(i) Differentiate \(y = x \cos 2x\) using the product rule: \(\frac{dy}{dx} = \cos 2x - 2x \sin 2x\). Set \(\frac{dy}{dx} = 0\) for maximum point: \(\cos 2x = 2x \sin 2x\). Rearrange to \(1 = 2x \tan 2x\).

(ii) Use the iterative formula \(x_{n+1} = \frac{1}{2} \arctan \left( \frac{1}{2x_n} \right)\) with \(x_1 = 0.4\):

\(x_2 = \frac{1}{2} \arctan \left( \frac{1}{2 \times 0.4} \right) = 0.4265\)

\(x_3 = \frac{1}{2} \arctan \left( \frac{1}{2 \times 0.4265} \right) = 0.4303\)

\(x_4 = \frac{1}{2} \arctan \left( \frac{1}{2 \times 0.4303} \right) = 0.4312\)

\(x_5 = \frac{1}{2} \arctan \left( \frac{1}{2 \times 0.4312} \right) = 0.4314\)

Final answer: 0.43

(iii) Use integration by parts for \(\int x \cos 2x \, dx\):

Let \(u = x\), \(dv = \cos 2x \, dx\). Then \(du = dx\), \(v = \frac{1}{2} \sin 2x\).

\(\int x \cos 2x \, dx = \frac{1}{2} x \sin 2x - \int \frac{1}{2} \sin 2x \, dx\)

\(= \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + C\)

Evaluate from \(0\) to \(\frac{1}{4} \pi\):

\(\left[ \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x \right]_0^{\frac{1}{4} \pi} = \frac{1}{8} \pi - \frac{1}{4}\)