(i) Use integration by parts to obtain:

\(axe^{-\frac{1}{2}x} + \int be^{-\frac{1}{2}x} \, dx\)

Obtain \(-8xe^{-\frac{1}{2}x} + 8 \int e^{-\frac{1}{2}x} \, dx\)

Obtain \(-8xe^{-\frac{1}{2}x} - 16e^{-\frac{1}{2}x}\)

Use limits correctly and equate to 9:

\(p = 2 \ln \left( \frac{8p + 16}{7} \right)\)

(ii) Use correct iteration formula:

Starting with \(p_0 = 3.5\), iterate:

\(p_1 = 3.6766\)

\(p_2 = 3.7398\)

\(p_3 = 3.7619\)

\(p_4 = 3.7696\)

\(p_5 = 3.7723\)

Final answer: \(p = 3.77\)

(i) Substitute \(x = u^2\), then \(dx = 2u \, du\). The integral becomes \(\int_0^{p^2} \cos(\sqrt{x}) \, dx = \int_0^p 2u \cos(u) \, du\).

Integrate by parts: let \(v = \cos(u)\) and \(dw = 2u \, du\), then \(dv = -\sin(u) \, du\) and \(w = u^2\).

\(\int 2u \cos(u) \, du = 2u \sin(u) + 2 \cos(u)\).

Using limits \(u = 0\) to \(u = p\), we have:

\(2p \sin(p) + 2 \cos(p) - (0 + 2) = 1\).

Simplifying gives \(2p \sin(p) = 3 - 2 \cos(p)\), hence \(\sin(p) = \frac{3 - 2 \cos(p)}{2p}\).

(ii) Using the iterative formula \(p_{n+1} = \sin^{-1} \left( \frac{3 - 2 \cos(p_n)}{2p_n} \right)\):

\(p_1 = 1.0000\)

\(p_2 = 1.2399\)

\(p_3 = 1.2497\)

\(p_4 = 1.2500\)

\(p_5 = 1.2500\)

Thus, \(p = 1.25\) correct to 2 decimal places.

(i) Use integration by parts to evaluate \(\int x \ln x \, dx\).

Let \(u = \ln x\) and \(dv = x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^2}{2}\).

Integration by parts gives:

\(\int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x \, dx\).

\(= \frac{x^2}{2} \ln x - \frac{1}{4} x^2\).

Substitute the limits from 1 to \(a\) and equate to 22:

\(\left[ \frac{a^2}{2} \ln a - \frac{1}{4} a^2 \right] - \left[ \frac{1}{2} \ln 1 - \frac{1}{4} \right] = 22\).

\(\frac{a^2}{2} \ln a - \frac{1}{4} a^2 + \frac{1}{4} = 22\).

Rearrange to find \(a\):

\(a = \sqrt{\frac{87}{2 \ln a - 1}}\).

(ii) Use the iterative formula \(a_{n+1} = \sqrt{\frac{87}{2 \ln a_n - 1}}\) with \(a_1 = 6\).

Calculate each iteration to 4 decimal places:

\(a_1 = 6\)

\(a_2 = 5.8030\)

\(a_3 = 5.8795\)

\(a_4 = 5.8491\)

\(a_5 = 5.8611\)

\(a_6 = 5.8564\)

The value of \(a\) correct to 2 decimal places is 5.86.

(i) To solve \(\int_1^a \frac{\ln x}{x^2} \, dx = \frac{2}{5}\), we use integration by parts. Let \(u = \ln x\) and \(dv = \frac{1}{x^2} \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = -\frac{1}{x}\).

Applying integration by parts:

\(\int \frac{\ln x}{x^2} \, dx = -\frac{\ln x}{x} + \int \frac{1}{x^2} \, dx = -\frac{\ln x}{x} - \frac{1}{x} + C\)

Evaluating from 1 to \(a\):

\(\left[ -\frac{\ln x}{x} - \frac{1}{x} \right]_1^a = \left( -\frac{\ln a}{a} - \frac{1}{a} \right) - \left( -0 - 1 \right) = -\frac{\ln a}{a} - \frac{1}{a} + 1\)

Equating to \(\frac{2}{5}\):

\(-\frac{\ln a}{a} - \frac{1}{a} + 1 = \frac{2}{5}\)

Rearranging gives:

\(1 - \frac{2}{5} = \frac{\ln a + 1}{a}\)

\(\frac{3}{5} = \frac{\ln a + 1}{a}\)

\(a = \frac{5}{3}(1 + \ln a)\)

(ii) Using the iteration formula \(a_{n+1} = \frac{5}{3}(1 + \ln a_n)\) with \(a_0 = 4\):

\(a_1 = \frac{5}{3}(1 + \ln 4) \approx 3.9772\)

\(a_2 = \frac{5}{3}(1 + \ln 3.9772) \approx 3.9676\)

\(a_3 = \frac{5}{3}(1 + \ln 3.9676) \approx 3.9636\)

\(a_4 = \frac{5}{3}(1 + \ln 3.9636) \approx 3.9619\)

The value of \(a\) correct to 2 decimal places is \(a \approx 3.96\).

(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 \(\frac{1}{4} \pi\):

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