(a) Calculate \(\cot \frac{1}{2}x - 3x\) at \(x = 0.5\) and \(x = 1\).

For \(x = 0.5\), \(\cot \frac{1}{2} \times 0.5 = \cot 0.25 \approx 3.92\) and \(3 \times 0.5 = 1.5\). So, \(3.92 - 1.5 > 0\).

For \(x = 1\), \(\cot \frac{1}{2} \times 1 = \cot 0.5 \approx 1.83\) and \(3 \times 1 = 3\). So, \(1.83 - 3 < 0\).

Since there is a change of sign, \(\alpha\) lies between 0.5 and 1.

(b) Rearrange \(\cot \frac{x}{2} = 3x\) to \(x = \frac{1}{3} \left( x + 4 \arctan \left( \frac{1}{3x} \right) \right)\).

The iterative formula \(x_{n+1} = \frac{1}{3} \left( x_n + 4 \arctan \left( \frac{1}{3x_n} \right) \right)\) is derived from this rearrangement.

If the sequence converges, it must converge to the root \(\alpha\).

(c) Start with an initial guess, say \(x_0 = 0.5\).

Calculate successive iterations:

\(x_1 = \frac{1}{3} \left( 0.5 + 4 \arctan \left( \frac{1}{3 \times 0.5} \right) \right) \approx 0.7623\)

\(x_2 = \frac{1}{3} \left( 0.7623 + 4 \arctan \left( \frac{1}{3 \times 0.7623} \right) \right) \approx 0.8037\)

\(x_3 = \frac{1}{3} \left( 0.8037 + 4 \arctan \left( \frac{1}{3 \times 0.8037} \right) \right) \approx 0.7921\)

\(x_4 = \frac{1}{3} \left( 0.7921 + 4 \arctan \left( \frac{1}{3 \times 0.7921} \right) \right) \approx 0.7951\)

\(x_5 = \frac{1}{3} \left( 0.7951 + 4 \arctan \left( \frac{1}{3 \times 0.7951} \right) \right) \approx 0.7943\)

\(x_6 = \frac{1}{3} \left( 0.7943 + 4 \arctan \left( \frac{1}{3 \times 0.7943} \right) \right) \approx 0.7945\)

The value converges to \(0.79\) to 2 decimal places.

(i) Consider the function \(f(x) = x - \frac{10}{e^{2x} - 1}\).

Calculate \(f(1) = 1 - \frac{10}{e^2 - 1}\) and \(f(2) = 2 - \frac{10}{e^4 - 1}\).

Since \(f(1) < 0\) and \(f(2) > 0\), by the Intermediate Value Theorem, \(\alpha\) lies between 1 and 2.

(ii) Assume the sequence \(x_n\) converges to \(L\). Then \(L = \frac{1}{2} \ln \left( 1 + \frac{10}{L} \right)\).

Rearranging gives \(L = \frac{10}{e^{2L} - 1}\), so \(L = \alpha\).

(iii) Use the iterative formula:

Start with \(x_1 = 1\).

Calculate \(x_2 = \frac{1}{2} \ln \left( 1 + \frac{10}{1} \right) = 1.1513\).

Calculate \(x_3 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1513} \right) = 1.1394\).

Calculate \(x_4 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1394} \right) = 1.1427\).

Calculate \(x_5 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1427} \right) = 1.1414\).

Calculate \(x_6 = \frac{1}{2} \ln \left( 1 + \frac{10}{1.1414} \right) = 1.1418\).

Since the values are converging, \(\alpha \approx 1.14\) to 2 decimal places.

(i) Start with \(x_1 = 3.5\).

Calculate \(x_2 = \frac{3.5(3.5^3 + 100)}{2(3.5^3 + 25)} = 3.683943\).

Calculate \(x_3 = \frac{3.683943(3.683943^3 + 100)}{2(3.683943^3 + 25)} = 3.684002\).

Calculate \(x_4 = \frac{3.684002(3.684002^3 + 100)}{2(3.684002^3 + 25)} = 3.684000\).

Calculate \(x_5 = \frac{3.684000(3.684000^3 + 100)}{2(3.684000^3 + 25)} = 3.684000\).

The sequence converges to \(\alpha = 3.6840\) correct to 4 decimal places.

(ii) The equation satisfied by \(\alpha\) is \(x = \frac{x(x^3 + 100)}{2(x^3 + 25)}\).

Solving for \(x\), we find \(\alpha = 3\sqrt{50}\).

(i) To find \(\alpha\), substitute \(x = -2\) into the equation \(y = x^4 + 2x^3 + 2x^2 - 4x - 16\):

\((-2)^4 + 2(-2)^3 + 2(-2)^2 - 4(-2) - 16 = 16 - 16 + 8 + 8 - 16 = 0\).

Thus, \(\alpha = -2\).

(ii) To show \(\beta\) satisfies \(x = \sqrt[3]{8 - 2x}\), factor the quartic equation:

\((x + 2)(x^3 + 2x - 8) = 0\).

Rearrange \(x^3 + 2x - 8 = 0\) to \(x = \sqrt[3]{8 - 2x}\).

(iii) Use the iterative formula \(x_{n+1} = \sqrt[3]{8 - 2x_n}\) starting with an initial guess, say \(x_0 = 1.5\):

\(x_1 = \sqrt[3]{8 - 2(1.5)} = 1.709975947\).

\(x_2 = \sqrt[3]{8 - 2(1.709975947)} = 1.671585\).

\(x_3 = \sqrt[3]{8 - 2(1.671585)} = 1.667680\).

\(x_4 = \sqrt[3]{8 - 2(1.667680)} = 1.667056\).

Thus, \(\beta \approx 1.67\) to 2 decimal places.

(i) Using the identity \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), substitute \(\tan x = t\) into the equation:

\(2 \left(\frac{2t}{1-t^2}\right) + 5t^2 = 0\)

Simplifying gives:

\(4t + 5t^2 - 5t^4 = 0\)

Factoring out \(t\), we get:

\(t(4 + 5t - 5t^3) = 0\)

Thus, \(t = 0\) or \(t = \sqrt[3]{(t + 0.8)}\).

(ii) Check the sign of \(t - \sqrt[3]{(t + 0.8)}\) at \(t = 1.2\) and \(t = 1.3\):

At \(t = 1.2\), \(1.2 - \sqrt[3]{(1.2 + 0.8)} = -0.06\)

At \(t = 1.3\), \(1.3 - \sqrt[3]{(1.3 + 0.8)} = 0.02\)

There is a change of sign, confirming a root between 1.2 and 1.3.

(iii) Using the iterative formula \(t_{n+1} = \sqrt[3]{(t_n + 0.8)}\):

Start with \(t_0 = 1.2\):

\(t_1 = \sqrt[3]{(1.2 + 0.8)} = 1.2765\)

\(t_2 = \sqrt[3]{(1.2765 + 0.8)} = 1.2760\)

\(t_3 = \sqrt[3]{(1.2760 + 0.8)} = 1.2760\)

The value of \(t\) correct to 3 decimal places is 1.276.

(iv) Solve \(2 \tan 2x + 5 \tan^2 x = 0\) using \(t = 1.276\):

\(\tan x = 1.276\), so \(x = \arctan(1.276)\)

\(x \approx 0.906\)

Considering the periodicity of \(\tan x\), the solutions are:

\(x = -2.24, 0.906, \pi\)