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江浦高级中学文昌校区2013届高三第五周周周练

江浦高级中学文昌校区2013届高三第五周周周练
江浦高级中学文昌校区2013届高三第五周周周练

文昌校区2013届高三第五周周周练

1.已知U 为实数集,集合{|02},{|M x x N x y =<<=,则()U M C N = ____ .

2.右边的流程图最后输出的n 的值是

3.符号函数??

?

??<-=>=0,10,00,1sgn x x x x ,则不等式2sgn

)1(>+x x 的解集是

4.复数2

(3)(,z a a i a R =--∈为纯虚数,则3

6+z a i

+=

5. 设双曲线22

1x y -=的两条渐近线与直线x =围成的三角形区域(包含边界)为D ,点(

,)P x y 为

D 内的一个动点,则目标函数2z x y =-的最小值为 .-

6.两个非零向量b a 与 ,定义||||sin a b a b θ?=

,其中θ为b a 与的夹角.若)6,3(-=+b a ,

)2,3(-=-b a ,则a b ?=

______.

7.若关于x 的不等式2

293x

x x kx ++-≥在[1,5]上恒成立,则实数k 的范围为 .

8.若直径为2的半圆上有一点P ,则点P 到直径两端点,A B 距离之和的最大值为 . 9.锐角三角形ABC 中,边长,a b 满足2a b ab +==,且2sin()0

A B +,则另一边长c = .

10.如图,已知某地一天从6时到14时的温度变化曲线近似满足sin()y A x ω?=++(02)?π≤<,则温度变化曲线的函数解析式为 .

11.

x x x f cos sin )(1+=,

'21

()()f x f x =,'32

()()f x f x =,…,)()('1

x f

x f n n -=)2*,(≥∈n N n ,

则122013()(()444

f f f πππ

+++= ________.

12.设等差数列

{}n a 的公差为d ,若7654321,,,,,,a a a a a a a 的方差为1,则d =_______.

13等差数列{}n a 的前n 项和为n S ,若存在正整数(),m n m n <,使得m n S S =,则0m n S +=.类比上述结论,设正项等比数列{}n b 的前n 项积为n T ,若存在正整数(),m n m n <,使得m n T T =,则m n T += .

14.P 是椭圆116

2522=+y x 上一点,A 和F 分别是椭圆的左顶点和右焦点,则?+?41

的最小

值为________.

15.已知,A B 是△ABC 的两个内角,sin 22

A B A B a i j +-=+

(其中,i j 是互相垂直的单位向

量),若||a =

(1)试问tanB tanA ?是否为定值,若是定值,请求出,否则说明理由;(2)求tan C

的最大值,并判断此时三角形的形状。

16.直棱柱

1

1

1A B C D A B C D -,ABCD 直角梯形,222A B

A D C D ===,

90BAD ADC ∠=∠= .(Ⅰ)求证:AC ⊥平面11BCC B ;(Ⅱ)在11A B 上是否存一点

P ,使得DP 与平面1BCB 与平面1ACB 都平行?证明你的结论.

17.过点

)0,1(-A 的动直线l 与圆C :4)3(22=-+y x 交于P 、Q 两点,M

是PQ 中点,l 与

直线m :063=++y x 相交于N .(1)求证:当l 与m 垂直时,l 必过圆心C ;(2)当3

2=PQ 时,求直线l 的方程;(3)探索?是否与直线l 的倾斜角有关,若无关,请求出其值;若有关,

请说明理由.

18.已知函数()ln 1f x x x =-+,()∞+∈

,0x .(1)求()f x 的单调区间和极值;

(2)设a ≥1,函数()2

2

325g x x ax a =-+-,若对于任意()001x ∈,,总存在()101x ∈,,使得

()()01x g x f =成立,

求a 的取值范围;(3)对任意()∞+∈

,0x ,求证:

111

ln 1x x x x

+<<+.

参考答案

{|01}x x <

<,5,}13{>-

22 ,6,

(],6-∞

310sin 208

4y x π

π??

=+

+ ?

??

12±,1,-9

15.解:(1)

2223||2cos sin 222A B A B a +-=+= ,1cos()3

1cos()22A B A B --+++= …4分

02sin sin cos cos sin sin cos cos =+-

-B A B A B A B A ;13tan tan 02

2A B

-=

1

tan tan 3

A B =(定值)……7分

(2)由(1)可知A 、B

为锐角

tan tan 3(tan tan )

tan tan()1tan tan 2A B A B C B A A B ++=-+=-

=-≤--

当且仅当tan tan 36

A B A B π

==

==时取得等号…………12分 所以tan C

的最大值为ABC 为钝角三角形。…………14分 16.证明:(Ⅰ) 直棱柱1111ABCD A B C D -中,BB 1⊥平面ABCD ,∴BB 1⊥AC .

又 ∠BAD =∠ADC =90°,222AB AD CD ===,

∴AC CAB =45°

,∴BC =∴ BC ⊥AC .………………5分 又1BB BC B = ,1,BB BC ?平面BB 1C 1C ,∴ AC ⊥平面BB 1C 1C . …7分 (Ⅱ)存在点P ,P 为A 1B 1的中点. ………………………………………8分 证明:由P 为A 1B 1的中点,有PB 1‖AB ,且PB 1=

1

2

AB .…………………9分 又∵DC ‖AB ,DC =

1

2

AB ,∴DC ∥PB 1,且DC = PB 1, ∴DC PB 1为平行四边形,从而CB 1∥DP .…………………………11分 又CB 1?面ACB 1,DP ?面ACB 1,∴DP ‖面ACB 1.……………13分 同理,DP ‖面BCB 1.………………………………………………14分

17.(1)∵l 与m 垂直,且3

1

-

=m k ,∴3l k =, 故直线l 方程为3(1)y x =+,即330x y -+=………2分

∵圆心坐标(0,3)满足直线l 方程,

∴当l 与m 垂直时,l 必过圆心C ………………… …4分

(2)①当直线l 与x 轴垂直时, 易知1-=x 符合题意…………………6分

②当直线l 与x 轴不垂直时, 设直线l 的方程为)1(+=x k y ,即0=+-k y kx , ∵32=PQ ,∴134=-=CM ,………………………………………8分

则由11

|3|2=++-=

k k CM ,得3

4

=

k , ∴直线l :0434=+-y x . 故直线l 的方程为1-=x 或0434=+-y x ………………………………………10分

(3)∵CM MN ⊥,∴ ()AM AN AC CM AN AC AN CM AN AC AN ?=+?=?+?=?

……12分 ① 当l 与x 轴垂直时,易得5(1,)3

N --,则5(0,)3

AN =-

,又(1,3)AC =

,

∴5AM AN AC AN ?=?=-

………………………………………………………14分

当l 的斜率存在时,设直线l 的方程为)1(+=x k y ,

则由?

??=+++=063)1(y x x k y ,得N (36,13k k --+k k 315+-),则55(

,)1313k

AN k k --=++ ∴AM AN AC AN ?=? =

51551313k

k k

--+=-++ 综上所述,?与直线l 的斜率无关,且5-=?.…………………16分 (本题若用其它解法,相应给分)

18. (1) ∵11()1x

f x x x

-'=

-=

…………………………………………………2分 ∴当x >1时,()f x '<0,当0<x <1时,()f x '>0. ∴()f x 的单调递增区间为()0,1,单调递减区间为()1,+∞,极大值为(1)0f =. ……4分 (2) ∵()23g x x a '=-(a ≥1)∴当)1,0(∈x 时,()230g x x a '=-<,)(x g 单调递减, 此时()g x 值域为2(234,a a --225)a -. …………………………………………6分 由(1)得,当)1,0(∈x 时,)(x f 值域为(),0-∞, ……………………………8分

由题意可得:225a -≤0,所以1≤a ≤2

. ………………………………10分 (3)令

1x t x +=,则11x t =

-,∵0x >,∴1t >,原不等式等价于1

1ln 1t t t

-<<- 由(1)知()ln 1f t t t =-+在()1,+∞上单调递减,∴()()10f t f <=,即ln 1t t <-……12分

令()1ln 1h t t t =-+,∵()22111

t h t t t t

-'=-=,当()1,t ∈+∞时,()0h t '>,

∴()1ln 1h t t t =-+在()1,+∞上单调递增,∴()()10h t h >=,即1

1ln t t -<……………14分

综上所述,对任意()∞+∈,0x ,恒有111ln 1x x x x

+<<+成立. ……………………16分

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