Question

求各位这道题怎样做啊.

谢谢啊啊

Answer 1

(1)
f(x)为奇函数，所以f(0)=0
将x=0代入，解得a=1

(2)
f(x)=1-2/(2^x+1)
f(x)在R上单调递增。
令x1＞x2
f(x1)-f(x2)=1-2/(2^x1+1)-[1-2/(2^x2+1)]
=2/(2^x2+1)-2/(2^x1+1)=2(1/(2^x2+1)-1/(2^x1+1))
2^x在R上递增，x1＞x2
所以，2^x1＞2^x2
2^x1+1＞2^x2+1
1/(2^x1+1)＜1/(2^x2+1)
所以，f(x1)-f(x2)＞0 x1＞x2
所以f(x)在R上单调递增

(3)
a=1，f(x)在R上单调递增
f(ax)＜f(x²-2)
x＜x²-2
x²-x-2＞0
(x-2)(x+1)＞0
x＜-1或x＞2