(a)(i) Composition \( (f \circ g)((x, y)) \):

\[ (f \circ g)((x, y)) = f(g((x, y))) = f((xy, x + y)) \quad (M1) \]

\[ f((xy, x + y)) = (xy + (x + y), xy – (x + y)) = (xy + x + y, xy – x – y) \quad (A1)(A1) \]

[3 marks]

(a)(ii) Composition \( (g \circ f)((x, y)) \):

\[ (g \circ f)((x, y)) = g(f((x, y))) = g((x + y, x – y)) \quad (M1) \]

\[ g((x + y, x – y)) = ((x + y)(x – y), (x + y) + (x – y)) = (x^2 – y^2, 2x) \quad (A1) \]

[2 marks]

(b) Commutativity of \( f \) and \( g \):

No, because \( f \circ g \neq g \circ f \quad (R1) \)

Note: Accept a counterexample, e.g., evaluating \( (f \circ g)((1, 1)) \neq (g \circ f)((1, 1)) \).

[1 mark]

(c) Inverse of \( f \):

\[ f((x, y)) = (a, b) \implies (x + y, x – y) = (a, b) \quad (M1) \]

Solve the system:

\[ \begin{cases} x + y = a \\ x – y = b \end{cases} \quad (M1) \]

\[ x = \frac{a + b}{2}, \quad y = \frac{a – b}{2} \quad (A1) \]

Thus, the inverse is:

\[ f^{-1}((x, y)) = \left( \frac{x + y}{2}, \frac{x – y}{2} \right) \quad (A1) \]

[3 marks]