\(f^{-1} : x \to \frac{2x + 3}{x - 1}, x \neq 1\)
\(f^{-1} : x \to \frac{x + 3}{x + 2}, x \neq -2\)
\(f^{-1} : x \to \frac{x - 1}{2x + 3}, x \neq -\frac{3}{2}\)
\(f^{-1}: x \to \frac{x - 2}{x + 3}, x \neq -3\)
Correct answer is A
\(f(x) = \frac{x + 3}{x - 2}\)
\(f(y) = \frac{y + 3}{y - 2}\)
Let f(y) = x,
\(x = \frac{y + 3}{y - 2}\)
\(x(y - 2) = y + 3\)
\(xy - y = 2x + 3 \implies y(x - 1) = 2x + 3\)
\(y = \frac{2x + 3}{x - 1}\)