Question

If $f:R-\left\{\frac{3}{5}\right\}\to R-\left\{\frac{3}{5}\right\}$ and
$f\left(x\right)=\frac{3x+1}{5x-3}$, then

• A. $f^{-1}\left(x\right)=2f\left(x\right)$
• B. $f^{-1}\left(x\right)=f\left(x\right)$
• C. $f^{-1}\left(x\right)=-f\left(x\right)$
• D. $f^{-1}\left(x\right)$ does not exist.

Answer 1

The correct option is B $f^{-1}\left(x\right)=f\left(x\right)$

$f\left(x\right)=\frac{3x+1}{5x-3}$
$f^{\prime }\left(x\right)=\frac{-14}{(5x-3{)}^2}<0,\forall x\in D_f$
$\Rightarrow f$ is one-one.

Let $y=\frac{3x+1}{5x-3}$
$\Rightarrow x=\frac{3y+1}{5y-3}\cdots \left(1\right)$
$x$ is defined for all $y\in R-\left\{\frac{3}{5}\right\}$
$\Rightarrow f$ is onto.
$\therefore f$ is invertible.

$y=f\left(x\right)\Rightarrow x=f^{-1}\left(y\right)$
From $\left(1\right),$ we get
$f^{-1}\left(y\right)=\frac{3y+1}{5y-3}$
$\Rightarrow f^{-1}\left(x\right)=\frac{3x+1}{5x-3}$
$\Rightarrow f^{-1}\left(x\right)=f\left(x\right)$