2f(x) = \(\frac{2+x}{2-x}\) ......(1)
f(x) = \(\lambda f(\frac{8x}{4+x^2})\)
f(1) = \(\lambda f(\frac{8}{5})\) ......(2)
2f(1) = \(\frac31=3\) From (i)
⇒ f(1) = log23
2f(\(\frac85\)) = \(\frac{2+\frac85}{2-\frac85}\)
= \(\frac{18}2 =9\)
\(\therefore\) f(\(\frac85\)) = log29 = log232 = 2log23
\(\therefore\) 2\(\lambda\) log23 = log23 From (2)
f\(\lambda\) = \(\frac12\)