(a) Sketch the graphs of \(y = \cot x\) and \(y = 2 - \cos x\) for \(0 < x \leq \frac{1}{2}\pi\). The intersection point of these graphs indicates the root of the equation \(\cot x = 2 - \cos x\). The graph shows one intersection in the given interval.

(b) Calculate \(\cot(0.6)\) and \(\cot(0.8)\) and compare with \(2 - \cos(0.6)\) and \(2 - \cos(0.8)\). For \(x = 0.6\), \(\cot(0.6) \approx 1.46\) and \(2 - \cos(0.6) \approx 1.34\). For \(x = 0.8\), \(\cot(0.8) \approx 1.03\) and \(2 - \cos(0.8) \approx 1.29\). The change in sign between these values indicates a root between 0.6 and 0.8.

(c) Use the iterative formula \(x_{n+1} = \arctan\left( \frac{1}{2 - \cos x_n} \right)\) starting with an initial guess, such as \(x_0 = 0.7\). Perform iterations:

\(x_1 = \arctan\left( \frac{1}{2 - \cos(0.7)} \right) \approx 0.6806\)

\(x_2 = \arctan\left( \frac{1}{2 - \cos(0.6806)} \right) \approx 0.6855\)

\(x_3 = \arctan\left( \frac{1}{2 - \cos(0.6855)} \right) \approx 0.6843\)

\(x_4 = \arctan\left( \frac{1}{2 - \cos(0.6843)} \right) \approx 0.6846\)

The iterations converge to approximately 0.68, confirming the root to 2 decimal places.

(i) Sketch the graphs of \(y = x^3\) and \(y = 3 - x\). The intersection of these graphs gives the solution to \(x^3 = 3 - x\). The graph of \(y = x^3\) is a cubic curve passing through the origin, and \(y = 3 - x\) is a straight line with a negative slope. They intersect at exactly one point, indicating one real root.

(ii) Given the iterative formula \(x_{n+1} = \frac{2x_n^3 + 3}{3x_n^2 + 1}\), if it converges, then \(x_{n+1} = x_n = x\). Substituting into the formula gives \(x = \frac{2x^3 + 3}{3x^2 + 1}\). Rearranging, we get \(x(3x^2 + 1) = 2x^3 + 3\), which simplifies to \(x^3 = 3 - x\), showing convergence to the root found in part (i).

(iii) Using the iterative formula, start with an initial guess, say \(x_0 = 1\). Calculate successive iterations to 5 decimal places:

\(x_1 = \frac{2(1)^3 + 3}{3(1)^2 + 1} = 1.25\)

\(x_2 = \frac{2(1.25)^3 + 3}{3(1.25)^2 + 1} = 1.21622\)

\(x_3 = \frac{2(1.21622)^3 + 3}{3(1.21622)^2 + 1} = 1.21305\)

\(x_4 = \frac{2(1.21305)^3 + 3}{3(1.21305)^2 + 1} = 1.21300\)

\(x_5 = \frac{2(1.21300)^3 + 3}{3(1.21300)^2 + 1} = 1.21300\)

The root correct to 3 decimal places is 1.213.

(i) Differentiate \(y = \frac{\ln x}{3 + x}\) using the quotient rule:

\(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

Let \(u = \ln x\) and \(v = 3 + x\). Then \(\frac{du}{dx} = \frac{1}{x}\) and \(\frac{dv}{dx} = 1\).

\(\frac{dy}{dx} = \frac{(3 + x) \cdot \frac{1}{x} - \ln x \cdot 1}{(3 + x)^2} = \frac{3 + x - x \ln x}{x(3 + x)^2}\)

Set \(\frac{dy}{dx} = 0\) for stationary points:

\(3 + x - x \ln x = 0\)

\(x \ln x = 3 + x\)

\(\ln x = 1 + \frac{3}{x}\)

(ii) Sketch the graphs of \(y = \ln x\) and \(y = 1 + \frac{3}{x}\). The intersection point represents the root. Since \(\ln x\) is a monotonically increasing function and \(1 + \frac{3}{x}\) is decreasing for \(x > 0\), they intersect at only one point.

(iii) Use the iterative formula \(x_{n+1} = \frac{3 + x_n}{\ln x_n}\):

Start with an initial guess, say \(x_1 = 4.5\).

Calculate successive iterations:

\(x_2 = \frac{3 + 4.5}{\ln 4.5} \approx 4.9686\)

\(x_3 = \frac{3 + 4.9686}{\ln 4.9686} \approx 4.9701\)

\(x_4 = \frac{3 + 4.9701}{\ln 4.9701} \approx 4.9700\)

Continue until the value stabilizes to 4 decimal places. The final answer is 4.97.

(i) Sketch the graphs of \(y = e^{2x}\) and \(y = 6 + e^{-x}\). The intersection of these graphs represents the solution to the equation \(e^{2x} = 6 + e^{-x}\). The graphs intersect at exactly one point, indicating one real root.

(ii) Calculate \(e^{2(0.5)} = e \approx 2.718\) and \(6 + e^{-0.5} \approx 6.606\). Since \(e^{2(0.5)} < 6 + e^{-0.5}\), check \(x = 1\): \(e^{2(1)} = e^2 \approx 7.389\) and \(6 + e^{-1} \approx 6.368\). Since \(e^{2(1)} > 6 + e^{-1}\), the root lies between 0.5 and 1.

(iii) The iterative formula \(x_{n+1} = \frac{1}{3} \ln(1 + 6e^{x_n})\) is derived from rearranging the original equation. If the sequence converges, it converges to the root of \(e^{2x} = 6 + e^{-x}\).

(iv) Using the iterative formula, start with an initial guess, say \(x_0 = 0.5\):

\(x_1 = \frac{1}{3} \ln(1 + 6e^{0.5}) \approx 0.91894\)

\(x_2 = \frac{1}{3} \ln(1 + 6e^{0.91894}) \approx 0.92754\)

\(x_3 = \frac{1}{3} \ln(1 + 6e^{0.92754}) \approx 0.92823\)

\(x_4 = \frac{1}{3} \ln(1 + 6e^{0.92823}) \approx 0.92828\)

\(x_5 = \frac{1}{3} \ln(1 + 6e^{0.92828}) \approx 0.92828\)

The root correct to 3 decimal places is 0.928.

(i) Sketch the graphs of \(y = e^{-\frac{1}{2}x}\) and \(y = 4 - x^2\). The intersection points of these graphs represent the roots of the equation. The graph \(y = e^{-\frac{1}{2}x}\) is an exponential decay curve, while \(y = 4 - x^2\) is a downward-opening parabola. The graphs intersect at one positive and one negative point, indicating one positive root and one negative root.

(ii) Calculate the values at \(x = -1\) and \(x = -1.5\):

For \(x = -1\):

\(e^{-\frac{1}{2}(-1)} = e^{0.5} \approx 1.6487\)

\(4 - (-1)^2 = 3\)

Since \(1.6487 > 3\), the function changes sign between \(-1\) and \(-1.5\).

For \(x = -1.5\):

\(e^{-\frac{1}{2}(-1.5)} = e^{0.75} \approx 2.1170\)

\(4 - (-1.5)^2 = 1.75\)

Since \(2.1170 > 1.75\), the function changes sign between \(-1\) and \(-1.5\).

(iii) Use the iterative formula \(x_{n+1} = -\sqrt{4 - e^{-\frac{1}{2}x_n}}\):

Start with an initial guess, say \(x_0 = -1\).

\(x_1 = -\sqrt{4 - e^{-\frac{1}{2}(-1)}} \approx -1.4142\)

\(x_2 = -\sqrt{4 - e^{-\frac{1}{2}(-1.4142)}} \approx -1.4106\)

\(x_3 = -\sqrt{4 - e^{-\frac{1}{2}(-1.4106)}} \approx -1.4100\)

\(x_4 = -\sqrt{4 - e^{-\frac{1}{2}(-1.4100)}} \approx -1.4100\)

The iterations converge to \(-1.41\) to 2 decimal places.