Question

已知定义在R上的函数f(x)=(-2的x次方+1)/(2的x次方+a)是奇函数

已知定义在R上的函数f(x)=(-2的x次方+1)/(2的x次方+a)是奇函数 (1)求a的值 (2)若对任意的t属于R，不等式f(t²-2t)+f(2t²-k)＜0恒成立，求k的取值范围

Answer 1

定义在R上的函数f(x)=(-2^x+1)/(2^x+a)是奇函数，则f(-x)=-f(x)
(-2^(-x)+1)/(2^(-x)+a) = -(-2^x+1)/(2^x+a)
-(2^(-x)+a) (-2^x+1) = (-2^(-x)+1)(2^x+a)
1+a*2^x-2^(-x)-a = -1-a*2^(-x)+2^x+a
(a-1)*2^x+(a-1)*2^(-x)+2(a-1) = 0
(a-1)*{2^x+2^(-x)+2 } = 0
2^x+2^(-x)+2 ＞0
∴a-1=0
∴a = 1

f(x) = (-2^x+1)/(2^x+1) = (-2^x-1+2)/(2^x+1) = -1 + 2/(2^x+1)
f(t²-2t)+f(2t²-k)＜0
- 1 + 2/[2^(t^2-2t)+1] - 1 + 2/[2^(2t^2-k)+1] ＜ 0
2/[2^(t^2-2t)+1] + 2/[2^(2t^2-k)+1] ＜ 2
1/[2^(t^2-2t)+1] + 1/[2^(2t^2-k)+1] ＜ 1
2^(t^2-2t)+1 ＞0， 2^(2t^2-k)+1 ＞0
∴两边同乘以[2^(t^2-2t)+1 ] [2^(2t^2-k)+1]：
2^(2t^2-k)+1 + 2^(t^2-2t)+1 ＜ [2^(t^2-2t)+1 ] [2^(2t^2-k)+1]
2^(2t^2-k) + 2^(t^2-2t) + 2 ＜ 2^(2t^2-k) * 2^(t^2-2t) + 2^(2t^2-k) + 2^(t^2-2t) + 1
1 ＜2^(2t^2-k) * 2^(t^2-2t)
2^(2t^2-k+t^2-2t)＞1
2^(3t^2-2t-k)＞1
3t^2-2t-k＞0
k＞3t^2-2t=3(t^2-2/3t)=3(t-1/3)^2-1/3
又：3(t-1/3)^2≥0，即3(t-1/3)^2-1/3≥-1/3
∴k＞-1/3