Question

已知奇函數f(x)=有最大值,且f(1)＞,其中實數P＞0,Q是正整數.

(1)求f(x)的解析式;

(2)令a_n=，證明a_n+1＞a_n（n是正整數）.

Answer 1

(1)由奇函數f(-x)=-f(x)可得r=0.                                                                 ?

x＞0時，由f(x)==≤=                                         ①?

以及f(1)=＞                                                                                ②?

可得到2q²-5q+2＜0,＜q＜2，只有q=1=p，?

∴f(x)=.                                                                                                        ?

(2)a_n===n+,則由a_n+1-a_n=(n+1+)-(n+)=1-＞0(n是正整數),?

可得所求證結論.