45

的直线过F 交该抛物线于,A B 两点,则

G

F

C

B ||AB =______.

(15) 实数y x ,满足??

?

??≤->≤≥,0),1(,1y x a a y x 若目标函数y x z +=取得最大值4,则实数a 的值为 .

(16)ABC ?中, 3,2AB AC BC ==,则.ABC ?面积的最大值为 .

三.解答题:解答应写出文字说明.证明过程或演算步骤. (17)(本小题满分12分)

设*

n N ∈,数列{}n a 的前n 项和为n S ,已知12n n n S S a +=++,125,,a a a 成等比数列.

(Ⅰ)求数列{}n a 的通项公式; (Ⅱ)若数列{}n b

满足

1n a n

n

b a +=,求数列{}n b 的前n 项和n T .

(18)(本小题满分12分)

某厂生产一种零件,其质量按测试指标划分为:指标大于或等于8为优质品,小于8大于等于4

若以上述测试中各组的频率作为相应的概率. (Ⅰ)试估计这种零件的平均质量指标;

(Ⅱ)生产一件零件,若是优质品可盈利40元,若是正品盈利20元,若是次品则亏损20元;若从大量的零件中随机抽取2件,其利润之和记为X (单位:元),求X 的分布列及数学期望.

(19) (本小题满分12分)

如图,四边形ABCD 是矩形,1,AB AD ==E 是AD 的中点,BE 与AC 交于点F , GF ⊥平面ABCD .

(Ⅰ)求证:AF ⊥面BEG ;

(Ⅱ)若AF FG =, 求直线EG 与平面ABG 所成角的正弦值.

(20)(本小题满分12分)

已知椭圆

22

22

:1(0)

x y

C a b

a b

+=>>,它的一个焦点为

1

(1,0)

F-,

且经过点(

M-

(Ⅰ)求椭圆C的方程;

(Ⅱ)已知圆的方程是2222

x y a b

+=+,过圆上任一点P作椭圆C的两条切线

1

l与2

l,求证

12

l l

⊥.

(21)(本小题满分12分)

已知函数()ln

f x x

=,()()

h x a x a R

=∈.

(Ⅰ)函数()

f x与()

h x的图象无公共点,试求实数a的取值范围;

(Ⅱ)是否存在实数m,使得对任意的

1

(,)

2

x∈+∞,都有函数()

m

y f x

x

=+的图象在()

x

e

g x

x

=

的图象的下方?若存在,请求出最大整数m的值;若不存在,请说理由.

(参考数据:ln20.6931

=,,ln

3 1.0986

=

1.3956

==).

请考生在第(22)、(23)、(24)题中任选一题作答,如果多做,则按所做的第一题记分,解答时请写清题号.

(22)(本题满分10分)选修4-1:几何证明选讲

E 如图,AF是圆E切线,F是切点, 割线ABC与圆E交于B、C,

的直径,EF交AC于D,AC

AB

3

1

=,0

30

=

∠EBC,2

MC=.

(Ⅰ)求线段AF 的长; (Ⅱ)求证:ED AD 3=.

(23)(本小题满分10分)选修4—4:极坐标与参数方程

在直角坐标系xOy 中,以原点O 为极点,x 轴的正半轴为极轴,建立极坐标系. 已知曲线1C :4cos ,3sin ,x t y t =+??

=-+? (t 为参数), 2C :6cos ,

2sin ,x y θθ=??=?

(θ为参数).

(Ⅰ)化1C ,2C 的方程为普通方程,并说明它们分别表示什么曲线; (Ⅱ)若1C 上的点P 对应的参数为2

t π

=-

,Q 为2C 上的动点,求线段PQ 的中点M 到直线

3:cos sin 8C ρθθ=+距离的最小值.

(24)(本小题满分 10 分)选修 4-5:不等式选讲

设函数()|23||1|.f x x x =++- (Ⅰ)解不等式()4f x >;

(Ⅱ)若存在3,12x ??∈-????

使不等式1()a f x +>成立,求实数a 的取值范围.

2016届高三调研测试数学(理科)试题

参考解答和评分标准

说明:

1.参考答案与评分标准给出了一种或几种解法供参考,如果考生的解法与参考答案不同,可根据试题主要考查的知识点和能力比照评分标准给以相应的分数.

2.对解答题中的计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的得分,但所给分数不得超过该部分正确解答应得分数的一半;如果后继部分的解答有较严重的错误,就不再给分.

3.解答右端所注分数,表示考生正确做到这一步应得的累加分数. 4.只给整数分数,选择题和填空题不给中间分. 一、选择题:BCDAC CCBBB DA

(1) 【解析】 由已知(1)1+11,(1)(1)222i i i z i i i +-=

==-+--所以2

z =

,选B.

(2) 【解析】2

2

cos 165sin 15-=

22cos 15sin 15cos30-==

C (3) 【解析】 命题p 为真,命题q 为假,故选D

(4)【解析】由

22

1(6)106x y m m m

+=<--知该方程表示焦点在x 轴上的椭圆,由22

1(59)59x y n n n

+=<<--知该方程表示焦点在y 轴上的双曲线,故只能选择答案A. (5) 【解析】 设圆柱的底面半径为R ,则三棱柱的底面边长为R 3,由3122)3(4

3

2=?R R 得2=R ,.162R R 2ππ=?=圆柱侧S 故选 C

(6) 【解析】因为已知随机变量ξ服从正态分布(1,1)N ,所以正态曲线关于直线1x =对称,又

(3)0.977P ξ<=,所以(3)10.9770.023P ξ>=-=,(13)P ξ-<<

所以1(1)(3)P P ξξ=-<-->12(3)10.0460.954P ξ=->=-=,故选C

(7) 【解析】考查算法的基本思想及程序框图。选C

(8) 【解析】解析:本题考查排列组合的应用.共可分为两类:选A 、B 、C 中的一门,其它7科

中选两门,有12

3763C C =; 不选A 、B 、C 中的一门,其它7科中选三门,有37

35C =;所以共有98种,故选B

(9)【解析】 因为BM 2MA = ,0AC BC ?=

11()()(())33CM CB CB MA AC CB BA AC CB AC CB AC ?=?+=?+=?++ 2133

BC ==

,选B

B

N

M

A

B

C

另解1:如图建立坐标系,由已知可得,(3,0)B ,设(0,)A y ,由2BM MA =

得,2(1,

)3

M y 所以,(3,0)CB = ,2

(1,)3CM y = ,3CB CM ?= ,选B

另解2 设MCN θ∠=,cos CB CM CB CM θ?=?

过M 作MN BC ⊥交BC 于N ,

则cos CM CN θ= ,由3BC =,2BM MA =

, 得1CN =,3CB CM ?=

(10)【解析】依题意,2ω=, ()sin(2)f x x ?=+,平移后得到的函数是2sin(2)3

y x π

?=++,其图象过(0,1),所以,2sin()=13π?+,因为0π?-<<,所以,6

π

?=-,()sin(2)6f x x π=-,

故选B

(11)【解析】设外接球的球心O ,M E ,分别是ACD BCD ??,的外心,⊥OE 平面BCD ,

⊥OM 平面ACD ,则2

22)3

3(

)2(+=R , 解得2

7

3

R =,故328

π=球表S 选.D

另解:设F 是BC 的中点,如图建立坐标系。则(2,1,0)B -

(0,1,0)C ,(0,1,0)D -,

设O ((,,)

x y z 是球心,球的半径为r ,由OA OB OC OD ====222222

222222

222222(2)(1)(1)(2)(1)(1)(2)(1)(x y z x y z x y z x y z

x y z x y z ?-+++=++

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

++? 解得 1,0,x y z === 所以,2222

7(3r x y z =++=

3

28π

=

球表S (12)【解析】:因为函数(1)y f x =-的图象关于直线1x =对称,所以()y f x =关于y 轴对称,

所以函数()y xf x =为奇函数.因为[()]

'()'(x f x f x x f x =+,所以当(,0)x ∈-∞

时,D A

B

O

E

F

M

[()]'()'()0xf x f x xf x =+<,函数()y xf x =单调递减,当(0,)x ∈+∞时,函数()y xf x =单调递

减. 110sin

22<<

,11ln 22>>=,121log 24=, 1211

0sin ln 2log 24

<<<,∴所以a b c >>,选 A.

(13) 【解析】因为(1)1f =,'

(1)3

f =

,所以f (1)+f ′(1)=53.

(14) 【解析】由题可知焦点(1,0)F ,直线AB 的方程1y x =-,设点12(,)A x y ,22(,)B x y 联

241

y x

y x ?=?

=-? 可得

261

x x -+= ,

126x x += ,12628AB x x p =++=+=

.

(15) 【解析】做出可行域,由题意可知可行域为ABC ?内部,z x y +-=,则z 的几何意义为直线在y 轴上的纵截距,将目标函数平移可知当直线经过点A 时,目标函数取得最大值4,此时A 点坐标为),(a a ,代入得a a a 24

=+=,所以2=a

.

(16) 解法一:设,2BC x AC x ==,则293cos ,sin 6x B

B x -== 即1sin 2AB

C S AB BC B ?=

==当13x <<时,三角形ABC 的面积的最大值为3.

解法二、设(1,0),(2,0),(,)A B C x y -,则点C 的轨迹方程为22(3)4x y -+=,当底边AB 上的高为半径2,三角形ABC 的面积的最大值为3.

(17)解:(Ⅰ)由12n n n S S a +=++得: *

12()n n a a n N +-=∈ …………………1分 所以数列{}n a 是以1a 为首项,2为公差的等差数列 ……………………………………3分 由125,,a a a 成等比数列.即2

111(2)(8)a a a +=+ 解得11a =…………………… ……4分 所以,*21

()n a n n N =-∈

………………………………………………………5分

(Ⅱ)由(Ⅰ)可得2(21)(21)2n n n b n n =-?=-,……………………………… ……6分 所以1231n n n T b b b b b -=+++++ ,即

123123252(21)2n n T n =?+?+?++-? ①. …………………………………………8分

23121232(23)2(21)2n n n T n n +=?+?++-?+-? ②. ……………………………10分

①—②可得 23122(222)(21)2n n n T n +-=++++-- ,

所以1(23)26n n T n +=-+. ………………………………………………………… ……12分

(18) 解:(1)平均质量指标为

2183325387209

6.32100

?+?+?+?+?= ……………3分

(2)由表可得任取一件零件为优质品的概率为

201

1005

=,任取一件零件为正品的概率为3238710010+=,任取一件零件为次品的概率为101

10010

=,……………………………6分 从大量的零件中随机抽取2件,其利润之和记为X ,则X 的可能取值为40,0,20,40,60,80-.

22211(40)()10100P X C =-==,12177(0)101050P X C ==??=,

12111(20)10525P X C ==??=,2

22749(40)()10100P X C ==?=,

12717(60)10525P X C ==??=,2

2111(80)5525

P X C ==??=.

…………………………………………………………………………………………………10分

故X

……………………………………………………………………………………………11分

X 的数学期望值为17149714002040608010050251002525

EX =-?

+?+?+?+?+? 40=

…………………………………………………………………………………………………12分 (19)证法1:

∵四边形ABCD 为矩形,∴AEF ?∽CBF ?, ∴

2

1

===BC AE BF EF CF AF ……………1分 又∵矩形ABCD 中,2,1==AD AB ,∴3,2

2

==

AC AE

在BEA Rt ?中,2622=

+=

AE AB BE ∴3

3

31==AC AF ,3632==BE BD

在ABF ?中,2222

21)3

6

()33(

AB BF AF ==+=+ ∴

90=∠AFB ,即BE AC ⊥ ……………3分

∵⊥GF 平面ABCD ,?AC 平面ABCD ∴GF AC ⊥ ……………4分 又∵F GF BE = ,?GF BE ,平面BCE ∴⊥AF 平面BEG ……………5分 证法2:(坐标法)证明1-=?BE AC K K ,得BE AC ⊥,往下同证法1. 证法3:(向量法)以,为基底,

∵AB AD BE AB AD AC -=+=2

1

,,0=?AB AD ∴)21()(-?+=

?-=01221=-?=

∴BE AC ⊥,往下同证法1. (2)在AGF Rt ?中,22GF AF AG +=

3

6

)33()33(

22=

+= 在BGF Rt ?中,22GF BF BG +=

1)3

3

()36(

22=+= ………… ……………7分 在ABG ?中,3

6

=

AG ,1==AB BG ∴2)6

6

(13621-??=

?ABG S 656303621=??

= ………………………………9分 设点E 到平面ABG 的距离为d ,则

GF S d S ABF ABG ?=???3131,∴ABG ABF S GF S d ??=10306

5

33

12221=?

??=

2

2

)66()33(

2222=

+=+=EF GF EG …………………… ……………11分 设直线EG 与平面ABG 所成角的大小为θ,则

EG d

=θsin .5152

2

1030

== ……………………………………………… ……………12分

另法:由(1)得FG BE AD ,,两两垂直,以点F 为原点,FG FE FA ,,所在直线分别为x 轴,y 轴,

z 轴建立如图所示的空间直角坐标系,………………………………………………6分

则???? ??0,0,33A ,???? ??-0,36,0B ,???? ??33,0,0G ,???? ??0,66,0E , ???? ??--=0,36,33AB ,????

??-=33,0,33AG , ???

?

??-=33,

66,0, …………8分 设),,(z y x n =是平面ABG 的法向量,则

????

?=?=?0

n AG ,即???

????

=+-=--03

333036

33z x y x , 取2=

x ,得)2,1,2(-=n ……………………………………………………………10分

设直线EG 与平面ABG 所成角的大小为θ,则

5

152123

1

610233

)1(6620sin =

++?++

?+-?-?=

=

θ ∴直线EG 与平面ABG 所成角的正弦值为.5

15

………………………………………12分

(20)(Ⅰ)一个焦点为2(1,0)F ,则

1222

a MF MF =+=

=> a ∴=2分

222312b a c ∴=-=-=.

G

F

E

D

C

B

A x

y

z

∴椭圆的标准方程是22

1.32

x y +

=………………………………………………………4分 (Ⅱ)设00(,)P x y ,若过点 的切线斜率都存在,设其方程为00()y y k x x -=- ,

由0022

()236

y y k x x x y -=-??+=? 得,222

0000(23)6()3()60k x k y kx x kx y ++-+--= ,……6分 直线与椭圆相切,0∴?= ,……………………………………………………………7分

2220000[6()]4(23)[3()6]0k y kx k kx y --+--= ,

整理得222

0000(3)220x k x y k y -++-= ,……………………………………………… 8分

椭圆 的两条切线的斜率分别为12,k k ,

2

122

23y k k x -∴?=- ,……………………………………………………………………… 9分 点 在圆 上,22005x y ∴+= ,即22

005y x =- , 222

00012222

000

22(5)3)

1333y x x k k x x x ----+∴?====---- 1

l ∴⊥

2

l (11)

分.

若过点P 的切线有一条斜率不存在,不妨设该直线为1l ,则1l

的方程为x =2l

的方程为

y =1l ⊥2l

综上,对任意满足题设的点P ,都有1l ⊥2l ……………………………………12分 (21) 解:(Ⅰ)函数()f x 与()h x 无公共点,等价于方程ln x

a x

=在(0,)+∞无解.…2分 令ln ()x t x

=

,则2

1ln '(),x

t x -=令'()0,t x =得x e = 因为x e =是唯一的极大值点,故max ()t t e e

==……………………………………4分

故要使方程

ln x

a x

=在(0,)+∞无解,当且仅当1a e > 故实数a 的取值范围为1

(,)e

+∞…………………………………………………………6分

(Ⅱ)假设存在实数m 满足题意,则不等式ln x

m e x x x

+<对1(,)2x ∈+∞恒成立.

即ln x

m e x x <-对1(,)2

x ∈+∞恒成立.……………………………………………6分

令()ln x r x e x x =-,则'()ln 1x r x e x =--,

令()ln 1x x e x ?=--,则1'()x

x e x

?=-,………………………………………7分

因为'()x ?在1(,)2+∞上单调递增,1

21

'()202

e ?=-<,'(1)10e ?=->,且'()x ?的图

象在1(,1)2上连续,所以存在01(,1)2x ∈,使得0'()0x ?=,即00

10x

e x -=,则00ln x x =-,…………………………………………………………………………9分

所以当01

(,)2

x x ∈时,()x ?单调递减;当0(,)x x ∈+∞时,()x ?单调递增,

则()x ?取到最小值000001()ln 11x

x e x x x ?=

--=+-110≥=>,

所以'()0r x >,即()r x 在区间1

(,)2

+∞内单调递增.………………………………11分

11

221111

()ln ln 2 1.995252222

m r e e ≤=-=+=,

所以存在实数m 满足题意,且最大整数m 的值为1. …… ………12分

(22)(本题满分10分)选修4-1:几何证明选讲 解:(Ⅰ)因为BM 是圆E 直径

所以, 0

90=∠BCM ,………………1分

又2MC =,0

30=∠EBC ,

所以

32=BC ,

........................ (2)

又,31

AC AB =

可知32

1

==BC AB ,所以33=AC …………………………………3分

根据切割线定理得:

93332=?=?=AC AB AF ,…………………………………………………4分

即3=AF …… …………………………………… …………………………………5分 (Ⅱ)过E 作BC EH ⊥于H ,……………………………………………………………6分

则ADF EDH ??~,……………………………… …………………………………7分

从而有

AF

EH

AD ED =,…………………………………………………………………8分 又由题意知,BC CH 32

1

==2=EB

所以1=EH , …………………………………9分 因此

3

1

=AD ED ,即ED AD 3= …………………………………10分

(23)(本小题满分10分)

解:(Ⅰ)221:(4)(3)1,C x y -++=,…………………………………………………1分

22

2:

1364

x y C += …………………………… ………………………………………2分 1C 为圆心是(4,3)-,半径是1的圆. ………………………………………3分 2C 为中心在坐标原点,焦点在x 轴上,长半轴长是6,短半轴长是2的椭圆.

…………………………………………………………4分

(Ⅱ)当2

t π

=

时,(4,4)P -,………………………………………………………5分 设(6cos ,2sin )Q θθ

则(23cos ,2sin )M θθ+-+, ………………………………………6分

3C

为直线(80x -+=,……………………………………7分

M 到3C

的距离d =

……………………8分

=

=

3)6

π

θ=+

………………………………………9分

从而当cos()1,6

π

θ+

=时,d

取得最小值3………………………………10分

∴实数a 的取值范围为3,2??

+∞ ???

…………………………………… …………………10分

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