Mark the correct alternative in the following question:

Solution:

$f: \mathbf{R}-\left\{\frac{3}{5}\right\} \rightarrow \mathbf{R}$ is defined by $f(x)=\frac{3 x+2}{5 x-3}$

$f \circ f(x)=f(f(x))$

$=f\left(\frac{3 x+2}{5 x-3}\right)$

$=\frac{3\left(\frac{3 x+2}{5 x-3}\right)+2}{5\left(\frac{3 x+2}{5 x-3}\right)-3}$

$=\frac{\left(\frac{9 x+6}{5 x-3}\right)+2}{\left(\frac{5 x+10}{5 x-3}\right)-3}$

$=\frac{\left(\frac{9 x+6+10 x-6}{5 x-3}\right)}{\left(\frac{15 x+10-15 x+9}{5 x-3}\right)}$

$=\frac{19 x}{19}$

$=x$

Let $y=\frac{3 x+2}{5 x-3}$

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

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

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

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

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

So, $f^{-1}(x)=\frac{3 x+2}{5 x-3}=f(x)$

Hence, the correct alternative is option (a).