(i) To express \(2x^2 - 12x + 11\) in the form \(a(x + b)^2 + c\), complete the square:

\(2x^2 - 12x + 11 = 2(x^2 - 6x) + 11\)

\(= 2((x - 3)^2 - 9) + 11\)

\(= 2(x - 3)^2 - 18 + 11\)

\(= 2(x - 3)^2 - 7\)

(ii) The function \(f(x) = 2x^2 - 12x + 11\) is a quadratic function that opens upwards. It is one-one on the interval where it is either increasing or decreasing. The vertex of the parabola is at \(x = 3\). Therefore, the largest value of \(k\) for which \(f\) is one-one is 3.

(iii) To find \(f^{-1}(x)\), start with \(y = 2(x - 3)^2 - 7\).

\(y + 7 = 2(x - 3)^2\)

\(\frac{y + 7}{2} = (x - 3)^2\)

\(x - 3 = \pm \sqrt{\frac{y + 7}{2}}\)

Since \(x \leq 3\), choose the negative root:

\(x = 3 - \sqrt{\frac{y + 7}{2}}\)

Thus, \(f^{-1}(x) = 3 - \sqrt{\frac{1}{2}(x + 7)}\).

The domain of \(f^{-1}\) is \(x \geq -7\).

(iv) With \(k = 1\), \(x + 3 \leq 1\) implies \(x \leq -2\). Therefore, the largest \(p\) is -2.

The composite function \(fg(x) = f(g(x)) = f(x + 3)\).

\(f(x + 3) = 2(x + 3)^2 - 12(x + 3) + 11\)

\(= 2(x^2 + 6x + 9) - 12x - 36 + 11\)

\(= 2x^2 + 12x + 18 - 12x - 36 + 11\)

\(= 2x^2 - 7\)