(i) Sketch the graphs of \(y = 2 \cot x\) and \(y = 1 + e^x\). The intersection of these graphs in the interval \(0 < x < \frac{1}{2} \pi\) indicates the root. The behavior of \(2 \cot x\) and \(1 + e^x\) suggests only one intersection point in this interval.

(ii) Evaluate \(2 \cot x - 1 - e^x\) at \(x = 0.5\) and \(x = 1.0\). The sign change between these points confirms a root exists between them.

(iii) Rearrange the original equation to show it is equivalent to \(x = \arctan\left(\frac{2}{1 + e^x}\right)\).

(iv) Use the iterative formula:

\(x_1 = 0.7\)

\(x_2 = \arctan\left(\frac{2}{1 + e^{0.7}}\right) = 0.6104\)

\(x_3 = \arctan\left(\frac{2}{1 + e^{0.6104}}\right) = 0.6100\)

\(x_4 = \arctan\left(\frac{2}{1 + e^{0.6100}}\right) = 0.6100\)

The root correct to 2 decimal places is 0.61.

(i) Sketch the graphs of \(y = \csc x\) and \(y = \frac{1}{2}x + 1\) over the interval \(0 < x < \frac{1}{2}\pi\). The intersection of these graphs indicates a root in this interval.

(ii) Evaluate \(\csc x - \frac{1}{2}x - 1\) at \(x = 0.5\) and \(x = 1\). If the signs are different, a root exists between these values.

(iii) Rearrange \(\csc x = \frac{1}{2}x + 1\) to \(x = \sin^{-1} \left( \frac{2}{x+2} \right)\) to show equivalence.

(iv) Use the iterative formula \(x_{n+1} = \sin^{-1} \left( \frac{2}{x_n+2} \right)\) starting with \(x_1 = 0.75\). Calculate successive iterations until the value stabilizes to 2 decimal places: \(x_2 = 0.8041\), \(x_3 = 0.8002\), \(x_4 = 0.8000\). The root is approximately 0.80.

(i) The area of triangle ONB is \(\frac{1}{2} r^2 \cos \alpha \sin \alpha\).

The area of sector OAB is \(\frac{1}{2} r^2 \alpha\).

Given that the area of triangle ONB is half the area of sector OAB, we have:

\(\frac{1}{2} r^2 \cos \alpha \sin \alpha = \frac{1}{2} \left( \frac{1}{2} r^2 \alpha \right)\)

Simplifying gives \(\sin 2\alpha = \alpha\).

(ii) Sketch the graphs of \(y = \sin 2x\) and \(y = x\) over the interval \(0 < x < \frac{1}{2}\pi\). The graphs intersect exactly once in this interval, indicating one root.

(iii) Using the iterative formula \(x_{n+1} = \sin(2x_n)\) with \(x_1 = 1\):

\(x_2 = \sin(2 \times 1) = \sin(2) \approx 0.909\)

\(x_3 = \sin(2 \times 0.909) \approx 0.973\)

\(x_4 = \sin(2 \times 0.973) \approx 0.929\)

\(x_5 = \sin(2 \times 0.929) \approx 0.964\)

\(x_6 = \sin(2 \times 0.964) \approx 0.945\)

\(x_7 = \sin(2 \times 0.945) \approx 0.955\)

\(x_8 = \sin(2 \times 0.955) \approx 0.949\)

\(x_9 = \sin(2 \times 0.949) \approx 0.953\)

The iterations converge to \(\alpha = 0.95\) correct to 2 decimal places.

(i) Sketch the graphs of \(y = \sec x\) and \(y = 3 - x^2\) for \(0 < x < \frac{1}{2}\pi\). The graph of \(y = \sec x\) has a vertical asymptote at \(x = \frac{1}{2}\pi\) and is positive in the interval. The graph of \(y = 3 - x^2\) is a downward-opening parabola with a positive y-intercept. The intersection of these graphs in the given interval shows exactly one root.

(ii) Assume \(\alpha = \cos^{-1} \left( \frac{1}{3-\alpha^2} \right)\). Rearrange to get \(\cos \alpha = \frac{1}{3-\alpha^2}\), which implies \(\sec \alpha = 3 - \alpha^2\). Thus, if the sequence converges, it converges to a root of the equation \(\sec x = 3 - x^2\).

(iii) Using the iterative formula \(x_{n+1} = \cos^{-1} \left( \frac{1}{3-x_n^2} \right)\) with \(x_1 = 1\):

1. Calculate \(x_2 = \cos^{-1} \left( \frac{1}{3-1^2} \right) \approx 1.047\).

2. Calculate \(x_3 = \cos^{-1} \left( \frac{1}{3-(1.047)^2} \right) \approx 1.032\).

3. Calculate \(x_4 = \cos^{-1} \left( \frac{1}{3-(1.032)^2} \right) \approx 1.030\).

The sequence converges to 1.03, correct to 2 decimal places.

(a) Sketch the graphs of \(y = \ln x\) and \(y = 3x - x^2\). The graph of \(y = \ln x\) passes through (1, 0) and has the y-axis as an asymptote. The graph of \(y = 3x - x^2\) is a parabola opening downwards, symmetric about the line \(x = 1.5\), passing through (0, 0) and (3, 0). The intersection point indicates one real root.

(b) Calculate \(\ln 2 \approx 0.693\) and \(3(2) - 2^2 = 2\). Since \(0.693 < 2\), the function changes sign between 2 and 2.8. Calculate \(\ln 2.8 \approx 1.029\) and \(3(2.8) - 2.8^2 = 1.04\). Since \(1.029 < 1.04\), the root lies between 2 and 2.8.

(c) Use the iterative formula \(x_{n+1} = \sqrt{3x_n - \ln x_n}\) starting with \(x_1 = 2.5\):

\(x_2 = \sqrt{3(2.5) - \ln(2.5)} \approx 2.6178\)

\(x_3 = \sqrt{3(2.6178) - \ln(2.6178)} \approx 2.6311\)

\(x_4 = \sqrt{3(2.6311) - \ln(2.6311)} \approx 2.6300\)

The root is approximately 2.63 to 2 decimal places.