If \(A=\mathrm{R}-\left\{\displaystyle\frac{3}{2}\right\}\) and the function \(f: A \to A\) is defined by \(f(x)=\displaystyle\frac{3x-2}{2x-3}\), then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(f^{-1}(x)=f(x)\)
Let us find \((f\circ f)(x) = f(f(x))\):
\( \displaystyle f(f(x)) = \frac{3\left(\frac{3x-2}{2x-3}\right) - 2}{2\left(\frac{3x-2}{2x-3}\right) - 3} \)
Multiplying the numerator and the denominator by \(2x - 3\):
\( \displaystyle = \frac{3(3x-2) - 2(2x-3)}{2(3x-2) - 3(2x-3)} \)
\( \displaystyle = \frac{9x - 6 - 4x + 6}{6x - 4 - 6x + 9} = \frac{5x}{5} = x \)
Since \(f(f(x)) = x\) for every x in A, f undoes itself: f is its own inverse, \(f^{-1}(x) = f(x)\).
Why the other options are incorrect
Option (2) \(f^{-1}(x)=-f(x)\) — Incorrect. \(f^{-1} = f\), and f(x) is not identically 0, so \(f^{-1} \ne -f\).
Option (3) \((f\circ f)(x)=-x\) — Incorrect. \((f\circ f)(x) = x\).
Option (4) \(f^{-1}(x)=\frac{3x+2}{2x+3}\) — Incorrect. At x = 1 it gives \(\frac{5}{5} = 1\), while \(f^{-1}(1) = f(1) = \frac{1}{-1} = -1\).