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(2)若當(dāng)時(shí).恒有.求實(shí)數(shù)a的取值范圍. 查看更多

 

題目列表(包括答案和解析)

已知函數(shù)(a∈R).

(Ⅰ)當(dāng)時(shí),求的極值;

(Ⅱ)當(dāng)時(shí),求單調(diào)區(qū)間;

(Ⅲ)若對(duì)任意,恒有

成立,求實(shí)數(shù)m的取值范圍.

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已知函數(shù)(a∈R).

(Ⅰ)當(dāng)時(shí),求的極值;

(Ⅱ)當(dāng)時(shí),求單調(diào)區(qū)間;

(Ⅲ)若對(duì)任意,恒有

成立,求實(shí)數(shù)m的取值范圍.

查看答案和解析>>

已知函數(shù)(a∈R).

(1)當(dāng)時(shí),求的極值;

(2)當(dāng)時(shí),求單調(diào)區(qū)間;

(3)若對(duì)任意,恒有

成立,求實(shí)數(shù)m的取值范圍.

 

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已知函數(shù)(a、b∈R),
(Ⅰ)若f(x)在R上存在最大值與最小值,且其最大值與最小值的和為2680,試求a 和b的值;
(Ⅱ)若f(x)為奇函數(shù):(1)是否存在實(shí)數(shù)b,使得f(x)在為增函數(shù),為減函數(shù),若存在,求出b的值,若不存在,請(qǐng)說(shuō)明理由;
(2)如果當(dāng)x≥0時(shí),都有f(x)≤0恒成立,試求b的取值范圍。

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已知函數(shù)(a∈R).
(Ⅰ)當(dāng)時(shí),求的極值;
(Ⅱ)當(dāng)時(shí),求單調(diào)區(qū)間;
(Ⅲ)若對(duì)任意,恒有
成立,求實(shí)數(shù)m的取值范圍.

查看答案和解析>>

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)

          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(mǎn)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)