1. Apply a vertical stretch by a factor of 2. This transformation changes the graph of \(y = f(x)\) to \(y = 2f(x)\), effectively doubling the y-values of all points on the graph.

2. Translate the graph vertically by \(\begin{pmatrix} 0 \\ -14 \end{pmatrix}\). This moves the graph down by 14 units, resulting in the graph of \(y = g(x)\).

(a)(i) For an arithmetic progression, the difference between consecutive terms is constant. Thus, \(k^2 - 4k = 8k - k^2\). Solving gives:

\(2k^2 = 12k\)

\(k(k - 6) = 0\)

Since \(k\) is non-zero, \(k = 6\).

(a)(ii) The first term \(a = 4k = 24\) and the common difference \(d = k^2 - 4k = 12\). The sum of the first 20 terms is:

\(S_{20} = \frac{20}{2} (2 \times 24 + 19 \times 12) = 2760\)

(b) Let the first term be \(a\) and the common ratio be \(r\). Then \(ar^3 = 36\) and \(ar^5 = 6\). Dividing gives:

\(r^2 = \frac{1}{6}\)

\(r = \frac{\sqrt{6}}{6}\)

\(a = \frac{36}{r^3} = \frac{1296}{\sqrt{6}}\)

The sum to infinity is:

\(S_\infty = \frac{a}{1 - r} = \frac{1296}{\sqrt{6} - 1}\)

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

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

Thus, \(a = 2\) and \(b = -2\).

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

Rearrange to solve for \(x\):

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

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

Since \(x \leq -2\), choose the negative root:

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

(b)(ii) To find \((gf)^{-1}(x)\), first find \(gf(x)\):

\(gf(x) = g(f(x)) = g((x+2)^2 - 2) = -(x+2)^2 - 2 - 4 = -(x+2)^2 - 6\).

Now, solve \(y = -(x+2)^2 - 6\) for \(x\):

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

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

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

Since \(x \leq -2\), choose the negative root:

\((gf)^{-1}(x) = -\sqrt{-x-2} - 2\).

First, solve the linear equation \(2x + y + 4 = 0\) for \(y\):

\(y = -2x - 4\).

Substitute \(y = -2x - 4\) into the curve equation \(2xy + 5y^2 = 24\):

\(2x(-2x - 4) + 5(-2x - 4)^2 = 24\).

Simplify and expand:

\(-4x^2 - 8x + 5(4x^2 + 16x + 16) = 24\).

\(-4x^2 - 8x + 20x^2 + 80x + 80 = 24\).

Combine like terms:

\(16x^2 + 72x + 56 = 0\).

Divide the entire equation by 8:

\(2x^2 + 9x + 7 = 0\).

Factor the quadratic equation:

\((2x + 7)(x + 1) = 0\).

Set each factor to zero and solve for \(x\):

\(2x + 7 = 0 \Rightarrow x = -\frac{7}{2}\).

\(x + 1 = 0 \Rightarrow x = -1\).

Substitute \(x = -\frac{7}{2}\) into \(y = -2x - 4\):

\(y = -2\left(-\frac{7}{2}\right) - 4 = 3\).

Substitute \(x = -1\) into \(y = -2x - 4\):

\(y = -2(-1) - 4 = -2\).

Thus, the points of intersection are \((-1, -2)\) and \(\left(-\frac{7}{2}, 3\right)\).

To find the coefficient of \(x^7\) in the expansion of \(\left( px^2 + \frac{4}{p}x \right)^5\), we use the binomial theorem. The general term in the expansion is given by:

\(\binom{5}{r} (px^2)^{5-r} \left( \frac{4}{p}x \right)^r\)

We need the power of \(x\) to be 7, so:

\(2(5-r) + r = 7\)

\(10 - 2r + r = 7\)

\(10 - r = 7\)

\(r = 3\)

Substitute \(r = 3\) into the general term:

\(\binom{5}{3} (px^2)^{5-3} \left( \frac{4}{p}x \right)^3\)

\(= \binom{5}{3} (px^2)^2 \left( \frac{4}{p}x \right)^3\)

\(= \binom{5}{3} p^2 x^4 \cdot \frac{64}{p^3} x^3\)

\(= \binom{5}{3} \cdot \frac{64p^2}{p^3} x^7\)

\(= 10 \cdot \frac{64}{p} x^7\)

The coefficient of \(x^7\) is \(\frac{640}{p}\), and we know this equals 1280:

\(\frac{640}{p} = 1280\)

\(640 = 1280p\)

\(p = \frac{640}{1280} = \frac{1}{2}\)

(a) Differentiate \(y = ax^{\frac{3}{2}} - 12x\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{3}{2}ax^{\frac{1}{2}} - 12\).

Using the chain rule, \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}\).

Substitute \(x = 9\) and \(\frac{dx}{dt} = 5\):

\(\frac{dy}{dt} = \left( \frac{3}{2}a \times 9^{\frac{1}{2}} - 12 \right) \times 5\).

\(\frac{dy}{dt} = 5 \left( \frac{9}{2}a - 12 \right) = \frac{45}{2}a - 60 = \frac{15}{2}(3a - 8)\).

(b) For a minimum point at \(x = \frac{1}{4}\), set \(\frac{dy}{dx} = 0\):

\(\frac{3}{2}a \left( \frac{1}{4} \right)^{\frac{1}{2}} - 12 = 0\).

\(\frac{3}{2}a \times \frac{1}{2} - 12 = 0\).

\(\frac{3}{4}a = 12\).

\(a = 16\).

(a) The function \(y = 4 \cos 2x + 3\) is based on the cosine function, which has a range of \([-1, 1]\). Therefore, the range of \(4 \cos 2x\) is \([-4, 4]\). Adding 3 shifts this range to \([-1, 7]\). Thus, the greatest value is 7 and the least value is -1.

(b) The graph of \(y = 4 \cos 2x + 3\) shows two complete cycles from \(0\) to \(2\pi\), starting at the maximum value of 7 and decreasing to the minimum value of -1, then repeating.

(c) To find the number of solutions to \(4 \cos 2x + 3 = 2x - 1\), we equate the expressions: \(4 \cos 2x + 3 = 2x - 1\). Rearranging gives \(4 \cos 2x = 2x - 4\). The number of intersections of the curve \(y = 4 \cos 2x + 3\) with the line \(y = 2x - 1\) within the interval \(0 \leq x \leq 2\pi\) is 3.

First, find the x-coordinate of the intersection point \(P\) by setting \(6 - 3x = 3\), which gives \(x = 1\).

Next, find the area under the curve from \(x = 0\) to \(x = 1\):

\(\int_0^1 \frac{9}{(5x+4)^{\frac{1}{2}}} \, dx = \left[ \frac{18}{5} (5x+4)^{\frac{1}{2}} \right]_0^1\)

Calculate the definite integral:

\(\frac{18}{5} \times 3 - \frac{18}{5} \times 2 = \frac{18}{5}\)

The line crosses the x-axis at \(x = 2\), forming a triangle with the x-axis and the line from \(x = 1\) to \(x = 2\). The area of this triangle is:

\(\frac{1}{2} \times 3 \times 1 = \frac{3}{2}\)

The total area of the shaded region is:

\(\frac{18}{5} + \frac{3}{2} = \frac{51}{10}\)

(a) Start with the left-hand side (LHS):

\(\frac{\tan \theta + 7}{\tan^2 \theta - 3}\)

Use \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\):

\(\frac{\frac{\sin \theta}{\cos \theta} + 7}{\frac{\sin^2 \theta}{\cos^2 \theta} - 3} = \frac{\sin \theta + 7 \cos \theta}{\sin^2 \theta - 3 \cos^2 \theta}\)

Use \(\sin^2 \theta + \cos^2 \theta = 1\) to rewrite the denominator:

\(\sin^2 \theta - 3 \cos^2 \theta = (1 - \cos^2 \theta) - 3 \cos^2 \theta = 1 - 4 \cos^2 \theta\)

Thus, the LHS becomes:

\(\frac{\sin \theta \cos \theta + 7 \cos^2 \theta}{1 - 4 \cos^2 \theta}\)

This matches the right-hand side (RHS), proving the identity.

(b) Using the identity from part (a), solve:

\(\frac{\tan \theta + 7}{\tan^2 \theta - 3} = \frac{5}{\tan \theta}\)

Cross-multiply to eliminate fractions:

\(\tan \theta (\tan \theta + 7) = 5 (\tan^2 \theta - 3)\)

\(\tan^2 \theta + 7 \tan \theta = 5 \tan^2 \theta - 15\)

Rearrange to form a quadratic equation:

\(4 \tan^2 \theta - 7 \tan \theta - 15 = 0\)

Factorize:

\((4 \tan \theta + 5)(\tan \theta - 3) = 0\)

Solutions for \(\tan \theta\):

\(\tan \theta = -1.25 \quad \text{and} \quad \tan \theta = 3\)

Find \(\theta\) in the range \(0^\circ \leq \theta \leq 180^\circ\):

\(\theta = 71.6^\circ\) and \(\theta = 128.7^\circ\).

(a) The equation of the circle is \((x - 7)^2 + (y + 4)^2 = 29\). The centre \(C\) is \((7, -4)\). The line \(y = -2\) intersects the circle, substituting \(y = -2\) into the circle's equation gives:

\((x - 7)^2 + (-2 + 4)^2 = 29\)

\((x - 7)^2 + 4 = 29\)

\((x - 7)^2 = 25\)

\(x - 7 = \pm 5\)

\(x = 12\) or \(x = 2\)

Thus, \(A = (2, -2)\) and \(B = (12, -2)\).

(b) Using the cosine rule in triangle \(ACB\):

\(\cos \theta = \frac{29 + 29 - 100}{2 \times \sqrt{29} \times \sqrt{29}} = \frac{-21}{29}\)

\(\theta = \cos^{-1}\left(\frac{-21}{29}\right)\)

\(\theta = 2.38\) radians (to 3 significant figures).

(c) The length of \(AC = BC = \sqrt{29}\). The area of the larger segment is:

Area of circle sector = \(\frac{1}{2} \times 29 \times (2\pi - 2.38) = 56.587\)

Area of triangle \(ACB = \frac{1}{2} \times \sqrt{29} \times \sqrt{29} \times \sin(2.38) = 10\)

Area of larger segment = \(\pi \times 29 - 34.52 + 10 = 66.6\)

(a) Integrate \(\frac{d^2y}{dx^2} = -\frac{24}{x^3}\) to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \int -\frac{24}{x^3} \, dx = \frac{12}{x^2} + c\).

Using the stationary point \((-2, 19)\), \(\frac{dy}{dx} = 0\) when \(x = -2\):

\(0 = \frac{12}{(-2)^2} + c \Rightarrow c = -3\).

Thus, \(\frac{dy}{dx} = \frac{12}{x^2} - 3\).

(b) Set \(\frac{dy}{dx} = 0\):

\(\frac{12}{x^2} - 3 = 0 \Rightarrow \frac{12}{x^2} = 3 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\).

Since \(x = -2\) is already given, the other point is \(x = 2\).

Determine the nature using \(\frac{d^2y}{dx^2} = -\frac{24}{x^3}\):

At \(x = 2\), \(\frac{d^2y}{dx^2} = -\frac{24}{8} = -3\), which is negative, indicating a maximum.

(c) Integrate \(\frac{dy}{dx} = \frac{12}{x^2} - 3\) to find \(y\):

\(y = \int \left( \frac{12}{x^2} - 3 \right) \, dx = -\frac{12}{x} - 3x + d\).

Using \((-2, 19)\):

\(19 = -\frac{12}{-2} - 3(-2) + d \Rightarrow 19 = 6 + 6 + d \Rightarrow d = 7\).

Thus, \(y = \frac{12}{x} - 3x + 7\).

(d) Find \(x\) where \(\frac{dy}{dx} = -\frac{9}{4}\):

\(\frac{12}{x^2} - 3 = -\frac{9}{4} \Rightarrow \frac{12}{x^2} = \frac{3}{4} \Rightarrow x^2 = 16 \Rightarrow x = 4\) (since \(x\) is positive).

Find \(y\) at \(x = 4\):

\(y = \frac{12}{4} - 3(4) + 7 = 3 - 12 + 7 = -2\).

The gradient of the normal is \(\frac{4}{9}\).

Equation of the normal: \(y + 2 = \frac{4}{9}(x - 4)\).

Rearrange to \(4x - 9y - 88 = 0\).