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**Hint:**From the options we can understand that we are asked to find the relation between the function $f(x)$ and its inverse ${f^{ - 1}}(x)$. To extract $x$ from the given expression, introduce another variable say $y$ dependent of $x$, such that, $f(x) = y$. This gives $x = {f^{ = 1}}(y)$ and so we can calculate ${f^{ - 1}}(x)$.

**Formula Used:**

For two variables $x,y$ and a function $f$,

$y = f(x) \Rightarrow {f^{ - 1}}(y) = {f^{ - 1}}f(x) = x$

Also, for any two constants $a,b$,

$f(ax + b) = af(x) + b$

**Complete step by step answer:**

Given, \[f(x) = \dfrac{{3x + 2}}{{5x - 3}}\]

We need to find the relation between $f(x)$ and its inverse ${f^{ - 1}}(x)$.

Let $f(x) = y$

\[ \Rightarrow \dfrac{{3x + 2}}{{5x - 3}} = y\]

Using this we can solve for $x$.

Cross multiplying the above equation we get,

$ \Rightarrow 3x + 2 = y(5x - 3)$

Simplifying we get,

$ \Rightarrow 3x + 2 = 5xy - 3y$

To find $x$, collect the terms with $x$ to one side of the equation.

$ \Rightarrow 3x - 5xy = - 3y - 2$

Taking $x$ common on the left side we get,

$ \Rightarrow x(3 - 5y) = - 3y - 2$

Simplifying we get,

$ \Rightarrow x = \dfrac{{ - 3y - 2}}{{3 - 5y}}$

Multiplying the numerator and denominator of the RHS by $ - 1$ we have,

$ \Rightarrow x = \dfrac{{3y + 2}}{{5y - 3}}$

To find ${f^{ - 1}}(x)$, using above equation we can take ${f^{ - 1}}$ on both sides,

$ \Rightarrow {f^{ - 1}}(x) = {f^{ - 1}}(\dfrac{{3y + 2}}{{5y - 3}})$

Also for any two constants $a,b$, we have the result,

$f(ax + b) = af(x) + b$

Using this we get,

$ \Rightarrow {f^{ - 1}}(x) = \dfrac{{3{f^{ - 1}}(y) + 2}}{{5{f^{ - 1}}(y) - 3}}$

We can observe that in the RHS, $y = f(x) \Rightarrow {f^{ - 1}}(y) = {f^{ - 1}}f(x) = x$

$ \Rightarrow {f^{ - 1}}(x) = \dfrac{{3x + 2}}{{5x - 3}}$

But we have ${f^{}}(x) = \dfrac{{3x + 2}}{{5x - 3}}$

So ${f^{ - 1}}(x) = f(x)$

**$\therefore $ Option A is correct.**

**Note:**

In this question, the inverse of the given function is the function itself. But in general it is not true. A function need not have inverse as well. For the existence of inverse, the function must be bijective that is, injective (one to one) as well as surjective (onto). Also the inverse may be negative of the function or a constant multiple like the other options.

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