Question

Let $f:R-\left\{\frac{3}{5}\right\}\to R$ be defined by $f\left(x\right)=\frac{3x+2}{5x-3}$. Then,

• A. $f^{-1}\left(x\right)=f\left(x\right)$
• B. $f^{-1}\left(x\right)=-f\left(x\right)$
• C. $f\circ f\left(x\right)=-x$
• D. $f^{-1}\left(x\right)=\frac{1}{19}f\left(x\right)$

Answer 1

Answer: (A)

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

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

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

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

$\Rightarrow$ $f\circ f\left(x\right)=f\left(f\right(x\left)\right)$

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

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

$=\frac{\left(\frac{9x+6}{5x-3}\right)+2}{\left(\frac{15x+10}{5x-3}\right)-3}$

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

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

$=x$

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

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

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

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

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

$\Rightarrow$ $f^{-1}\left(y\right)=\frac{3y+2}{5y-3}$ [ From ( 1 ) ]

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