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 (     )

    A.      B.      C.            D.

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一.選擇題

1―5  CBABA   6―10  CADDA

二.填空題

11.       12.()       13.2          14.         15.

16.(1,4)

三.解答題

數(shù)學(xué)理數(shù)學(xué)理17,解:①         =2(1,0)                      (2分)             

        ?,                                        (4分)

?

        cos              =

 

        由,  ,    即B=              (6分)

                                               (7分)

                                                        (9分)

                                                        (11分)

的取值范圍是(,1                                                      (13分)

18.解:①設(shè)雙曲線方程為:  ()                                 (1分)

由橢圓,求得兩焦點(diǎn),                                           (3分)

,又為一條漸近線

, 解得:                                                     (5分)

                                                    (6分)

②設(shè),則                                                      (7分)

      

?                             (9分)

,  ?              (10分)

                                                (11分)

  ?

?                                        (13分)

    <s id="kmtfb"></s>

    <strike id="kmtfb"><source id="kmtfb"><optgroup id="kmtfb"></optgroup></source></strike>

        <s id="kmtfb"></s>

        單減區(qū)間為[]        (6分)

       

      ②(i)當(dāng)                                                      (8分)

      (ii)當(dāng),

      ,  (),

      則有                                                                     (10分)

                                                     (11分)

        在(0,1]上單調(diào)遞減                     (12分)

                                                       (13分)

      20.解:①       

                                                              (2分)

      從而數(shù)列{}是首項(xiàng)為1,公差為C的等差數(shù)列

        即                                (4分)

       

         即………………※              (6分)

      當(dāng)n=1時(shí),由※得:c<0                                                    (7分)

      當(dāng)n=2時(shí),由※得:                                                 (8分)

      當(dāng)n=3時(shí),由※得:                                                 (9分)

      當(dāng)

          (

                                                (11分)

                               (12分)

      綜上分析可知,滿足條件的實(shí)數(shù)c不存在.                                    (13分)

      21.解:①設(shè)過(guò)A作拋物線的切線斜率為K,則切線方程:

                                                                      (2分)

          即

                                                                                                         (3分)

      ②設(shè)   又

           

                                                               (4分)

      同理可得 

                                                      (5分)

      又兩切點(diǎn)交于  ,

                                     (6分)

      ③由  可得:

       

                                                      (8分)

                        (9分)

       

      當(dāng) 

      當(dāng) 

                                                           (11分)

      當(dāng)且僅當(dāng),取 “=”,此時(shí)

                                             (12分)

      22.①證明:由,    

        即證

        ()                                    (1分)

      當(dāng)  

            即:                          (3分)

        ()    

      當(dāng)   

         

                                                               (6分)

      ②由      

      數(shù)列

                                                    (8分)

      由①可知, 

                          (10分)

      由錯(cuò)位相減法得:                                       (11分)

                                          (12分)

       

       


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