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高一假期作业3

高一假期作业3
高一假期作业3

大港中学高一数学假期作业3

班级: 姓名:

一、填空题:本大题共14小题,每小题5分,共计70分. 1.已知直线l 经过点(2,0)A -,(5,3)B -,则直线l 的倾斜角为 .

2.在ABC ?

中,已知AB =1AC =,30A = ,则ABC ?的面积为 . 3.不等式(1)0x x ->的解集为 .

4.经过点(1,2)-,且与直线052=-+y x 平行的直线方程为 .

5.在ABC ?中,角A ,B ,C 所对的边分别为a ,b ,c ,若bc a c b =-+2

2

2

,则角A

的大小为 .

6.在数列{}n a 中,已知11a =,且1n n a a n +=+,*n ∈N ,则9a 的值为 .

7.三角形三边成公差为2的等差数列,且最大角的正弦值为

2

3

,则此三角形的面积为____ 8.已知直线1:(2)10l ax a y +++=,2:20l x ay ++=,若21l l ⊥,则实数a 的值为 . 9.若实数x ,y 满足条件,4,3120y x x y x y ??

+??-+?

≥≥≥,则2z x y =+的最大值为 .

10.在等比数列{}n a 中,已知22a =,832a =,则5a 的值为 .

11.已知实数x ,y 满足42=-y x ,则y

x

??

?

??+214的最小值为 .

12.已知正数x 、y 满足x +2y =2,则x +8y

xy 的最小值为_____

13.设n S 为数列{}n a 的前n 项和,已知11,1,1,2,

n n a n n =?=?+?≥*n ∈N ,则n S

n 的最小值为

______ .

14.已知直线l 的方程为0=++c by ax ,其中a ,b ,c 成等差数列,则原点O 到直线l 距

离的最大值为 ____ .

二、解答题:本大题共6小题,共计90分.

15.已知点()()2,3,3,2P Q -,直线l :(2)(12)(12)0()a x a y a a R --+++=∈; (1)求当直线l 与直线PQ 平行时实数a 的值;

(2)求直线l 所过的定点(与a 的值无关的点)M 的坐标; (3)直线l 与线段PQ (包含端点)相交,求实数a 的取值范围.

16.已知{n a }是首项为1,公比为q 的等比数列,且465,,a a a 成等差数列. (Ⅰ)求{n a }的前n 项和n S ;

(Ⅱ)设{n b }是以2为首项, q 为公差的等差数列,其前n 项和为n T ,当2n ≥时,比较

n T 与n b 的大小,并说明理由.

17.已知直线l 的方程为210x my m +--=,m ∈R 且0m ≠.

(1) 若直线l 在x 轴,y 轴上的截距之和为6,求实数m 的值;

(2) 设直线l 与x 轴,y 轴的正半轴分别交于A ,B 两点,O 为坐标原点,求AOB ?面

积最小时直线l 的方程.

18.如图,洪泽湖湿地为拓展旅游业务,现准备在湿地内建造一个观景台P ,已知射线AB ,AC 为湿地两边夹角为120 的公路(长度均超过2千米),在两条公路AB ,AC 上分别设

立游客接送点M ,N ,从观景台P 到M ,N 建造两条观光线路PM ,PN ,测得2AM =千米,2AN =千米.

(1) 求线段MN 的长度;

(2) 若60MPN ∠= ,求两条观光线路PM 与PN 之和的最大值.

M

B

N

C P

(第18题)

19.已知函数22()24f x x ax a =-+-,8)(22-+-=a x x x g ,a ∈R .

(1) 当1a =时,解不等式()0f x <;

(2) 若对任意0>x ,都有)()(x g x f >成立,求实数a 的取值范围;

(3) 若对任意[]1,01∈x ,总存在[]1,02∈x ,使得不等式)()(21x g x f >成立,求实数a

的取值范围.

20.在等差数列{}n a 中,已知11a =,公差0d ≠,且1a ,2a ,5a 成等比数列,数列{}n b 的 前n 项和为n S ,11b =,22b =,且243n n S S +=+,*n ∈N .

(1) 求n a 和n b ;

(2) 设(1)n n n c a b =- ,数列{}n c 的前n 项和为n T ,若(1)n λ-≤2(3)n n T n +-对任意

*n ∈N 恒成立,求实数λ的取值范围.

大港中学高一数学假期作业3

一、填空题 1.34p

2. 3.(0,1) 4.20x y += 5.3

p

6.37

7.

8.0a =或3a =- 9. 18 10. 8± 11.8 12. 9 13.

23

4

二、解答题 15.

16. 解:(Ⅰ)由题设6452a a a =+ .012,021=--∴≠q q a

.21

1-=∴或q

……4分

若1q =则n S n =; ……5分

若12q =-则12123

n n S ????-- ? ? ????

?= ……6分 (Ⅱ)若1q =则232

n n n

T +=

当2n ≥时,()()1122

n n n n n T b T --+-==

n n T b > ……10分

若12q =-时,294

n n n

T -+=

当2n ≥时,()()

11104

n n n n n T b T ----==

故对于,n N +∈当29n ≤≤时,n n T b >;当10n =时,=n n T b ;

当11n ≥时,n n T b < ……15分

17.(1)令0=x ,得m

y 12+

=. 令0=y ,得12+=m x . ………2分

由题意知,1

2126m m

+++

=. ……4分 即01322

=+-m m , 解得1

2

m =或1=m . ………6分 (2)方法一:

由(1)得 )1

2,0(),0,12(m

B m A +

+, 由210,1

20.m m +>???+>??

解得0m >. ……8分 BO AO S ABC ?=

?21)1

2)(12(21121221m

m m m ++=+?+= ………10分 )1

2)(21(m

m ++

= 1

222242m m

=++

+=≥, ……12分 当且仅当m m 212=

,即2

1

=m 时,取等号. ……13分 此时直线l 的方程为042=-+y x . ……14分 方法二:

由210x my m +--=,得0)2()1(=-+-y m x .

所以1020x y -=??-=?,解得12

x y =??=?.

所以直线l 过定点)2,1(P . ……8分 设)0,0)(,0(),0,(>>b a b B a A ,则直线l 的方程为:)0,0(1>>=+b a b

y

a x . 将点)2,1(代人直线方程,得12

1=+b

a . ……10分

由基本不等式得

12a b +≥8ab ≥. ……12分 当且仅当

b

a 2

1=,即4,2==b a 时,取等号. ……13分 所以1

42

ABC S ab ?=

≥, 当AOB ?面积最小时,直线l 的方程为042=-+y x . ……14分 18.(1)在AMN ?中,由余弦定理得,

2222cos120MN AM AN AM AN =+-? ……2分

=12)2

1(222222

2=-???-+,

所以32=MN 千米. ……4分 (2)设α=∠PMN ,因为60MPN ∠= ,所以120PNM α∠=-

在PMN ?中,由正弦定理得,

0sin sin(120)sin MN PM PN

MPN αα

==∠- . ……6分

因为

sin MN

MPN

∠4==, 所以ααsin 4),120sin(40

=-=PN PM ……8分

因此ααsin 4)120sin(40+-=+PN PM ……10分

=αααsin 4)sin 2

1

cos 23(

4++ =ααcos 32sin 6+=)30sin(340+α ………13分

因为0120α<< ,所以3030150α<+< .

所以当009030=+α,即0

60=α时,PN PM +取到最大值34.………15分

答:两条观光线路距离之和的最大值为34千米. ………………………………16分 19.(1)当1=a 时,032)(2<--=x x x f

所以(23)(1)0x x -+<,…………………………………………………………………2分 解得3

12

x -<<

. ………………………………………………3分 所以当1=a 时,不等式0)(

??

???<

<-231|x x . ……………4分 (2)由)()(x g x f >,得8422

2

2

2

-+->-+-a x x a ax x ,

即2(1)40x a x --+>.

所以2(1)4a x x -<+,因为0x >,所以4

1a x x

-<+.……………………………6分 因为44x x +

≥,当且仅当x

x 4

=,即2=x 时,取等号. 所以5a <,

所以实数a 的取值范围为)5,(-∞ . …………………………………………8分 (3)由题意知,min min )()(x g x f >. ………………………………………10分

因为4

33)2

1

()(2

2-

+-=a x x g ,

当[]1,0∈x 时,4

33

)2

1()(2

min -

==a g x g . ………………………………12分 又因为22()24f x x ax a =-+-48

7)4(22

2-+-=a a x

当0

3342

2->-a a 成立,

所以0a <时,min min )()(x g x f > …………………………………………13分 当04a ≤≤时,2min 7()()448

a f x f a ==-, 由

4

33

48722->-a a ,解得34a 时,2)1()(2min --==a a f x f ,

因为4

3322

2

-

>--a a a ,解得425

44a << ………15分

综上,a 的取值范围为)4

25

,(-∞ . ………16分

20. (1)设数列{}n a 的公差为d ,由题设可得2(1)1(14)d d +=?+.

解得 d=0(舍)或d=2,所以21n a n =-. ………2分 由2

43n n S

S +=+,可得214(1)n n S S ++=+ ………4分

又因为11b =,22b =,所以112S +=,214S +=. 当n 为奇数时,12n

n S +=; 当n 为偶数时,12n n S +=.

所以12,n

n S n *

+=∈N ……6分 当2n ≥时,1

12n n n n b S S --=-=,

所以1

2

,n n b n -*=∈N . ……8分

(2)因为 1

1(21)(2

1)(21)2(21)n n n c n n n --=--=-?--,

则0

1

2

2

12123252(23)2

(21)2n n n T n n n --=?+?+?++-+-- . 设0

1

2

2

1123252(23)2(21)2n n n M n n --=?+?+?++-+-

则1

2

1

21232(23)2

(21)2n n n M n n -=?+?++-+-

两式相减,得2

3

1222(21)2(23)23n

n

n

n M n n -=++++--=--+ 所以(23)23n

n M n =--.

所以2

(23)23n n T n n =--- ……12分 令 2

(3)(23)2n

n n e n T n n n =+-=-, 由1n n e e +< ,得1(23)2(1)(21)2n n n n n n +-<+- 即(23)2(1)(21)n n n n -<+-,解得对任意*n ∈N 成立,

即数列{}n e 为单调递增数列. …14分 当n 为奇数时,12e λ-=-≤,所以2λ≥; 当n 为偶数时,28e λ=≤,

所以28λ≤≤. ……16分

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