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2015.1 海淀区高二年级第一学期期末试卷及答案(文科)

2015.1 海淀区高二年级第一学期期末试卷及答案(文科)
2015.1 海淀区高二年级第一学期期末试卷及答案(文科)

海淀区高二年级第一学期期末练习

数学(文科)

2015.1

本试卷共100分.考试时间90分钟.

一、选择题:本题共8小题, 每题4分,共32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.直线2y x =+的倾斜角是( )

A.

π6 B.π4 C. 2π3 D.3π4

2. 焦点在x 轴上的椭圆22

13x y

m +=的离心率是12,则实数m 的值是( )

A.4

B.94

C. 1

D.3

4

3. 一个空间几何体的三视图如右图所示,该几何体的体积为( ) A. 8 B.

83 C. 163

D. 6 4. 已知圆22:1O x y +=,直线:3430l x y +-=,则直线l 被圆O 所截的弦长为( ) A.

65 B. 1 C.8

5

D.2 5. 命题“0k ?>,使得直线2y kx =-的图象经过第一象限”的否定是( ) A. 0k ?>,使得直线2y kx =-的图象不经过第一象限 B. 0k ?≤,使得直线2y kx =-的图象经过第一象限 C. 0k ?>,使得直线2y kx =-的图象不经过第一象限 D. 0k ?≤,使得直线2y kx =-的图象不经过第一象限

6.已知等差数列{}n a ,则“21a a >”是“数列{}n a 为单调递增数列”的( ) A. 充分而不必要条件 B.必要而不充分条件 C. 充分必要条件 D. 既不充分也不必要条件

7.已知正四面体A BCD -的棱长为2,点E 是AD 的中点,则下面四个命题中正确的是( ) A. F BC ?∈,EF AD ⊥ B. F BC ?∈,EF AC ⊥ C. F BC ?∈

,EF D. F BC ?∈,EF AC ∥

8.

已知曲线||1W y =, 则曲线W 上的点到原点距离的最小值是( ) A. 12

C.2

1

二、填空题:本大题共6小题, 每小题4分,共24分.把答案填在题中横线上. 9.已知直线10x ay --=与直线y ax =平行,则实数___.a =

10.双曲线22

1169x y -=的渐近线方程为________________.

11.椭圆22

12516

x y +=上一点P 到一个焦点的距离为4,则P 到另一个焦点的距离是_______.

12.已知椭圆22

221(0)x y C a b a b

+=>>:的左右焦点分别为12,F F ,若等边12P F F △的一个顶点P 在椭圆C 上,

则椭圆C 的离心率为______.

13. 已知平面αβ⊥,且l αβ= ,在l 上有两点,,A B 线段AC α?, 线段BD β?,, AC l ⊥, BD l ⊥ 4, 3, 12, AB AC BD === 则线段CD 的长为_____.

14. 已知点(1,0)A -, 抛物线24y x =的焦点为F ,点(,)P x y

在抛物线上,且|||AP PF =, 则||___.OP =

三、解答题:本大题共4小题,共44分. 解答应写出文字说明,证明过程或演算步骤. 15.(本小题共10分)

已知点(0,2)A , 圆22

:1O x y +=.

( I ) 求经过点A 且与圆O 相切的直线方程;

( II ) 若点P 是圆O 上的动点,求OA AP ?

的取值范围.

16.(本小题共12分)

已知直线:l y x t =+与椭圆22

:22C x y +=交于,A B 两点.

( I ) 求椭圆C 的长轴长和焦点坐标; ( II )

若||AB =,求t 的值.

17.(本小题共12分)

如图所示的几何体中,直线AF ⊥平面ABCD ,且ABCD 为正方形,ADEF 为梯形,DE AF ∥,又

1AB =,2=2AF DE a =.

( I ) 求证:直线C E ∥平面ABF ; ( II ) 求证:直线BD ⊥平面AC F (Ⅲ) 若直线AE CF ⊥,求a 的值.

18.(本小题共10分)

已知椭圆22

143

x y +=,经过点(0,3)A 的直线与椭圆交于,P Q 两点. ( I ) 若||||PO PA =,求点P 的坐标; ( II ) 若=OAP OPQ S S △△,求直线PQ 的方程.

海淀区高二年级第一学期期末练习数学(文科)

参考答案及评分标准2015.1

一. 选择题:本大题共8小题, 每小题4分,共32分.

二.填空题:本大题共6小题, 每小题4分,共24分.

9. 1或1- 10.34y x =

或34

y x =- 11. 6 12.1

2 13. 1

3 14.

说明:9,10题每个答案两分,丢掉一个减两分 三.解答题:本大题共4小题,共44分. 15. (本小题满分10分)

解:(I )由题意知道,所求直线的斜率存在,

设切线方程为2y kx =+,即20kx y -+=,-------------1分 所以圆心O 到直线的距离为

d =

,-------------3分

所以

1d =

=,解得k =, -------------4分

所求的直线方程为2y +或2y =+. -------------5分 (II )设点(,)P x y ,

所以 (0,2)OA = ,(,2)AP x y =-

,-------------6分 所以 2(2)OA AP y ?=-

.-------------7分

又因为22=1x y +,所以11y -≤≤, -------------9分

所以[6,2]OA AP ?∈--

. -------------10分

16.(本小题满分12分)

解:(I )因为2

2

22x y +=,所以2

212

x y +=,-------------1分

所以1a b ==,所以1c =, -------------3分

所以长轴为2a =焦点坐标分别为12(1,0),(1,0)F F -. -------------4分

(II )设点1122(,),(,)A x y B x y .

因为22220x y y x t ?+-=?=+?, 消元化简得2234+220x tx t +-=, -------------6分

所以222122121612(22)=24804+3223t t t t x x t x x ?

??=--->?

-?

=??

?-=

??

-------------8分

所以12||AB x x -=

-------------10分

又因为||=

3AB

,解得1t =±. -------------12分

17.(本小题满分12分)

解: (I )因为ABCD 为正方形,所以AB CD . -------------1分 又DE AF ∥,且,AB AF A CD DE D == .

所以平面ABF ∥平面DCE . -------------3分 而CE ?平面EDC , 所以CE ∥平面ABF . -------------4分 (II) 因为ABCD 为正方形,所以AC BD ⊥-------------5分 因为直线AF ⊥平面ABCD ,所以AF BD ⊥,-------------6分

因为AF AC A = ,所以直线BD ⊥平面ACF . -------------8分 (Ⅲ)连接FD .

因为直线AF ⊥平面ABCD ,所以AF CD ⊥, 又CD AD ⊥, AD AF A =

所以CD ⊥平面ADEF ,-------------9分 所以CD AE ⊥.

又AE CF ⊥, FC CD C = ,所以AE ⊥平面FCD ,所以AE FD ⊥. -------------11分 所以π2EAD FDA ∠+∠=,所以11

tan 1tan 2a EAD EAD a

∠===∠

解得a =

-------------12分

18.(本小题满分10分)

解:(I )设点11(,)P x y , 由题意||||PO PA =, 所以点P 在OA 的中垂线上,而OA 的中垂线为32y =

, 所以有13

2

y =.-------------2分 把其代入椭圆方程,求得11x =±.

所以3

(1,)2P 或3(1,)2

P -. -------------4分 (II) 设22(,)Q x y .根据题意,直线PQ 的斜率存在, 设直线PQ 的方程为3y kx =+,

所以22341203x y y kx ?+-=?=+?

.

消元得到22(34)24240k x kx +++=,

所以22122

122(24)96(34)024+342434k k k x x k x x k ?

??=-+>?

-?

=?+?

?

=?+?

-------------6分 因为=OAP OPQ S S △△,

所以=2OAQ OPQ S S △△, 即

1211

||||=2||||22

OA x OA x ?-------------7分 所以有12||=2||x x ,-------------8分 因为122

24

034x x k =

>+,

所以12x x ,同号,所以122x x =.

所以12122

122224342434x x k x x k x x k ?

?=?

-?

+=?+?

?

=?+?

,-------------9分 解方程组得到3

2

k =±

, 经检验,此时0?>, 所以直线PQ 的方程为332y x =+,或3

32

y x =-+. -------------10分

法二:设22(,)Q x y ,

因为=OAP OPQ S S △△,所以||||AP PQ =. -------------6分 即点P 为线段OQ 的中点,

所以2121=2, 23x x y y =-. -------------7分 把点,P Q 的坐标代入椭圆方程得到

22

1122

11143

(2)(23)14

3x y x y ?+=???-?+=??-------------8分 解方程组得到11132x y =???=??或者111

32

x y =-??

?=??,

即3

(1,)2P , 或者3(1,)2

P -. -------------9分

所以直线PQ 的斜率为32k =或者3

2k =-, 所以直线PQ 的方程为332y x =+,3

32

y x =-+. -------------10分

说明:解答题有其它正确解法的请酌情给分.

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