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

\(2(x^2 + 6x) + 11 = 2((x + 3)^2 - 9) + 11 = 2(x + 3)^2 - 18 + 11 = 2(x + 3)^2 - 7\).

Thus, \(a = 3\) and \(b = -7\).

(b) Start with \(y = 2(x + 3)^2 - 7\). Solve for \(x\):

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

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

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

Since \(x \leq -4\), choose the negative root:

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

The domain of \(f^{-1}\) is \(x > -5\) or \([-5, \infty)\).

(c) Given \(k = -1\), solve \(fg(x) = 193\):

\(fg(x) = 2((2x - 3) + 3)^2 - 7 = 8x^2 - 7\)

\(8x^2 - 7 = 193\)

\(8x^2 = 200\)

\(x^2 = 25\)

\(x = -5\) only (since \(x \leq -4\)).

(d) The largest value of \(k\) for \(fg\) to be defined is \(\frac{1}{2}\).