摘要:11.椭圆短轴长是2.长轴长是短轴长的2倍.则椭圆中心到其准线距离为 .

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一.选择题

1―5  CBABA   6―10  CADDA

二.填空题

11.       12.()       13.2          14.         15.

16.(1,4)

三.解答题

数学理数学理17,解:①         =2(1,0)                      (2分)             

        ?,                                        (4分)

?

        cos              =

 

        由,  ,    即B=              (6分)

②                                               (7分)

                                                        (9分)

又,                                                         (11分)

故的取值范围是(,1                                                      (13分)

18.解:①设双曲线方程为:  ()                                 (1分)

由椭圆,求得两焦点,                                           (3分)

,又为一条渐近线

, 解得:                                                     (5分)

                                                    (6分)

②设,则                                                      (7分)

      

?                             (9分)

又,  ?              (10分)

又                                                (11分)

  ?

?                                        (13分)

  单减区间为[]        (6分)

 

②(i)当                                                      (8分)

(ii)当,

令,  (),,

则有                                                                     (10分)

又,

由                                               (11分)

又  在(0,1]上单调递减                     (12分)

                                                 (13分)

20.解:①       

即                                                        (2分)

从而数列{}是首项为1,公差为C的等差数列

  即                                (4分)

② 

   即………………※              (6分)

当n=1时,由※得:c<0                                                    (7分)

当n=2时,由※得:                                                 (8分)

当n=3时,由※得:                                                 (9分)

当

令    ()

                                          (11分)

则

又                          (12分)

综上分析可知,满足条件的实数c不存在.                                    (13分)

21.解:①设过A作抛物线的切线斜率为K,则切线方程:

由                                                                 (2分)

    即

                                                                                                   (3分)

②设   又

     

又                                                         (4分)

同理可得 

又                                                (5分)

又两切点交于  ,

                               (6分)

③由  可得:

 

又

                                                (8分)

                  (9分)

令 

则

当 

当 

                                                     (11分)

当且仅当,取 “=”,此时

                                       (12分)

22.①证明:由,   故 

  即证

令  ()                                    (1分)

当  

      即:                          (3分)

又

令  ()    

当   

   

                                                         (6分)

②由      

数列

                                              (8分)

由①可知, 

                    (10分)

令

由错位相减法得:                                       (11分)

                                    (12分)

 

 

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