A self inverse function $f(x)=\frac{ax+1}{x-b}$ 
If  $$f(x)=\frac{ax+1}{x-b} \forall x \in\mathbb{R}-b,ab\neq1,a\neq1$$ is a self inverse function such that $$\frac{f(4)}{4}=\frac{f(12)}{12}={f\left(\frac{1+b}{1-a}\right)}$$The question is to find out $a$ and $b$

For a self inverse function $f(f(x))=x$.So I tried to put $f(x)$ in place of $x$ and solve the resulting equation but it didnot helped me .Is there a more logical way to solve this problem in limited time?Any ideas?Thanks.


Is there a more logical way to solve this problem in limited time?

Hint: if you take the premise to mean that there do in fact exist $a,b$ such as $f=f^{-1}$ then, for a shortcut, you can simply plug in $x=0$ and determine that $f\big(f(0)\big)=f(\frac{-1}{b})=0 \implies a=b\,$. Then use the second condition to determine $a\,$ or, as is the case, prove that no solutions exist.
