Browse Questions

# Let $f:R-\{\frac{3}{5}\}\rightarrow R$ be defined by $f(x)=\Large {\frac{3x+2}{5x-3}}$ .Then

\begin{array}{1 1}(A)\;f^{-1}(x)=f(x) & (B)\;f^{-1}(x)=-f(x)\\(C)\;(f\;o\;f) x=-x & (D)\;f^{-1}(x)=\frac{1}{19}f(x)\end{array}

Toolbox:
• A function g is the inverse function $f:R -\{\frac{3}{5}\} \to R$ if
• $gof=fog=I;g(x)=f^{-1}(x)$
$f:R -\{\frac{3}{5}\} \to R$

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

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

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

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

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

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

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

we define $g(y)=\frac{3y+2}{5y-3}$ then

$fog(y)=f \bigg(\frac{3y+2}{5y-3}\bigg)$

$=\frac{3\bigg(\large\frac{3y+2}{5y-3}\bigg)+2}{5 \bigg(\frac{3y+2}{5y-3}\bigg)-3}=y$

$=\large\frac{9y+6+10y-6}{15y+10-15y-9}=\frac{y}{1}=y$

Hence $g=f^{-1}$

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

'A' option is correct