Problem Statement
Let $f(x) = \frac{3x + k}{2x - 3}$ for $x \neq \frac{3}{2}$, where $k$ is a constant.
**(a)** [4 marks] Find an expression for $f^{-1}(x)$ in terms of $x$ and $k$.
**(b)** [2 marks] Show that $f(x)$ is a self-inverse function for all values of $k$.
Verified Solution & Marking Scheme
Part (a) [M1]
Let $y = \frac{3x + k}{2x - 3}$. Swap variables: $x = \frac{3y + k}{2y - 3}$
Part (a) [M1]
Multiply by denominator: $x(2y - 3) = 3y + k \implies 2xy - 3x = 3y + k$
Part (a) [M1]
Isolate $y$: $2xy - 3y = 3x + k \implies y(2x - 3) = 3x + k$
Part (a) [A1]
$f^{-1}(x) = \frac{3x + k}{2x - 3}$
Part (b) [R1]
Since $f^{-1}(x) = \frac{3x + k}{2x - 3} = f(x)$ for all $x \neq \frac{3}{2}$ and for all $k$.
Part (b) [A1]
Hence, $(f \circ f)(x) = x$, proving $f$ is self-inverse.