回覆 3# kobelian 的帖子
第 2 題
設\(f(x)\)為定義在區間\((0,\infty)\)上的嚴格遞增函數,且對任意的\(x>0\),\(\displaystyle f(x)\cdot f(\frac{1}{x}+f(x))=1\)均成立,試求\(f(1)=\)?
[解答]
f(x) * f(1/x + f(x)) = 1
f(1) * f(1 + f(1)) = 1
令 f(1) = k
k * f(1 + k) = 1
f(1 + k) = 1/k
f(1 + k) * f(1/(1 + k) + f(1 + k)) = 1
f(1/(1 + k) + 1/k) = k = f(1)
1/(1 + k) + 1/k = 1
k = (1 + √5)/2 或 (1 - √5)/2
當 k = (1 + √5)/2,f(1 + k) = 2/(1 + √5) < 1 < k = f(1),不合