Question

$F:R-\{-\frac{3}{2}\}\to R-\left\{\frac{3}{2}\right\},F\left(x\right)=\frac{3x+2}{2x+3}$. So find $F^{-1}$.

• A. $\frac{2-3x}{2x-3}$
• B. $\frac{2+3x}{2x-3}$
• C. $\frac{2-3x}{2x+3}$
• D. $\frac{2+3x}{2x+3}$

Answer 1

The correct option is A $\frac{2-3x}{2x-3}$

Now Inverse of Function $F$ defined by $F^{-1}:R-\left\{\frac{3}{2}\right\}\to R-\left\{\frac{-3}{2}\right\}$

let $f\left(x\right)=y$

$y=\frac{3x+2}{2x+3}$

$y(2x+3)=3x+2$

$2xy+3y-3x-2=0$

$x(2y-3)+3y-2=0$

$x=\frac{2-3y}{2y-3}$

so, $f^{-1}\left(y\right)=x$

i.e., $F^{-1}\left(x\right)=\frac{2-3x}{2x-3}$

now,

$FO{F}^{-}=F\left(\frac{2-3x}{2x-3}\right)$

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

$=\frac{6-9\times +4x-6}{4-6x+6x-9}$

$=\frac{-5x}{-5}$

$=x$

Now, $FO{F}^{-1}\left(x\right)$ gives $x$. Hence it proof that $F^{-1}$ is inverse of $F$