Mark the correct alternative in the following question:
Let $f: \mathrm{R}-\left\{\frac{3}{5}\right\} \rightarrow \mathrm{R}$ be defined by $f(x)=\frac{3 x+2}{5 x-3} .$ Then,
(a) $f^{-1}(x)=f(x)$
(b) $f^{-1}(x)=-f(x)$
(c) fof $f(x)=-x$
(d) $f^{-1}(x)=\frac{1}{19} f(x)$
$f: \mathbf{R}-\left\{\frac{3}{5}\right\} \rightarrow \mathbf{R}$ 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).