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高三数学限时训练(教师用)4,5

高三数学限时训练(教师用)4,5
高三数学限时训练(教师用)4,5

数学限时作业(4)

1.已知=U R ,集合23|

02x M x x

-??

=>??+??

,则U M =e . ]23,2[-

2.非负实数x 、y 满足?

??≤-+≤-+030

42y x y x ,则3x y +的最大值为 . 9

3.在ABC ?中,若2,3,4===c b a ,则ABC ?的外接圆半径长为 .

1515

8

4.已知函数g (x )=|x -1|-|x -2|,(x ∈R),若关于x 的不等式g (x )≤a 恒成立,则实数a 的取值范围是________.a

≥1

5.一堆除颜色外其他特征都相同的红白两种颜色的球若干个,已知红球的个数比白球的多, 但比白球的2倍少,若把每一个白球都记作数值2,每一个红球都记作数值3,则所有球的

数值的总和等于60.现从中任取一个球,则取到红球的概率等于________.14

23

6.如图在正四棱锥S -ABCD 中,E 是BC 的中点,P 点在侧面△SCD 内及其边界上运动,并

且总是保持PE ⊥AC ,则动点P 的轨迹与△SCD 组成的相关图形是 .D

7.设向量i 、j 为直角坐标系的x 轴、y 轴正方向上的单位向量,若向量a =(x +1)i +yj ,

b =(x -1)i +yj ,且|a |-|b |=1,则满足上述条件的点P (x ,y )的轨迹方程是

x 21

4

-y 2

34

=1(x ≥0) 8.设有一组圆k C :2

2

4

(1)(3)2x k y k k -++-= (k 属于正整数集).下列四个命题: ①存在一条定直线与所有的圆均相切 ②存在一条定直线与所有的圆均相交 ③存在一条定直线与所有的圆均不相交 ④所有的圆均不经过原点

其中真命题的代号是__________.(写出所有真命题的序号) ②④

9.已知平面向量)1),(sin(x a -=π,)cos ,3(x b =,函数b a x f ?=)(. (1)写出函数)(x f 的单调递减区间; (2)设1)6

()(+-=π

x f x g ,求直线2=y 与)(x g y =在闭区间],0[π上的图像的所有

交点坐标. 答案:(1))6

sin(2cos )sin(3)(π

π+

=+-=

x x x x f ,…4分

单调递减区间)](3

42,3

2[Z k k k ∈+

ππ

π; …… 6分 (2)1sin 21)6

()(+=+-

=x x f x g π

,…………………………… 8分

解2)(=x g ,即21sin =

x ,],0[π∈x 得65,6ππ=x ,…………12分 所以交点坐标为:)2,6

5(

),2,6(π

π. ……14分 10.定义:如果数列{}n a 的任意连续三项均能构成一个三角形的三边长,则称{}n a 为“三角形”数列.对于“三角形”数列{}n a ,如果函数()=y f x 使得()n n b f a =仍为一个“三角形”数列,则称()=y f x 是数列{}n a 的“保三角形函数”,(n N*)∈.

(1)已知{}n a 是首项为2,公差为1的等差数列,若(),(1)x

f x k k =>是数列{}n a 的“保三

角形函数”,求k 的取值范围;

(2)已知数列{}n c 的首项为2010,n S 是数列{}n c 的前n 项和,且满足1438040+-=n n S S ,证明{}n c 是“三角形”数列;

(3)根据“保三角形函数”的定义,对函数2

()2h x x x =-+,[1,]∈x A ,和数列1,1+d ,

12+d (0>d )提出一个正确的命题,并说明理由.

答案:(1)显然1n a n =+,12n n n a a a +++>对任意正整数都成立, 即{}n a 是三角形数列. …… 2分

因为k>1,显然有12()()()n n n f a f a f a ++<<

12n n n k k k +++>,解得15

2

k +<

.

所以当15

(1,

)2

+∈k 时, ()x f x k =是数列{}n a 的“保三角形函数”. …… 5分

(2) 由1438040+-=n n S S 得1438040--=n n S S ,两式相减得1430+-=n n c c

所以,1

320104-??

= ?

??

n n c ,

经检验,此通项公式满足1438040+-=n n S S ……7分 显然12++>>n n n c c c ,因为1

1

12

3321320102010201044164+-++??????

+=+=?> ? ?

?????

??

n n n n n n c c c ,

所以{}n c 是“三角形”数列. …… 10分

(3) 探究过程: 函数2()2h x x x =-+,[1,]x A ∈是数列1,1+d ,1+2d (0)d > 的“保三角形函数”,必须满足三个条件:

①1,1+d ,1+2d (0)d >是三角形数列,所以1112d d ++>+,即01d <<. ②数列中的各项必须在定义域内,即12+≤d A . ③(1),(1),(12)++h h d h d 是三角形数列.

由于2

()2h x x x =-+,[1,]x A ∈是单调递减函数,所以(1)(12)(1)h d h d h +++>,解得

505

d <<

兴泰高补中心数学限时作业(5) 2010.9 2

1.设i 为虚数单位,则复数

_________1=-i

i

.i 2121+-

2.函数3cos 6sin 2)(2++=x x x f 的最大值为_______.9

3.不等式1|2|≤-x 的解集是 . [1,3]

4.在ABC ?中,0

60=∠A ,,5=AB 且35=?S ,则BC 的长为 . 21 5.在等差数列{a n }中,满足3a 4=7a 7,且a 1>0,S n 是数列{a n }前n 项的和,若S n 取得最大值,则

n = . 9

6.已知双曲线

)0(122

2

2>=-b b y x 的左、右焦点分别为21,F F ,其一条渐近线方程为x y =,点),3(0y P 在该双曲线上,则________

21=?PF PF .0 7.棱长为a 的正方体1111ABCD A B C D -的8个顶点都在球O 的表面上,E 、F 分别是棱1AA 、

1DD 的中点,则直线EF 被球O 截得的线段长是__________.a 2

8.在平面直角坐标系中,定义点()11,y x P 、()22,y x Q 之间的“直角距离”为

.),(2121y y x x Q P d -+-=若()y x C ,到点()3,1A 、()9,6B 的“直角距离”相等,其中实

数x 、y 满足100≤≤x 、100≤≤y ,则所有满足条件的点C 的轨迹的长度之和为 .

()

125+.

9.已知函数2()1

x

f x x -=

+; (1)证明:函数()f x 在(1,)-+∞上为减函数;

(2)是否存在负数0x ,使得00()3x

f x =成立,若存在求出0x ;若不存在,请说明理由. 答案:(1)任取12,(1,)x x ∈-+∞,且12x x < (1分)

∵1221

1212122233()()011(1)(1)

x x x x f x f x x x x x ----=

-=>++++ (4分) ∴函数()f x 在(1,)-+∞上为减函数 (1分) (2)不存在 (1分)

假设存在负数0x ,使得00()3x

f x =成立, (1分) 则000,031x

x <∴<< (1分) 即00()1f x <<

∴0

02011

x x -<

<+ (1分) 000000

12122110112x x x x x x -<

=>-+??<<->??+??或 01

22

x =>

<< (2分) 与00x <矛盾, (1分)

所以不存在负数0x ,使得00()3x

f x =成立。 (1分)

10.设数列{}n a 中,若)(,21*++∈+=N n a a a n n n ,则称数列{}n a 为“凸数列”.

(1)设数列{}n a 为“凸数列”,若2,121-==a a ,试写出该数列的前6项,并求出该6项

之和;

(2)在“凸数列”{}n a 中,求证:*+∈=N n a a n n ,6;

(3)设b a a a ==21,,若数列{}n a 为“凸数列”,求数列前n 项和n S . 答案:(1)2,121-==a a ,1,343-=-=a a ,3,265==a a ,

06=∴S 。 …………………………………………………………4分

(2)由条件得???+=+=+++++31221n n n n n n a a a a a a ,n n a a -=∴+3,………………………6分

n n n a a a =-=∴++36,即n n a a =+6。………………………………………8分

(3)b a a b a a a a b a b a a a -=-=-=-===654321,,,,,。

06=∴S 。 …………………………………………………………10分

由(2)得6,,1,,6 =∈=*

+k N n S S k k n 。………………………………12分

??????????

?∈+=-+=-+=+=++===∴*N k k n a

b k n a b k n b k n b

a k n a k n S n ,5

6462362261660………………………………………14分

2013届高三数学考点限时训练11

2013届高三数学考点大扫描限时训练011 1. 命题“x ?∈R ,20x ≥”的否定是 . 2. 若关于x 的不等式2260ax x a -+<的解集为(1, m ),则实数m = . 3. 已知()*3211 n a n n =∈-N ,数列{}n a 的前n 项和为n S ,则使0n S >的n 的最小值是 . 4. 某商品的单价为5000元,若一次性购买超过5件,但不超过10件时,每件优惠500元;若一次性购买超过10件,则每件优惠1000元. 某单位购买x 件(*,15x x ∈≤N ),设最低的购买费用是()f x 元,则()f x 的解析式是 . 5. 如图,A 、B 是单位圆O 上的动点,C 是圆与x 轴正半轴的交点,设CO A α∠=. (1)当点A 的坐标为()34,55时,求sin α的值; (2)若π02α≤≤,且当点A 、B 在圆上沿逆时针方向移动时,总有π3 AOB ∠=,试求BC 的取值范围. 6. 设实数x , y 同时满足条件:224936x y -=,且0xy <. (1)求函数()y f x =的解析式和定义域; (2)判断函数()y f x =的奇偶性,并证明.

参考答案: 1.2,0x x ?∈,所以33x x ><-或. ………………2分 因为0xy < ,所以3,() 3.x f x x <-=??>? ………………6分 函数()y f x =的定义域为()(),33,.-∞-+∞ ………………8分 (2)当3x <-时,3x ->,所以()f x - = =()f x =-. ………10分 同理,当3x >时,有()()f x f x -=-. ………………12分 综上,任意取()(),33,x ∈-∞-+∞ ,都有()()f x f x -=-,故()f x 是奇函数.…14分

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