Consider the function f in $A = R - \{\frac{2}{3}\}$ defined as $f (x) = \frac{4 x + 3}{6 x - 4}, then f^{- 1}$ is equal to

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$\frac{6 x - 4}{3 + 4 x}$

$\frac{9 + 2 x}{6 x - 4}$

$\frac{3 + 4 x}{6 x - 4}$

$\frac{3 - 4 x}{6 x - 4}$

answer is A.

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Detailed Solution

Given $f (x) = \frac{4 x + 3}{6 x - 4}$

Let $y = \frac{4 x + 3}{6 x - 4}$

$\begin{array}{l} \Rightarrow 6 x y 4 y - 4 x + 3 \Rightarrow x (6 y 4) - 3 + 4 y \\ \Rightarrow x - \frac{3 + 4 y}{6 y - 4} \\ f^{- 1} (x) = \frac{3 + 4 x}{6 x - 4} \end{array}$

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