Question:
Mark (√) against the correct answer in the following:
Let $\mathrm{f}: \mathrm{R}-\left\{\frac{-4}{3}\right\} \rightarrow-\left\{\frac{4}{3}\right\}: \mathrm{f}(\mathrm{x})=\frac{4 \mathrm{x}}{(3 \mathrm{x}+4)}$. Then $\mathrm{f}^{-1}(\mathrm{y})=$ ?
A. $\frac{4 y}{(4-3 y)}$
B. $\frac{4 y}{(4 y+3)}$
C. $\frac{4 y}{(3 y-4)}$
D. None of these
Solution:
$f(x)=\frac{4 x}{3 x+4}$
$\Rightarrow y=\frac{4 x}{3 x+4}$
$x \Longleftrightarrow y$
$\Rightarrow \mathrm{x}=\frac{4 y}{3 y+4}$
$\Rightarrow 3 y x+4 x=4 y$
$\Rightarrow y(3 x-4)=-4 x$
$\Rightarrow y=\frac{4 x}{4-3 x}$
$x \Longleftrightarrow y$
$\Rightarrow \mathrm{X}=\frac{4 y}{4-3 y}$