(a) To find \(ff(x)\), apply \(f\) twice:

\(f(x) = a - 2x\)

\(ff(x) = f(f(x)) = f(a - 2x) = a - 2(a - 2x)\)

\(= a - 2a + 4x = 4x - a\)

To find \(f^{-1}(x)\), solve \(y = a - 2x\) for \(x\):

\(y = a - 2x\)

\(2x = a - y\)

\(x = \frac{a-y}{2}\)

Thus, \(f^{-1}(x) = \frac{a-x}{2}\).

(b) Given \(ff(x) = f^{-1}(x)\):

\(4x - a = \frac{a-x}{2}\)

Multiply through by 2 to clear the fraction:

\(8x - 2a = a - x\)

\(8x + x = a + 2a\)

\(9x = 3a\)

\(x = \frac{a}{3}\)

Given \(a = \frac{7}{2}\), we need to solve \(gf(x) = 0\).

First, find \(f(x)\):

\(f(x) = \left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\)

For \(x \leq -\frac{7}{2}\), \(f(x) = \left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\).

Now, find \(gf(x)\):

\(gf(x) = g(f(x)) = 2f(x) - 1\)

Set \(gf(x) = 0\):

\(2f(x) - 1 = 0\)

\(2\left(\left(x + \frac{7}{2}\right)^2 - \frac{7}{2}\right) - 1 = 0\)

\(2\left(x + \frac{7}{2}\right)^2 - 7 - 1 = 0\)

\(2\left(x + \frac{7}{2}\right)^2 = 8\)

\(\left(x + \frac{7}{2}\right)^2 = 4\)

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

\(x = -\frac{7}{2} \pm 2\)

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

\(x = -\frac{7}{2} - 2 = -\frac{11}{2}\)

Since \(x \leq -\frac{7}{2}\), the valid solution is \(x = -\frac{11}{2}\).

(a) To find \(b\), use \(gg(2) = 10\):

\(g(x) = 3x + b\), so \(g(2) = 3(2) + b = 6 + b\).

Then \(gg(2) = g(g(2)) = g(6 + b) = 3(6 + b) + b = 18 + 3b + b = 18 + 4b\).

Set \(18 + 4b = 10\), solving gives \(b = -2\).

To find \(a\), use \(f^{-1}(2) = 14\):

\(f(x) = \frac{1}{2}x - a\), so \(f^{-1}(x) = 2(x + a)\).

Set \(2(x + a) = 14\), solving gives \(a = 5\).

(b) Find \(gf(x)\):

\(gf(x) = g(f(x)) = g\left(\frac{1}{2}x - 5\right)\).

\(g(x) = 3x + b\), so \(g\left(\frac{1}{2}x - 5\right) = 3\left(\frac{1}{2}x - 5\right) - 2\).

\(= \frac{3}{2}x - 15 - 2 = \frac{3}{2}x - 17\).

(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 \(\frac{1}{2}\).

(ii) Substitute \(k = -9\) into \(g(x)\):

\(g(x) = 2x + 9\)

Set \(f(x) < g(x)\):

\(2x^2 + 8x + 1 < 2x + 9\)

Rearrange to form a quadratic inequality:

\(2x^2 + 6x - 8 < 0\)

Factorize:

\((x + 4)(x - 1) < 0\)

Solution: \(-4 < x < 1\)

(iii) Find \(g^{-1}(x)\):

\(y = 2x + 1 \Rightarrow x = \frac{y - 1}{2}\)

So, \(g^{-1}(x) = \frac{x - 1}{2}\)

Substitute \(f(x)\) into \(g^{-1}(x)\):

\(g^{-1}f(x) = \frac{2x^2 + 8x + 1 - 1}{2} = \frac{2x^2 + 8x}{2} = x^2 + 4x\)

Solve \(g^{-1}f(x) = 0\):

\(x^2 + 4x = 0\)

\(x(x + 4) = 0\)

Solutions: \(x = 0, -4\)

(iv) Complete the square for \(f(x)\):

\(f(x) = 2(x^2 + 4x) + 1\)

\(= 2((x + 2)^2 - 4) + 1\)

\(= 2(x + 2)^2 - 8 + 1\)

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

Least value of \(f(x)\) is \(-7\) when \(x = -2\)