(i) Sketch the graphs of \(y = \csc \frac{1}{2}x\) and \(y = \frac{1}{3}x + 1\). The intersection of these graphs in the interval \(0 < x \leq \pi\) indicates a root.

(ii) Calculate the values of the expressions at \(x = 1.4\) and \(x = 1.6\). If there is a sign change, the root lies between these values.

(iii) Rearrange the iterative formula \(x_{n+1} = 2 \sin^{-1} \left( \frac{3}{x_n + 3} \right)\) to show it is equivalent to \(\csc \frac{1}{2}x = \frac{1}{3}x + 1\). This shows convergence to the root.

(iv) Use the iterative formula starting with an initial guess. Calculate successive values to 5 decimal places until they stabilize to 3 decimal places. The final answer is 1.471.

(i) To show that the equation \(5e^{-x} = \sqrt{x}\) has one root, sketch the graphs of \(y = 5e^{-x}\) and \(y = \sqrt{x}\). The intersection point of these graphs represents the root. The exponential function \(5e^{-x}\) decreases rapidly, while \(\sqrt{x}\) increases slowly. They intersect at one point, indicating one root.

(ii) The iterative formula is \(x_{n+1} = \frac{1}{2} \ln\left(\frac{25}{x_n}\right)\). Rearrange this to show it is equivalent to \(5e^{-x} = \sqrt{x}\). If the sequence converges, it must satisfy the original equation, thus converging to the root.

(iii) Using the iterative formula with \(x_1 = 1\):

\(x_2 = \frac{1}{2} \ln\left(\frac{25}{1}\right) = 1.6094\)

\(x_3 = \frac{1}{2} \ln\left(\frac{25}{1.6094}\right) = 1.4332\)

\(x_4 = \frac{1}{2} \ln\left(\frac{25}{1.4332}\right) = 1.4292\)

\(x_5 = \frac{1}{2} \ln\left(\frac{25}{1.4292}\right) = 1.4286\)

The sequence converges to 1.43, correct to 2 decimal places.

(i) Sketch the curve \(y = \ln(x + 1)\), which is an increasing curve passing through the origin for \(x \geq 0\). The second curve is \(y = 40 - x^3\), which is a decreasing curve for \(x > 0\). The intersection of these curves indicates the root of the equation \(x^3 + \ln(x + 1) = 40\).

(ii) Calculate \(x^3 + \ln(x + 1) - 40\) at \(x = 3\) and \(x = 4\):

For \(x = 3\), \(3^3 + \ln(4) - 40 = 27 + 1.386 - 40 = -11.614\).

For \(x = 4\), \(4^3 + \ln(5) - 40 = 64 + 1.609 - 40 = 25.609\).

Since the sign changes between \(x = 3\) and \(x = 4\), the root lies between these values.

(iii) Using the iterative formula \(x_{n+1} = \sqrt[3]{40 - \ln(x_n + 1)}\):

Start with \(x_1 = 3.5\).

\(x_2 = \sqrt[3]{40 - \ln(3.5 + 1)} = 3.377\).

Continue iterations until the value stabilizes to 3 decimal places: \(x \approx 3.377\).

(iv) For \((e^y - 1)^3 + y = 40\), solve for \(y\):

\(y = \ln(x + 1) \approx 1.48\).

(i) Sketch the graphs of \(y = \csc x\) and \(y = x(\pi - x)\) for \(0 < x < \pi\). The graph of \(y = \csc x\) has vertical asymptotes at \(x = 0\) and \(x = \pi\), and the graph of \(y = x(\pi - x)\) is a downward-opening parabola with roots at \(x = 0\) and \(x = \pi\). The intersection points of these graphs represent the roots of the equation \(\csc x = x(\pi - x)\). By analyzing the graphs, we find exactly two intersection points in the interval \(0 < x < \pi\).

(ii) Start with \(\csc x = x(\pi - x)\), which is \(\frac{1}{\sin x} = x(\pi - x)\). Rearrange to get \(x = \frac{1 + x^2 \sin x}{\pi \sin x}\).

(iii)(a) Use the iterative formula \(x_{n+1} = \frac{1 + x_n^2 \sin x_n}{\pi \sin x_n}\) starting with an initial guess. Iterate until the value stabilizes to 4 decimal places. The iterations converge to \(\alpha = 0.66\).

(iii)(b) From the mark scheme, \(\beta = 2.48\).

(i) Sketch the graphs of \(y = \sec x\) and \(y = 3 - \frac{1}{2}x^2\) over the interval \(0 < x < \frac{1}{2}\pi\). The intersection of these graphs indicates a root. The sketch should show that the graphs intersect, confirming a root exists in the interval.

(ii) Calculate \(\sec 1 - (3 - \frac{1}{2} \times 1^2)\) and \(\sec 1.4 - (3 - \frac{1}{2} \times 1.4^2)\). If the signs of these values are different, a root exists between 1 and 1.4.

(iii) Rearrange \(\sec x = 3 - \frac{1}{2}x^2\) to \(x = \cos^{-1}\left( \frac{2}{6 - x^2} \right)\) to show equivalence.

(iv) Use the iterative formula \(x_{n+1} = \cos^{-1}\left( \frac{2}{6 - x_n^2} \right)\) starting with an initial guess. Iterate until the value stabilizes to 4 decimal places. The iterations yield \(x \approx 1.13\).