If $f\left( x \right) = \dfrac{{\left[ {ax + b} \right]}}{{\left[ {cx + d} \right]}}$, then $fof(x) = x$ provided that
1.$d = - a$
2. $d = a$
3. $a = b = c = d = 1$
4. $a = b = 1$

Hint: The process of combining two or more functions into a single function is known as composition of function. \[f\left( {g\left( x \right)} \right)\]or $fog(x)$ represents the composition of functions $f\left( x \right)$ and $g\left( x \right)$, where $g\left( x \right)$ acts first. We always simplify anything within brackets first using BODMAS.

Formula used:
Composite function –
$fof = f\left( {f\left( x \right)} \right)$

Complete step by step solution:
Given that,
$f\left( x \right) = \dfrac{{\left[ {ax + b} \right]}}{{\left[ {cx + d} \right]}}$
And $fof(x) = x$
$ \Rightarrow f\left( {f\left( x \right)} \right) = x$
$ \Rightarrow \dfrac{{\left[ {a\left( {\dfrac{{\left[ {ax + b} \right]}}{{\left[ {cx + d} \right]}}} \right) + b} \right]}}{{\left[ {c\left( {\dfrac{{\left[ {ax + b} \right]}}{{\left[ {cx + d} \right]}}} \right) + d} \right]}} = x$
$ \Rightarrow \dfrac{{a\left( {ax + b} \right) + b\left( {cx + d} \right)}}{{c\left( {ax + b} \right) + d\left( {cx + d} \right)}} = x$
$ \Rightarrow a\left( {ax + b} \right) + b\left( {cx + d} \right) = x\left( {c\left( {ax + b} \right) + d\left( {cx + d} \right)} \right)$
$ \Rightarrow {a^2}x + ab + bcx + bd = ac{x^2} + bcx + dc{x^2} + {d^2}x$
$ \Rightarrow {a^2}x + ab + bcx + bd - ac{x^2} - bcx - dc{x^2} - {d^2}x = 0$
$ \Rightarrow (ac + dc){x^2} + ({d^2} - {a^2})x - ab - bd = 0$
$ \Rightarrow (a + d)c = 0,{d^2} - {a^2} = 0,(a + d)b = 0$
$ \Rightarrow a + d = 0$
Therefore, $d = - a$
Hence, Option (1) is the correct answer i.e., $d = - a$.

Note: The key concept involved in solving this problem is the good knowledge of composition of function. Students must remember that the output of one function inside the parenthesis becomes the input of the outside function in a function composition.