(i) To express \(f(x) = 2x^2 - 8x + 11\) in the form \(a(x + b)^2 + c\), complete the square:

\(f(x) = 2(x^2 - 4x) + 11\)

\(= 2((x - 2)^2 - 4) + 11\)

\(= 2(x - 2)^2 - 8 + 11\)

\(= 2(x - 2)^2 + 3\)

(ii) The range of \(f\) is \(\geq 3\) because the vertex of the parabola is at its minimum point \((2, 3)\).

(iii) \(f\) does not have an inverse because it is not one-to-one. It is a quadratic function, which means it is symmetric and has two x-values for each y-value.

(iv) The largest value of \(A\) for which \(g\) has an inverse is \(A = 2\), as this makes the function one-to-one on the interval \(x \leq 2\).

(v) To find \(g^{-1}(x)\), start with \(y = 2(x - 2)^2 + 3\) and solve for \(x\):

\(y - 3 = 2(x - 2)^2\)

\(\frac{y - 3}{2} = (x - 2)^2\)

\(x - 2 = \pm \sqrt{\frac{y - 3}{2}}\)

Since \(x \leq 2\), choose the negative root:

\(x = 2 - \sqrt{\frac{y - 3}{2}}\)

Thus, \(g^{-1}(x) = 2 - \sqrt{\frac{x - 3}{2}}\).

The range of \(g^{-1}\) is \(\leq 2\) because the inverse function reflects the domain of \(g\).