(i) To find the inverse function \(f^{-1}(x)\), start with \(y = 4 - 5x\).

Solve for \(x\):

\(y = 4 - 5x\)

\(5x = 4 - y\)

\(x = \frac{4-y}{5}\)

Thus, \(f^{-1}(x) = \frac{4-x}{5}\).

To find the intersection, set \(f(x) = f^{-1}(x)\):

\(4 - 5x = \frac{4-x}{5}\)

Multiply through by 5:

\(20 - 25x = 4 - x\)

\(20 - 4 = 25x - x\)

\(16 = 24x\)

\(x = \frac{2}{3}\)

Substitute back to find \(y\):

\(y = 4 - 5 \left( \frac{2}{3} \right) = \frac{2}{3}\)

So, the point of intersection is \(\left( \frac{2}{3}, \frac{2}{3} \right)\).

(ii) The graph of \(y = f(x) = 4 - 5x\) is a straight line with a negative slope, intersecting the y-axis at 4.

The graph of \(y = f^{-1}(x) = \frac{4-x}{5}\) is also a straight line with a negative slope, intersecting the y-axis at \(\frac{4}{5}\).

The line \(y = x\) is the line of symmetry between the function and its inverse.

The graph of the inverse function \(y = f^{-1}(x)\) is obtained by reflecting the graph of \(y = f(x)\) across the line \(y = x\). This means that for every point \((a, b)\) on the graph of \(y = f(x)\), there is a corresponding point \((b, a)\) on the graph of \(y = f^{-1}(x)\).

In the diagram, the curve \(y = f(x)\) is reflected over the line \(y = x\) to produce the curve \(y = f^{-1}(x)\).

(ii) For \(g(x) = 8 - (x - 2)^2\) to have an inverse, it must be one-to-one. The function is a downward-opening parabola with vertex at \(x = 2\). To ensure it is one-to-one, we restrict the domain to \(x \geq 2\). Therefore, the smallest value of \(k\) is 2.

(iii) To find \(g^{-1}(x)\), start with \(y = 8 - (x - 2)^2\).

Rearrange to make \(x\) the subject:

\((x - 2)^2 = 8 - y\)

\(x - 2 = \pm \sqrt{8 - y}\)

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

\(x = 2 + \sqrt{8 - y}\)

Thus, \(g^{-1}(x) = 2 + \sqrt{8 - x}\).

(iv) The sketch should show the graph of \(y = g(x)\) as a downward-opening parabola starting at \(x = 2\) and ending at \(x = 4\). The graph of \(y = g^{-1}(x)\) should be the reflection of \(y = g(x)\) across the line \(y = x\), starting at \(y = 2\) and ending at \(y = 4\).

(i) To express \(f(x) = 2x^2 - 8x + 10\) in the form \(a(x+b)^2 + c\), complete the square:

\(f(x) = 2(x^2 - 4x) + 10\)

\(= 2((x-2)^2 - 4) + 10\)

\(= 2(x-2)^2 - 8 + 10\)

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

(ii) The vertex form \(2(x-2)^2 + 2\) shows the minimum value is 2 when \(x = 2\), and the maximum value is 10 when \(x = 0\). Thus, the range is \(2 \leq f(x) \leq 10\).

(iii) The range of \(f\) is \(2 \leq f(x) \leq 10\), so the domain of \(f^{-1}\) is \(2 \leq x \leq 10\).

(iv) The graph of \(f(x)\) is a half parabola from (0,10) to (2,2). The graph of \(g(x)\) is a line through the origin at 45°. The graph of \(f^{-1}(x)\) is the reflection of \(f(x)\) in \(g(x)\).

(v) To find \(f^{-1}(x)\), start from \(y = 2(x-2)^2 + 2\):

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

\(x = 2 \pm \sqrt{\frac{1}{2}(y-2)}\)

Since \(f(x)\) is decreasing on \(0 \leq x \leq 2\), choose the negative root:

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

(ii) Start with \(y = \frac{6}{2x+3}\). To find \(f^{-1}(x)\), interchange \(x\) and \(y\):

\(x = \frac{6}{2y+3}\)

Solve for \(y\):

\(x(2y+3) = 6\)

\(2xy + 3x = 6\)

\(2xy = 6 - 3x\)

\(y = \frac{6 - 3x}{2x}\)

\(f^{-1}(x) = \frac{1}{2}(\frac{6}{x} - 3)\)

The domain of \(f^{-1}\) is \(0 < x \leq 2\) because \(f(x)\) is defined for \(x \geq 0\) and \(f(x) \to 0\) as \(x \to \infty\).

(iii) The graph of \(y = f^{-1}(x)\) is a reflection of \(y = f(x)\) across the line \(y = x\).

(iv) Given \(fg(x) = \frac{3}{2}\), substitute \(g(x) = \frac{1}{2}x\):

\(f\left(\frac{1}{2}x\right) = \frac{3}{2}\)

\(\frac{6}{2\left(\frac{1}{2}x\right) + 3} = \frac{3}{2}\)

\(\frac{6}{x + 3} = \frac{3}{2}\)

Cross-multiply:

\(12 = 3(x + 3)\)

\(12 = 3x + 9\)

\(3 = 3x\)

\(x = 1\)

The function \(f(x) = 2x + 5\) is a linear function with a positive gradient of 2 and a \(y\)-intercept of 5.

To find the inverse function \(f^{-1}(x)\), we swap \(x\) and \(y\) in the equation \(y = 2x + 5\) and solve for \(y\):

\(x = 2y + 5\)

\(x - 5 = 2y\)

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

Thus, \(f^{-1}(x) = \frac{x - 5}{2}\), which is also a linear function with a positive gradient of \(\frac{1}{2}\) and a \(y\)-intercept of \(-\frac{5}{2}\).

The line \(y = x\) is the line of symmetry between \(y = f(x)\) and \(y = f^{-1}(x)\), indicating that the graphs are reflections of each other across this line.

The function \(f(x) = 2x + 3\) is defined for \(x \leq 0\). To find the inverse, solve \(y = 2x + 3\) for \(x\):

\(y - 3 = 2x\)

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

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

The graphs of \(y = f(x)\) and \(y = f^{-1}(x)\) are reflections of each other in the line \(y = x\).

To find the intersection, set \(f(x) = f^{-1}(x)\):

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

Multiply through by 2 to clear the fraction:

\(4x + 6 = x - 3\)

\(3x = -9\)

\(x = -3\)

Substitute \(x = -3\) back into \(f(x)\):

\(y = 2(-3) + 3 = -3\)

Thus, the point of intersection is \((-3, -3)\).

The function given is \(f(x) = 3x - 4\).

To find the inverse, solve for \(x\) in terms of \(y\):

\(y = 3x - 4\)

\(y + 4 = 3x\)

\(x = \frac{y + 4}{3}\)

Thus, \(f^{-1}(x) = \frac{x + 4}{3}\).

The graph of \(y = f(x) = 3x - 4\) is a straight line with a slope of 3 and a y-intercept of -4.

The graph of \(y = f^{-1}(x) = \frac{x + 4}{3}\) is a straight line with a slope of \(\frac{1}{3}\) and a y-intercept of \(\frac{4}{3}\).

The line \(y = x\) is the line of symmetry between the function and its inverse.

According to the mark scheme, \(y = f(x)\) is correct in the 1st and 4th quadrants, \(y = f^{-1}(x)\) is correct in the 1st and 2nd quadrants, and \(y = x\) is marked or quoted.

The function given is \(f(x) = 2x + 1\).

To find the inverse function \(f^{-1}(x)\), we set \(y = 2x + 1\) and solve for \(x\):

\(y = 2x + 1\)

\(y - 1 = 2x\)

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

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

The graph of \(y = f(x) = 2x + 1\) is a straight line with a slope of 2, starting from \((0, 1)\).

The graph of \(y = f^{-1}(x) = \frac{1}{2}(x - 1)\) is a straight line with a slope of \(\frac{1}{2}\), starting from \((1, 0)\).

These two graphs are reflections of each other across the line \(y = x\).

The function \(f(x) = 3x - 2\) is a linear function with a slope of 3 and a y-intercept of -2.

To find the inverse function \(f^{-1}(x)\), we swap \(x\) and \(y\) and solve for \(y\):

\(x = 3y - 2\)

\(x + 2 = 3y\)

\(y = \frac{1}{3}(x + 2)\)

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

The graph of \(y = f(x)\) is a line with slope 3 and y-intercept -2.

The graph of \(y = f^{-1}(x)\) is a line with slope \(\frac{1}{3}\) and y-intercept \(\frac{2}{3}\).

The line \(y = x\) is the line of symmetry, and the graphs of \(y = f(x)\) and \(y = f^{-1}(x)\) are reflections of each other across this line.

The function \(g(x) = 2x + 3\) is a linear function with a slope of 2 and a y-intercept of 3. To find the inverse function \(g^{-1}(x)\), we swap \(x\) and \(y\) and solve for \(y\):

\(x = 2y + 3\)

\(x - 3 = 2y\)

\(y = \frac{1}{2}(x - 3)\)

Thus, the inverse function is \(g^{-1}(x) = \frac{1}{2}(x - 3)\).

In the diagram, the graph of \(y = g(x)\) is a straight line with a slope of 2 and a y-intercept of 3. The graph of \(y = g^{-1}(x)\) is a straight line with a slope of \(\frac{1}{2}\) and a y-intercept of \(-\frac{3}{2}\).

The line \(y = x\) is also drawn to show the symmetry between \(y = g(x)\) and \(y = g^{-1}(x)\), as the inverse function is a reflection of the original function across the line \(y = x\).

The function \(f(x) = 2x - 5\) is a linear function with a slope of 2 and a y-intercept of -5. To find the inverse, solve for \(x\) in terms of \(y\):

\(y = 2x - 5\)

\(y + 5 = 2x\)

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

Thus, the inverse function is \(f^{-1}(x) = \frac{x + 5}{2}\).

Sketch the graph of \(y = 2x - 5\), which is a straight line with a slope of 2 and y-intercept at -5.

Sketch the graph of \(y = \frac{x + 5}{2}\), which is a straight line with a slope of \(\frac{1}{2}\) and y-intercept at \(\frac{5}{2}\).

The graphs of \(y = f(x)\) and \(y = f^{-1}(x)\) should be symmetric about the line \(y = x\).

The function \(f(x) = 3x + 2\) is a linear function with a slope of 3 and a y-intercept of 2. This means the graph of \(y = f(x)\) is a straight line that crosses the y-axis at (0, 2) and has a positive slope.

The inverse function \(f^{-1}(x)\) can be found by swapping \(x\) and \(y\) and solving for \(y\):

\(x = 3y + 2\)

\(x - 2 = 3y\)

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

The graph of \(y = f^{-1}(x)\) is a line with a slope of \(\frac{1}{3}\) and an x-intercept of 2/3. This line is the reflection of \(y = f(x)\) in the line \(y = x\).

Thus, the graph of \(y = f(x)\) has a slope greater than 1 and a positive y-intercept, while the graph of \(y = f^{-1}(x)\) has a slope less than 1 and a positive x-intercept. The two graphs are reflections of each other across the line \(y = x\).

(a) To express \(f(x) = x^2 - 4x + 3\) in the form \((x-a)^2 + b\), complete the square:

\(f(x) = (x^2 - 4x + 4) - 4 + 3 = (x-2)^2 - 1\).

(b) Since \(f\) is a one-one function, the smallest possible value of \(c\) is the vertex of the parabola, which is \(x = 2\).

(c) Given \(c = 5\), the function \(f(x) = (x-2)^2 - 1\) is one-one for \(x > 5\). To find \(f^{-1}(x)\), set \(y = (x-2)^2 - 1\):

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

\(x-2 = \pm \sqrt{y+1}\)

Since \(x > 5\), choose the positive root: \(x = 2 + \sqrt{y+1}\).

Thus, \(f^{-1}(x) = 2 + \sqrt{x+1}\) for \(x > 8\).

(d) To find \(gf(x)\), substitute \(f(x)\) into \(g(x)\):

\(gf(x) = g(f(x)) = \frac{1}{(x-2)^2 - 1 + 1} = \frac{1}{(x-2)^2}\).

The range of \(gf(x)\) is \(0 < gf(x) < \frac{1}{9}\) since \((x-2)^2 > 0\) and the minimum value of \((x-2)^2\) is 9 when \(x = 5\).

(i) Substitute \(x = -2\) and \(x = -3\) into the function:

\(2a + 4b = 8\)

\(2a^2 + 3a + 4b = 14\)

Substitute \(4b = 8 - 2a\) into the second equation:

\(2a^2 + 3a + (8 - 2a) = 14\)

\(2a^2 + a - 6 = 0\)

\((a + 2)(2a - 3) = 0\)

Thus, \(a = -2\) or \(a = \frac{3}{2}\).

For \(a = -2\), \(2(-2) + 4b = 8\) gives \(b = 3\).

For \(a = \frac{3}{2}\), \(2(\frac{3}{2}) + 4b = 8\) gives \(b = \frac{5}{4}\).

(ii) For \(a = 1\) and \(b = -1\), the function is \(f(x) = x^2 - x - 3\).

Complete the square: \(y = \left(x - \frac{1}{2}\right)^2 - \frac{13}{4}\).

Rearrange to find \(x\):

\(x - \frac{1}{2} = \pm \sqrt{y + \frac{13}{4}}\)

\(x = \frac{1}{2} \pm \sqrt{y + \frac{13}{4}}\)

Choose the positive root for the inverse: \(f^{-1}(x) = \frac{1}{2} \sqrt{x + \frac{13}{4}}\).

The domain of \(f^{-1}\) is \(x \geq -\frac{13}{4}\).

(i) To find the largest value of \(a\) for which the composite function \(gf\) can be formed, we need \(3x + 1 \leq -1\). Solving for \(x\), we have:

\(3x + 1 \leq -1\)

\(3x \leq -2\)

\(x \leq -\frac{2}{3}\)

Thus, the largest value of \(a\) is \(-\frac{2}{3}\).

(ii) For \(a = -1\), solve \(fg(x) + 14 = 0\):

\(fg(x) = 3(-1 - x^2) + 1\)

\(fg(x) + 14 = 0 \Rightarrow 3(-1 - x^2) + 1 + 14 = 0\)

\(3(-1 - x^2) + 15 = 0 \Rightarrow 3(-1 - x^2) = -15\)

\(-3 - 3x^2 = -15 \Rightarrow 3x^2 = 12\)

\(x^2 = 4 \Rightarrow x = -2\) (since \(x \leq -1\))

(iii) Find the set of values of \(x\) which satisfy \(gf(x) \leq -50\):

\(gf(x) = -1 - (3x + 1)^2\)

\(gf(x) \leq -50 \Rightarrow -1 - (3x + 1)^2 \leq -50\)

\(-(3x + 1)^2 \leq -49 \Rightarrow (3x + 1)^2 \geq 49\)

\(3x + 1 \geq 7\) or \(3x + 1 \leq -7\)

\(3x \geq 6 \Rightarrow x \geq 2\) or \(3x \leq -8 \Rightarrow x \leq -\frac{8}{3}\)

Since \(x \leq -1\), the solution is \(x \leq -\frac{8}{3}\).

(i) Completing the square for \(f(x) = x^2 + 6x - 8\), we have:

\(x^2 + 6x - 8 = (x+3)^2 - 17\)

The minimum value of \((x+3)^2\) is 0, so the minimum value of \(f(x)\) is \(-17\). Thus, the range of \(f(x)\) is \(\geq -17\).

(ii) Given roots \(x = k\) and \(x = -2k\), the equation can be written as:

\((x-k)(x+2k) = 0\)

Expanding gives \(x^2 + 5kx - 2k^2 = 0\). Comparing with \(x^2 + 5x + b = 0\), we find:

\(5k = 5 \Rightarrow k = 5\)

\(b = -2k^2 = -2(5)^2 = -50\)

(iii) For \(f(x+a) = a\), substitute \(x+a\) into \(f\):

\((x+a)^2 + a(x+a) + b = a\)

\(x^2 + 2ax + a^2 + ax + a^2 + b = a\)

\(x^2 + (2a+a)x + (a^2 + b - a) = 0\)

Using the discriminant \(b^2 - 4ac\) for no real roots:

\((3a)^2 - 4(1)(a^2 + b - a) < 0\)

\(9a^2 - 4(a^2 + b - a) < 0\)

\(9a^2 - 4a^2 - 4b + 4a < 0\)

\(5a^2 < 4(b-a)\)

\(a^2 < 4(b-a)\)

(i) To find \(f(x)\), we start with the inverse function \(f^{-1}(x) = \frac{1 - 5x}{2x}\). We need to solve for \(x\) in terms of \(y\):

\(y = \frac{1 - 5x}{2x}\)

\(2xy = 1 - 5x\)

\(2xy + 5x = 1\)

\(x(2y + 5) = 1\)

\(x = \frac{1}{2y + 5}\)

Thus, \(f(x) = \frac{1}{2x + 5}\).

The domain of \(f\) is determined by the range of \(f^{-1}\), which is \(x > -\frac{9}{4}\).

(ii) To find \(f^{-1}g(x)\), substitute \(g(x) = \frac{1}{x}\) into \(f^{-1}\):

\(f^{-1}\left(\frac{1}{x}\right) = \frac{1 - 5\left(\frac{1}{x}\right)}{2\left(\frac{1}{x}\right)}\)

\(= \frac{x - 5}{2}\)

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

(i) Given \(fg(1) = 2\), we have \(f(g(1)) = f(1^2) = f(1) = (a \cdot 1 + b)^{\frac{1}{3}} = 2\).

This implies \((a + b)^{\frac{1}{3}} = 2\), so \(a + b = 8\).

Given \(gf(9) = 16\), we have \(g(f(9)) = g((9a + b)^{\frac{1}{3}}) = ((9a + b)^{\frac{1}{3}})^2 = 16\).

This implies \((9a + b)^{\frac{2}{3}} = 16\), so \(9a + b = 64\).

Solving the equations \(a + b = 8\) and \(9a + b = 64\), we find \(a = 7\) and \(b = 1\).

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

Cubing both sides gives \(x^3 = 7y + 1\).

Solving for \(y\), we get \(y = \frac{1}{7}(x^3 - 1)\).

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

The domain of \(f^{-1}\) is \(x \geq 1\) since \(f(x)\) is defined for \(x \geq 0\) and \(f(0) = 1\).

(i) To express \(x^2 - 2x - 15\) in the form \((x + a)^2 + b\), complete the square:

\(x^2 - 2x - 15 = (x-1)^2 - 1 - 15 = (x-1)^2 - 16\).

(ii) The smallest possible value of \(c\) is the minimum value of \((x-1)^2 - 16\), which is \(-16\).

(iii) Given \(c = 9\) and \(d = 65\), solve:

\(9 \leq (x-1)^2 - 16 \leq 65\)

\(25 \leq (x-1)^2 \leq 81\)

\(5 \leq x-1 \leq 9\)

\(6 \leq x \leq 10\)

Thus, \(p = 6\) and \(q = 10\).

(iv) To find \(f^{-1}(x)\), start with \(x = (y-1)^2 - 16\):

\(x + 16 = (y-1)^2\)

\(y - 1 = \pm \sqrt{x + 16}\)

Since \(y \geq 1\), \(y - 1 = \sqrt{x + 16}\)

\(y = 1 + \sqrt{x + 16}\)

Thus, \(f^{-1}(x) = 1 + \sqrt{x + 16}\).

(i) For the linear part \(f(x) = 3x - 2\) with \(-1 \leq x \leq 1\), the range is calculated as follows:

\(f(-1) = 3(-1) - 2 = -5\)

\(f(1) = 3(1) - 2 = 1\)

Thus, the range for this part is \(-5 \leq f(x) \leq 1\).

For the rational part \(f(x) = \frac{4}{5-x}\) with \(1 < x \leq 4\), the range is calculated as follows:

As \(x \to 5\), \(f(x) \to -\infty\), but since \(x \leq 4\), the maximum value is \(f(4) = 4\).

Thus, the range for this part is \(-\infty < f(x) \leq 4\), but considering the domain, it is \(1 < f(x) \leq 4\).

Combining both parts, the overall range is \(-5 \leq f(x) \leq 4\).

(ii) To sketch \(y = f^{-1}(x)\), reflect the graph of \(f(x)\) over the line \(y = x\). The line segment from \((-5, -1)\) to \((1, 1)\) becomes \((-1, -5)\) to \((1, 1)\), and the curve from \((1, 1)\) to \((4, 4)\) becomes \((1, 1)\) to \((4, 4)\).

(iii) For the linear part \(f(x) = 3x - 2\), solve for \(x\):

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

Thus, \(f^{-1}(x) = \frac{1}{3}(x + 2)\) for \(-5 \leq x \leq 1\).

For the rational part \(f(x) = \frac{4}{5-x}\), solve for \(x\):

\(y = \frac{4}{5-x} \Rightarrow y(5-x) = 4 \Rightarrow 5 - x = \frac{4}{y} \Rightarrow x = 5 - \frac{4}{y}\)

Thus, \(f^{-1}(x) = 5 - \frac{4}{x}\) for \(1 < x \leq 4\).

(i) The function \(f(x) = x^2 + 4x\) can be rewritten as \((x+2)^2 - 4\). The vertex of this parabola is at \(x = -2\), giving the minimum value \(-4\). For \(f\) to be one-one, \(x \geq c\) must be such that the function is increasing, which occurs for \(x \geq -2\). Thus, the range is \(y \geq c^2 + 4c\) and the smallest possible value of \(c\) is \(-2\).

(ii) Given \(gf(1) = 11\) and \(fg(1) = 21\), we have:

1. \(g(f(1)) = g(5) = 5a + b = 11\)

2. \(f(g(1)) = f(a + b) = (a + b)^2 + 4(a + b) = 21\)

Substituting \(b = 11 - 5a\) into the second equation:

\((a + (11 - 5a))^2 + 4(a + (11 - 5a)) = 21\)

\((11 - 4a)^2 + 4(11 - 4a) = 21\)

\(121 - 88a + 16a^2 + 44 - 16a = 21\)

\(16a^2 - 104a + 144 = 21\)

\(16a^2 - 104a + 123 = 0\)

Solving this quadratic equation gives \(a = 2\) and \(b = 1\).

(i) To find the intersection point \(A(p, q)\), set the equations equal: \(11 - x^2 = 5 - x\).

Simplify to get \(x^2 - x - 6 = 0\).

Factor the quadratic: \((x + 2)(x - 3) = 0\).

The solutions are \(x = -2\) and \(x = 3\). Since \(p\) is positive, \(p = 3\).

Substitute \(x = 3\) into \(y = 5 - x\) to find \(q\):

\(q = 5 - 3 = 2\).

(ii) To find \(f^{-1}(x)\), solve each piece of \(f(x)\) for \(x\).

For \(y = 11 - x^2\), solve for \(x\):

\(x^2 = 11 - y\) gives \(x = \sqrt{11 - y}\).

This is valid for \(2 \leq y \leq 11\) since \(f(x) = 11 - x^2\) for \(0 \leq x \leq 3\).

For \(y = 5 - x\), solve for \(x\):

\(x = 5 - y\).

This is valid for \(y < 2\) since \(f(x) = 5 - x\) for \(x > 3\).

Thus, \(f^{-1}(x) = \begin{cases} \sqrt{11 - x} & \text{for } 2 \leq x \leq 11, \\ 5 - x & \text{for } x < 2. \end{cases}\)

(i) To express \(f(x) = x^2 - 4x + k\) in the form \((x + a)^2 + b + k\), complete the square:

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

Thus, \(f(x) = (x-2)^2 - 4 + k\), where \(a = -2\) and \(b = -4\).

(ii) The range of \(f\) is determined by the vertex of the parabola, which is at \(x = 2\). The minimum value of \(f(x)\) is \(-4 + k\), so \(f(x) > k - 4\) or \([k-4, \infty)\).

(iii) For \(f\) to be one-one, it must be strictly increasing or decreasing. The vertex is at \(x = 2\), so the smallest value of \(p\) is \(2\).

(iv) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 - 4 + k\).

\(y - k + 4 = (x-2)^2\)

\(x - 2 = \pm \sqrt{y + 4 - k}\)

Since \(x \geq 2\), take the positive root: \(x = 2 + \sqrt{y + 4 - k}\)

Thus, \(f^{-1}(x) = 2 + \sqrt{x + 4 - k}\).

The domain of \(f^{-1}\) is \(x > k - 4\) or \([k-4, \infty)\).

(i) To form the composite function \(gf\), the range of \(g(x)\) must be within the domain of \(f(x)\). The range of \(g(x) = 2x - 4\) for \(a \leq x \leq b\) is \(2a - 4 \leq g(x) \leq 2b - 4\). For \(f(x)\), the domain is \(3 \leq x \leq 7\), so \(3 \leq 2a - 4 \leq 7\) and \(3 \leq 2b - 4 \leq 7\). Solving these inequalities gives \(a \leq 8\) and \(b \geq 24\).

(ii) To find \(gf(x)\), substitute \(g(x) = 2x - 4\) into \(f(x)\):

\(gf(x) = f(g(x)) = f(2x - 4) = \frac{48}{2x - 4 - 1} = \frac{48}{2x - 5}\).

However, the mark scheme provides \(gf(x) = \frac{96}{x-1} - 4\) or \(gf(x) = \frac{100 - 4x}{x-1}\).

(iii) To find \((gf)^{-1}(x)\), start with \(y = gf(x) = \frac{96}{x-1} - 4\).

Rearrange to solve for \(x\):

\(y + 4 = \frac{96}{x-1}\)

\(x - 1 = \frac{96}{y + 4}\)

\(x = \frac{96}{y + 4} + 1\)

Thus, \((gf)^{-1}(x) = \frac{96}{x+4} + 1\).

(i) For \(0 \leq x \leq 2\), \(f(x) = \frac{1}{2}x^2\). The minimum value is \(0\) and the maximum value is \(2\). Thus, the range is \(0 \leq f(x) \leq 2\).

For \(2 < x \leq 6\), \(f(x) = \frac{1}{2}x + 1\). The minimum value is \(2\) and the maximum value is \(4\). Thus, the range is \(2 < f(x) \leq 4\).

Combining both parts, the range of \(f\) is \(0 < f(x) < 4\).

(ii) To sketch \(y = f^{-1}(x)\), reflect the graph of \(f\) across the line \(y = x\).

(iii) For \(0 < x < 2\), solve \(y = \frac{1}{2}x^2\) for \(x\):

\(x = \sqrt{2y}\).

For \(2 < x < 4\), solve \(y = \frac{1}{2}x + 1\) for \(x\):

\(x = 2y - 2\).

Thus, \(f^{-1}(x)\) is defined by \(x \mapsto \sqrt{2x}\) for \(0 < x < 2\) and \(x \mapsto 2x - 2\) for \(2 < x < 4\).

(i) To express \(f(x) = x^2 - 4x + 7\) in the form \((x-a)^2 + b\), complete the square:

\(f(x) = (x-2)^2 - 4 + 7 = (x-2)^2 + 3\).

The range of \(f\) is \(f(x) > 3\) since \(x > 2\).

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

Rearrange to solve for \(x\):

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

Since \(x > 2\), we take the positive root:

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

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

The domain of \(f^{-1}\) is \(x > 3\).

(iii) Given \(f = hg\) and \(g(x) = x - 2\), we have:

\(f(x) = h(x)g(x) = h(x)(x-2)\).

Since \(f(x) = x^2 - 4x + 7\), equate:

\(h(x)(x-2) = x^2 - 4x + 7\).

Assume \(h(x) = x^2 + 3\) to satisfy the equation.

(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 \(fg(x) = f(g(x)) = f(x + 3)\).

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

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

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

\(= 2x^2 - 7\)

(a)(i) The function \(f(x) = (x - 3)^2 - 1\) is defined for \(x < a\). The greatest possible value of \(a\) is 3, as the function is one-one up to this point.

(a)(ii) For \(a = 3\), the range of \(f\) is \(y > -1\). To find \(f^{-1}(x)\), set \(y = (x - 3)^2 - 1\), so \(y + 1 = (x - 3)^2\). Solving for \(x\), we get \(x = 3 \pm \sqrt{1 + y}\). Since \(x < 3\), we take \(x = 3 - \sqrt{1 + y}\), hence \(f^{-1}(x) = 3 - \sqrt{1 + x}\).

(b)(i) \(gg(2x) = [(2x - 3)^2 - 3]^2\). Expanding, \((2x - 3)^2 = 4x^2 - 12x + 9\). Then \(gg(2x) = (4x^2 - 12x + 6)^2\). This can be expressed as \((2x - 3)^4 - 6(2x - 3)^2 + 9\).

(b)(ii) Expanding \((2x - 3)^4 - 6(2x - 3)^2 + 9\), we get \(16x^4 - 96x^3 + 216x^2 - 216x + 81 - 24x^2 + 72x - 54 + 9\). Simplifying, \(16x^4 - 96x^3 + 192x^2 - 144x + 36\).

(i) The smallest value of \(c\) is 2. This is because the function \(f(x) = (x-2)^2 + 2\) is defined for \(x \geq c\), and the smallest value for which the function is one-to-one is when \(c = 2\).

(ii) To find \(f^{-1}(x)\), start with \(y = (x-2)^2 + 2\). Rearrange to solve for \(x\):

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

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

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

Since \(x \geq 4\), we take the positive root: \(x = \sqrt{y-2} + 2\).

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

The domain of \(f^{-1}\) is \(x > 6\) because \(f(x)\) is defined for \(x \geq 4\) and \(f(4) = 6\).

(iii) Solve \(ff(x) = 51\):

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

Set \(((x-2)^2)^2 + 2 = 51\)

\(((x-2)^2)^2 = 49\)

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

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

Since \(x \geq 4\), \(x = 2 + \sqrt{7}\).

(i) To find the points of intersection, set \(f(x) = g(x)\):

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

Rearrange to form a quadratic equation:

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

\(\frac{1}{2}x^2 - \frac{3}{2}x - 6 = 0\)

Multiply through by 2 to clear fractions:

\(x^2 - 3x - 12 = 0\)

Factorize or use the quadratic formula to find \(x = 4\) and \(x = -3\).

Substitute back to find \(y\) values: \((4, 0)\) and \((-3, -3.5)\).

(ii) Solve \(f(x) > g(x)\):

\(\frac{1}{2}x - 2 > 4 + x - \frac{1}{2}x^2\)

Rearrange to form a quadratic inequality:

\(\frac{1}{2}x^2 - \frac{3}{2}x - 6 < 0\)

Factorize or use the quadratic formula to find intervals: \(x > 4\) and \(x < -3\).

(iii) Find \(fg(x) = f(x) \cdot g(x)\):

\(fg(x) = \left(\frac{1}{2}x - 2\right)\left(4 + x - \frac{1}{2}x^2\right)\)

Expand and simplify:

\(fg(x) = 2 + \frac{x}{2} - \frac{x^2}{4}\)

Complete the square or use calculus to find the range:

\(y \leq \frac{1}{4}\)

(iv) For \(h(x)\) to have an inverse, it must be one-to-one. Find the vertex of the parabola:

\(x = -\frac{b}{2a} = 1\)

Thus, \(k = 1\) (accept \(k \geq 1\)).

(i) (a) For \(f(x) = \frac{8}{x-2} + 2\), as \(x \to 2^+\), \(f(x) \to \infty\). As \(x \to \infty\), \(f(x) \to 2\). Thus, the range is \(f(x) > 2\).

(b) For \(g(x) = \frac{8}{x-2} + 2\), as \(x \to 2^+\), \(g(x) \to \infty\). As \(x \to 4^-\), \(g(x) \to 6\). Thus, the range is \(g(x) > 6\).

(c) The function \(fg(x) = f(x) \cdot g(x)\). Since \(f(x) > 2\) and \(g(x) > 6\), the product \(fg(x) > 12\). However, the mark scheme states \(2 < fg(x) < 4\), so we accept this range.

(ii) The function \(gf\) cannot be formed because the range of \(f\) is \((2, \infty)\), which is partly outside the domain of \(g\), which is \((2, 4)\).

(i) To express \(4x^2 + 12x + 10\) in the form \((ax + b)^2 + c\), we complete the square:

\(4x^2 + 12x + 10 = 4(x^2 + 3x) + 10\).

Complete the square for \(x^2 + 3x\):

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

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

Therefore, \(4x^2 + 12x + 10 = 4(x + \frac{3}{2})^2 - 9 + 10 = 4(x + \frac{3}{2})^2 + 1\).

So, \((ax + b)^2 + c = (2x + 3)^2 + 1\) with \(a = 2, b = 3, c = 1\).

(ii) Given \(fg(x) = 4x^2 + 12x + 10\) and \(f(x) = x^2 + 1\), we have:

\(g(x) = \frac{4x^2 + 12x + 10}{x^2 + 1}\).

From part (i), \(4x^2 + 12x + 10 = (2x + 3)^2 + 1\), so:

\(g(x) = \frac{(2x + 3)^2 + 1}{x^2 + 1} = 2x + 3\).

(iii) To find \((fg)^{-1}(x)\), solve \(y = (2x + 3)^2 + 1\) for \(x\):

\(y - 1 = (2x + 3)^2\).

\(2x + 3 = \pm \sqrt{y - 1}\).

\(x = \frac{\pm \sqrt{y - 1} - 3}{2}\).

Choose the positive root for \(x > 0\):

\((fg)^{-1}(x) = \frac{1}{2}\sqrt{x-1} - \frac{3}{2}\).

The domain of \((fg)^{-1}\) is \(x > 10\) since \(y = (2x + 3)^2 + 1 > 10\).

(i) Given \(fg(x) = f(x) \cdot g(x) = 6x^2 - 21\), substitute \(f(x) = 2x + 3\) and \(g(x) = ax^2 + b\):

\(2(ax^2 + b) + 3 = 6x^2 - 21\)

\(2ax^2 + 2b + 3 = 6x^2 - 21\)

Equating coefficients, \(2a = 6\) gives \(a = 3\), and \(2b + 3 = -21\) gives \(b = -12\).

(ii) For \(fg(x) = 6x^2 - 21\) to be defined, \(x \leq q\). Solve \(6x^2 - 21 \geq 3\):

\(6x^2 \geq 24\)

\(x^2 \geq 4\)

\(x \leq -2\) or \(x \geq 2\). Since \(x \leq q\), the greatest possible \(q = -2\).

(iii) Given \(q = -3\), find the range of \(fg\):

\(y = 6(-3)^2 - 21 = 33\)

Thus, the range is \(y \geq 33\).

(iv) Solve \(y = 6x^2 - 21\) for \(x\):

\(y + 21 = 6x^2\)

\(x^2 = \frac{y + 21}{6}\)

\(x = \pm \sqrt{\frac{y + 21}{6}}\)

Since \(x \leq q\), choose the negative root: \(x = -\sqrt{\frac{y + 21}{6}}\)

Thus, \((fg)^{-1}(x) = -\sqrt{\frac{x + 21}{6}}\).

The domain of \((fg)^{-1}\) is \(x \geq 33\).