(i) Integrate by parts to reach \(ax^{\frac{3}{2}} \ln x + b \int x^{\frac{3}{2}} \frac{1}{x} \, dx\).

Obtain \(\frac{2}{3} x^{\frac{3}{2}} \ln x - \frac{2}{3} \int x^{\frac{1}{2}} \, dx\).

Obtain integral \(\frac{2}{3} x^{\frac{3}{2}} \ln x - \frac{4}{9} x^{\frac{3}{2}}\).

Substitute limits correctly and equate to 2 to obtain the given answer.

(ii) Evaluate the expression at \(x = 2\) and \(x = 4\) to show \(a\) lies between these values.

(iii) Use the iterative formula:

1. Start with an initial guess, say \(a_1 = 3\).

2. Calculate \(a_2 = \left( \frac{7 + 2 \times 3^{\frac{3}{2}}}{3 \ln 3} \right)^{\frac{2}{3}}\).

3. Continue iterations until the value stabilizes to 5 decimal places.

4. Final answer is \(a = 3.031\) to 3 decimal places.

(i) Differentiate \(y = x^2 \cos 2x\) using the product rule: \(y' = 2x \cos 2x - 2x^2 \sin 2x\). Set \(y' = 0\) to find the maximum point. Solving \(2x \cos 2x - 2x^2 \sin 2x = 0\) gives \(\cos 2x = x \sin 2x\). Rearranging, \(\tan 2x = \frac{1}{x}\). At \(x = p\), \(2p = \arctan \left( \frac{1}{p} \right)\), so \(p = \frac{1}{2} \arctan \left( \frac{1}{p} \right)\).

(ii) Use the iterative formula \(p_{n+1} = \frac{1}{2} \arctan \left( \frac{1}{p_n} \right)\). Start with \(p_0 = 0.5\):

\(p_1 = 0.55357\)

\(p_2 = 0.53261\)

\(p_3 = 0.54070\)

\(p_4 = 0.53755\)

Final answer: \(p = 0.54\) to 2 decimal places.

(iii) Integrate \(y = x^2 \cos 2x\) by parts. Let \(u = x^2\) and \(dv = \cos 2x \, dx\), then \(du = 2x \, dx\) and \(v = \frac{1}{2} \sin 2x\). The integration by parts gives:

\(\int x^2 \cos 2x \, dx = \frac{1}{2} x^2 \sin 2x - \int x \sin 2x \, dx\).

Integrate \(\int x \sin 2x \, dx\) by parts again. Let \(u = x\) and \(dv = \sin 2x \, dx\), then \(du = dx\) and \(v = -\frac{1}{2} \cos 2x\). This gives:

\(\int x \sin 2x \, dx = -\frac{1}{2} x \cos 2x + \frac{1}{2} \int \cos 2x \, dx\).

Complete the integration to obtain:

\(\frac{1}{2} x^2 \sin 2x + \frac{1}{2} x \cos 2x - \frac{1}{4} \sin 2x\).

Substitute limits \(x = 0\) and \(x = \frac{1}{4} \pi\):

\(\frac{1}{32}(\pi^2 - 8)\).

(i) Integrate \(x \cos x\) using integration by parts:

Let \(u = x\) and \(dv = \cos x \, dx\). Then \(du = dx\) and \(v = \sin x\).

\(\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C\).

Evaluate from 0 to \(a\):

\([x \sin x + \cos x]_0^a = a \sin a + \cos a - (0 \cdot \sin 0 + \cos 0) = a \sin a + \cos a - 1\).

Set equal to 0.5:

\(a \sin a + \cos a - 1 = 0.5\)

\(a \sin a = 1.5 - \cos a\)

\(\sin a = \frac{1.5 - \cos a}{a}\)

(ii) Verify \(a > 1\):

Consider \(a = 1\):

\(1 \sin 1 + \cos 1 - 1 \approx 0.8415 + 0.5403 - 1 = 0.3818\)

\(0.3818 < 0.5\), so \(a > 1\).

(iii) Iterative formula:

Start with \(a_1 = 1.2\).

\(a_2 = \sin^{-1} \left( \frac{1.5 - \cos 1.2}{1.2} \right) \approx 1.2464\)

\(a_3 = \sin^{-1} \left( \frac{1.5 - \cos 1.2464}{1.2464} \right) \approx 1.2461\)

\(a_4 = \sin^{-1} \left( \frac{1.5 - \cos 1.2461}{1.2461} \right) \approx 1.2461\)

Final answer: \(a \approx 1.2461\)

(a) The area of the major sector is \(\frac{1}{2} r^2 (2\pi - x)\).

The area of the shaded segment is \(\frac{1}{2} r^2 x - \frac{1}{2} r^2 \sin x\).

Given that the area of the major sector is 3 times the area of the shaded region:

\(\frac{1}{2} r^2 (2\pi - x) = 3 \left( \frac{1}{2} r^2 x - \frac{1}{2} r^2 \sin x \right)\)

Simplifying gives:

\(2\pi - x = 3x - 3\sin x\)

\(2\pi = 4x - 3\sin x\)

\(x = \frac{3}{4} \sin x + \frac{1}{2} \pi\)

(b) Calculate the values at \(x = 2\) and \(x = 2.5\):

For \(x = 2\), \(\frac{3}{4} \sin 2 + \frac{1}{2} \pi \approx 2.2528\)

For \(x = 2.5\), \(\frac{3}{4} \sin 2.5 + \frac{1}{2} \pi \approx 2.0197\)

Since \(2.2528 > 2\) and \(2.0197 < 2.5\), the root lies between 2 and 2.5.

(c) Using the iterative formula:

\(x_{n+1} = \frac{3}{4} \sin x_n + \frac{1}{2} \pi\)

Starting with \(x_1 = 2\):

\(x_2 = \frac{3}{4} \sin 2 + \frac{1}{2} \pi \approx 2.2528\)

\(x_3 = \frac{3}{4} \sin 2.2528 + \frac{1}{2} \pi \approx 2.1530\)

\(x_4 = \frac{3}{4} \sin 2.1530 + \frac{1}{2} \pi \approx 2.1972\)

\(x_5 = \frac{3}{4} \sin 2.1972 + \frac{1}{2} \pi \approx 2.1784\)

\(x_6 = \frac{3}{4} \sin 2.1784 + \frac{1}{2} \pi \approx 2.1866\)

\(x_7 = \frac{3}{4} \sin 2.1866 + \frac{1}{2} \pi \approx 2.1830\)

\(x_8 = \frac{3}{4} \sin 2.1830 + \frac{1}{2} \pi \approx 2.1846\)

The root is approximately 2.18 to 2 decimal places.

(i) To show that \(\cos 2θ = \frac{2 \sin 2θ - π}{4θ}\), we start by considering the geometry of the circle and the shaded area. The area of the circle is \(πr^2\), so half the area is \(\frac{1}{2}πr^2\).

The area of sector OAB is \(\frac{1}{2}r^2θ\), and the area of triangle OAB is \(\frac{1}{2}r^2 \sin θ\).

The area of the segment ABC is \(\frac{1}{2}r^2(θ - \sin θ)\).

Equating the shaded area to half the circle's area, we have:

\(\frac{1}{2}r^2θ + \frac{1}{2}r^2(θ - \sin θ) = \frac{1}{2}πr^2\)

Simplifying gives:

\(r^2θ - \frac{1}{2}r^2 \sin θ = \frac{1}{2}πr^2\)

Dividing through by \(r^2\) and rearranging gives:

\(θ - \frac{1}{2} \sin θ = \frac{1}{2}π\)

Using the double angle identity \(\cos 2θ = 1 - 2 \sin^2 θ\), we can express \(\sin θ\) in terms of \(θ\) and solve for \(\cos 2θ\).

(ii) Using the iterative formula \(θ_{n+1} = \frac{1}{2} \cos^{-1} \left( \frac{2 \sin 2θ_n - π}{4θ_n} \right)\), start with \(θ_1 = 1\).

Calculate successive iterations:

\(θ_2 = \frac{1}{2} \cos^{-1} \left( \frac{2 \sin 2(1) - π}{4(1)} \right)\)

Continue iterating until the value stabilizes to 4 decimal places.

The final value of \(θ\) correct to 2 decimal places is 0.95.