The function $f$ defined by $f\left(x\right)=\frac{ax+b}{cx+d}$, where a,b,c, and d are nonzero real numbers, has the properties$f\left(19\right)=19$ , $f\left(97\right)=97$ ,  and $f\left(f\right(x\left)\right)=x$ for all values except $\frac{-d}{c}$ .

Column I		Column II
1	$\frac{a}{c}$	A	0
2	$\text{a}+\text{d}$	B	58
3	$\raisebox{1ex}{$-d$}\!\left/ \!\raisebox{-1ex}{$c$}\right.$	C	1
4	$f\left(f\right(x\left)\right)=x$ for all values except	D	5

Given:
$f\left(x\right)=\frac{ax+b}{cx+d}$
We know:
$f\left(f\right(x\left)\right)=x$
This implies that f is its own inverse function except at  $x=\frac{-d}{c}$.
Substituting $f\left(x\right)$  into itself:
$f\left(f\left(x\right)\right)=f\left(\frac{ax+b}{cx+d}\right)=x$
Substituting the function definition:
$f\left(\frac{ax+b}{cx+d}\right)=\frac{a\left(\frac{ax+b}{cx+d}\right)+b}{c\left(\frac{ax+b}{cx+d}\right)+d}=x$
Simplify the expression:
$\begin{array}{l}\frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)}=x\\ \frac{a^{2}x+ab+bcx+bd}{acx+bc+dcx+d^2}=x\\ \frac{(a^2+bc)x+(ab+bd)}{(ac+dc)x+(bc+d^2)}=x\end{array}$
For this fraction to reduce to x, the coefficients of x and the constants must match:
$\begin{array}{l}{a}^2+bc=x(ac+dc)\\ ab+bd=bc+d^2\end{array}$
In order for these equations to hold for all x, we must have:
$\begin{array}{l}{a}^2+bc=ac+dc\\ ab+bd=bc+d^2\end{array}$
For the function to be an inverse of itself, certain conditions must be met:
$a=-1$
Given the conditions:
$\begin{array}{l}f\left(19\right)=19\\ f\left(97\right)=97\end{array}$
Substituting $x=19$  and $x=97$  into the function:
$\begin{array}{l}\frac{a\left(19\right)+b}{c\left(19\right)+d}=19\\ \frac{a\left(97\right)+b}{c\left(97\right)+d}=97\end{array}$
Setting up the equations:
$\begin{array}{l}19a+b=19(19c+d)\\ 97a+b=97(97c+d)\end{array}$
Solving these equations simultaneously:
$\begin{array}{l}19a+b=361c+19d\\ 97a+b=9409c+97d\end{array}$
Subtracting the first equation from the second to eliminate b:
$\begin{array}{l}97a+b-(19a+b)=9409c+97d-(361c+19d)\\ 78a=9048c+78d\\ a=116c+d\end{array}$
Since we know  $a=-d$, substituting  $d=-a$:
$\begin{array}{l}a=116c-a\\ 2a=116c\\ a=58c\end{array}$
So, the value that is not in the range of this function can be found by:
$\frac{a}{c}=58$
Hence, the unique number that is not in the range of $f$  is:      $58$