(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 = \(\pi \times 29 - 34.52 + 10 = 66.6\)