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.

Write this as

$$\begin{pmatrix}
a && 1 \\
1 && -b
\end{pmatrix}\begin{pmatrix}x \\ 1\end{pmatrix}$$

Then this is self inverse iff
$$\begin{pmatrix}
a && 1 \\
1 && -b
\end{pmatrix}^2= \lambda I_2$$
i.e.
$$\begin{pmatrix}
a^2+1 && a-b \\
a-b && 1+b^2
\end{pmatrix}= \lambda I_2$$
So $a=b$.
The transformation is projective so all we need now is to check $a,b$ so that your conditions follow.
$$12f(4)=4f(12)\iff 3(4a+1)/(4-a)=(12a+1)/(12-a)$$
$$\iff 3(4a+1)(12-a)=(12a+1)(4-a)$$
$$\iff 143a+36=47a+4$$
$$\iff 96a=-32\iff a=-1/3.$$
Finally
$$\displaystyle f\left({1+b\over1-a}\right)=f(1/2)=1$$
$$\ne -1/52=f(4)/4.$$
So no such transformation exists.
