(ii) To express \(f(x) = 2x^2 - 8x + 14\) in the form \(a(x+b)^2 + c\), complete the square:

\(2x^2 - 8x + 14 = 2(x^2 - 4x) + 14\)

\(= 2((x-2)^2 - 4) + 14\)

\(= 2(x-2)^2 - 8 + 14\)

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

Thus, \(a = 2, b = -2, c = 6\).

(iii) The vertex form \(2(x-2)^2 + 6\) shows the minimum value is 6, so the range is \(f \geq 6\).

(iv) For \(g(x) = 2x^2 - 8x + 14\) to have an inverse, it must be one-to-one. The function is one-to-one for \(x \geq 2\) (the vertex of the parabola).

(v) To find \(g^{-1}(x)\), solve \(y = 2x^2 - 8x + 14\) for \(x\):

\(y = 2(x-2)^2 + 6\)

\(y - 6 = 2(x-2)^2\)

\(\frac{y-6}{2} = (x-2)^2\)

\(x-2 = \pm \sqrt{\frac{y-6}{2}}\)

Since \(x \geq 2\), choose the positive root:

\(x = \sqrt{\frac{y-6}{2}} + 2\)

Thus, \(g^{-1}(x) = \sqrt{\frac{x-6}{2}} + 2\).

(i) To express \(f(x) = 2x^2 - 12x + 7\) in the form \(a(x-b)^2 - c\), we complete the square:

First, factor out the 2 from the quadratic terms: \(2(x^2 - 6x) + 7\).

Complete the square inside the parentheses: \(x^2 - 6x = (x-3)^2 - 9\).

Thus, \(2(x^2 - 6x) = 2((x-3)^2 - 9) = 2(x-3)^2 - 18\).

Adding the constant term: \(2(x-3)^2 - 18 + 7 = 2(x-3)^2 - 11\).

So, \(f(x) = 2(x-3)^2 - 11\).

(ii) The range of \(f\) is determined by the vertex form \(2(x-3)^2 - 11\). The minimum value occurs at \(x = 3\), giving \(f(3) = -11\). Therefore, the range is \(f \geq -11\).

(iii) To find the set of values of \(x\) for which \(f(x) < 21\), solve:

\(2x^2 - 12x + 7 < 21\).

Rearrange: \(2x^2 - 12x + 7 - 21 < 0\) or \(2x^2 - 12x - 14 < 0\).

Complete the square: \(2(x-3)^2 - 11 < 21\).

Rearrange: \(2(x-3)^2 < 32\).

Divide by 2: \((x-3)^2 < 16\).

Take the square root: \(-4 < x-3 < 4\).

Thus, \(-1 < x < 7\).

(i) To express \(f(x) = 2x^2 - 12x + 13\) in the form \(a(x+b)^2 + c\), complete the square:

\(f(x) = 2(x^2 - 6x) + 13\)

\(= 2((x-3)^2 - 9) + 13\)

\(= 2(x-3)^2 - 18 + 13\)

\(= 2(x-3)^2 - 5\)

Thus, \(a = 2\), \(b = -3\), \(c = -5\).

(ii) The line of symmetry is at \(x = 3\). Since \(0 \leq x \leq A\), \(A = 6\) for symmetry about \(x = 3\).

(iii) The vertex form \(2(x-3)^2 - 5\) shows the minimum value is \(-5\) at \(x = 3\). The maximum value is \(f(0) = 13\). Thus, the range is \(-5 \leq y \leq 13\).

(iv) The function \(g\) is defined for \(x \geq 4\), which is a one-to-one function because it is strictly increasing (as \(4 > 3\)).

(v) To find \(g^{-1}(x)\), start with \(y = 2x^2 - 12x + 13\) and solve for \(x\):

\(y = 2(x-3)^2 - 5\)

\(y + 5 = 2(x-3)^2\)

\(\frac{y+5}{2} = (x-3)^2\)

\(x-3 = \pm \sqrt{\frac{y+5}{2}}\)

Since \(x \geq 4\), take the positive root:

\(x = \sqrt{\frac{y+5}{2}} + 3\)

Thus, \(g^{-1}(x) = \sqrt{\frac{x+5}{2}} + 3\).

(iii) To express \(6x - x^2\) in the form \(a - (x-b)^2\), we complete the square:

\(6x - x^2 = -(x^2 - 6x)\)

\(= -(x^2 - 6x + 9 - 9)\)

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

Thus, \(6x - x^2 = 9 - (x-3)^2\).

(iv) To find \(h^{-1}(x)\), start with \(y = 6x - x^2\).

Rearrange to express \(x\) in terms of \(y\):

\(y = 9 - (x-3)^2\)

\((x-3)^2 = 9 - y\)

\(x-3 = \pm \sqrt{9-y}\)

Since \(x \geq 3\), we take the positive root:

\(x = 3 + \sqrt{9-y}\)

Thus, \(h^{-1}(x) = 3 + \sqrt{9-x}\).

(ii) To express \(f(x) = x^2 - 3x\) in the form \((x-a)^2 - b\), complete the square:

\(x^2 - 3x = (x - \frac{3}{2})^2 - \frac{9}{4}\).

Thus, \(a = \frac{3}{2}\) and \(b = \frac{9}{4}\).

(iii) The minimum value of \(f(x) = (x - \frac{3}{2})^2 - \frac{9}{4}\) is \(-\frac{9}{4}\), so the range is \(f(x) \geq -\frac{9}{4}\).

(iv) The function \(f(x) = x^2 - 3x\) is not one-to-one because it is a quadratic function, which is symmetric about its vertex. Therefore, \(f\) does not have an inverse.

(v) Solve \(g(x) = x - 3\sqrt{x} = 10\).

Let \(y = \sqrt{x}\), then \(x = y^2\).

Substitute into the equation: \(y^2 - 3y = 10\).

Rearrange to form a quadratic: \(y^2 - 3y - 10 = 0\).

Factorize: \((y - 5)(y + 2) = 0\).

Thus, \(y = 5\) or \(y = -2\).

Since \(y = \sqrt{x} \geq 0\), \(y = 5\).

Therefore, \(x = y^2 = 25\).