(i) Calculate \(\cot(2.5)\) and \(1 - 2.5 = -1.5\). Since \(\cot(2.5) > -1.5\), \(\alpha > 2.5\).

(ii) Assume \(x = \pi + \arctan \left( \frac{1}{1-x} \right)\). Rearrange to \(\cot x = 1 - x\), showing convergence to \(\alpha\).

(iii) Using the iterative formula:

1. \(x_1 = 3.00000\)

2. \(x_2 = 2.57732\)

3. \(x_3 = 2.57605\)

4. \(x_4 = 2.57600\)

5. \(x_5 = 2.57600\)

Thus, \(\alpha = 2.576\) correct to 3 decimal places.

(i) Differentiate both equations: \(y = x \cos x\) gives \(\frac{dy}{dx} = \cos x - x \sin x\) and \(y = \frac{k}{x}\) gives \(\frac{dy}{dx} = -\frac{k}{x^2}\). Equate the derivatives at \(x = a\):

\(\cos a - a \sin a = -\frac{k}{a^2}\).

Since the curves touch, \(a \cos a = \frac{k}{a}\). Substitute \(k = a^2 \cos a\) into the derivative equation:

\(\cos a - a \sin a = -\frac{a^2 \cos a}{a^2} = -\cos a\).

Rearrange to get \(\tan a = \frac{2}{a}\).

(ii) Use the iterative formula \(a_{n+1} = \arctan \left( \frac{2}{a_n} \right)\) starting with an initial guess, say \(a_0 = 1\):

\(a_1 = \arctan \left( \frac{2}{1} \right) = 1.10715\)

\(a_2 = \arctan \left( \frac{2}{1.10715} \right) = 1.07989\)

\(a_3 = \arctan \left( \frac{2}{1.07989} \right) = 1.07735\)

\(a_4 = \arctan \left( \frac{2}{1.07735} \right) = 1.07708\)

\(a_5 = \arctan \left( \frac{2}{1.07708} \right) = 1.07703\)

Thus, \(a \approx 1.077\) to 3 decimal places.

(iii) Use \(a \cos a = \frac{k}{a}\) with \(a = 1.077\):

\(k = a^2 \cos a = (1.077)^2 \cos(1.077) \approx 0.55\).

(i) Evaluate the function at \(x = 1\) and \(x = 2\):

\(f(1) = 1^5 - 3(1)^3 + 1^2 - 4 = 1 - 3 + 1 - 4 = -5\)

\(f(2) = 2^5 - 3(2)^3 + 2^2 - 4 = 32 - 24 + 4 - 4 = 8\)

Since \(f(1) < 0\) and \(f(2) > 0\), there is a root between 1 and 2.

(ii) Rearrange the equation:

Start with \(x^5 - 3x^3 + x^2 - 4 = 0\).

Add 4 to both sides: \(x^5 - 3x^3 + x^2 = 4\).

Divide by \(x^2\): \(x^3 - 3x + \frac{1}{x} = \frac{4}{x^2}\).

Rearrange to: \(x^3 = 3x + \frac{4}{x^2} - 1\).

Take the cube root: \(x = \sqrt[3]{3x + \frac{4}{x^2} - 1}\).

(iii) Use the iterative formula \(x_{n+1} = \sqrt[3]{3x_n + \frac{4}{x_n^2} - 1}\).

Start with \(x_0 = 1.5\).

\(x_1 = \sqrt[3]{3(1.5) + \frac{4}{(1.5)^2} - 1} = 1.8041\)

\(x_2 = \sqrt[3]{3(1.8041) + \frac{4}{(1.8041)^2} - 1} = 1.7817\)

\(x_3 = \sqrt[3]{3(1.7817) + \frac{4}{(1.7817)^2} - 1} = 1.7800\)

\(x_4 = \sqrt[3]{3(1.7800) + \frac{4}{(1.7800)^2} - 1} = 1.7800\)

The root correct to 2 decimal places is 1.78.

(i) Evaluate \(x^3 - x^2 - 6\) for \(x = 2\) and \(x = 3\):

For \(x = 2\), \(2^3 - 2^2 - 6 = 8 - 4 - 6 = -2\).

For \(x = 3\), \(3^3 - 3^2 - 6 = 27 - 9 - 6 = 12\).

Since the sign changes between \(x = 2\) and \(x = 3\), \(\alpha\) lies between 2 and 3.

(ii) The iterative formula is \(x_{n+1} = \sqrt{x_n + \frac{6}{x_n}}\).

If the sequence converges, then \(x_{n+1} = x_n = \alpha\).

Substitute into the formula: \(\alpha = \sqrt{\alpha + \frac{6}{\alpha}}\).

Square both sides: \(\alpha^2 = \alpha + \frac{6}{\alpha}\).

Multiply through by \(\alpha\): \(\alpha^3 = \alpha^2 + 6\).

Rearrange to \(\alpha^3 - \alpha^2 - 6 = 0\), which is the original equation, confirming convergence to \(\alpha\).

(iii) Use the iterative formula starting with an initial guess, say \(x_0 = 2.5\):

\(x_1 = \sqrt{2.5 + \frac{6}{2.5}} = 2.31662\)

\(x_2 = \sqrt{2.31662 + \frac{6}{2.31662}} = 2.22192\)

\(x_3 = \sqrt{2.22192 + \frac{6}{2.22192}} = 2.21936\)

\(x_4 = \sqrt{2.21936 + \frac{6}{2.21936}} = 2.21900\)

\(x_5 = \sqrt{2.21900 + \frac{6}{2.21900}} = 2.21900\)

The value converges to \(\alpha = 2.219\) correct to 3 decimal places.

(a) To find the maximum point, we need to differentiate \(y = x^2 \cos 3x\) and set the derivative to zero. Using the product rule:

\(\frac{d}{dx}(x^2 \cos 3x) = \frac{d}{dx}(x^2) \cdot \cos 3x + x^2 \cdot \frac{d}{dx}(\cos 3x)\).

\(= 2x \cos 3x - 3x^2 \sin 3x\).

Setting the derivative to zero for maximum point:

\(2x \cos 3x - 3x^2 \sin 3x = 0\).

\(2x \cos 3x = 3x^2 \sin 3x\).

\(\frac{2}{3x} = \tan 3x\).

\(x = \frac{1}{3} \arctan \left( \frac{2}{3x} \right)\).

At maximum point \(x = a\), so \(a = \frac{1}{3} \arctan \left( \frac{2}{3a} \right)\).

(b) Using the iterative formula \(a_{n+1} = \frac{1}{3} \arctan \left( \frac{2}{3a_n} \right)\), start with an initial guess and iterate:

1. \(a_1 = 0.5\)

2. \(a_2 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.5} \right) \approx 0.3435\)

3. \(a_3 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.3435} \right) \approx 0.3826\)

4. \(a_4 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.3826} \right) \approx 0.4264\)

5. \(a_5 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.4264} \right) \approx 0.4740\)

6. \(a_6 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.4740} \right) \approx 0.3017\)

7. \(a_7 = \frac{1}{3} \arctan \left( \frac{2}{3 \times 0.3017} \right) \approx 0.3989\)

Continue iterating until the value stabilizes to 2 decimal places: \(a \approx 0.36\).