(a) Calculate \(f(1) = \frac{e^{2} + 1}{e^{2} - 1}\) and \(f(1.5) = \frac{e^{3} + 1}{e^{3} - 1}\). Verify that \(f(1) < 1\) and \(f(1.5) > 1.5\), confirming \(a\) is between 1 and 1.5.

(b) Use the iterative formula \(x_{n+1} = \frac{e^{2x_n} + 1}{e^{2x_n} - 1}\). Start with \(x_1 = 1\):

\(x_2 = 1.2311\)

\(x_3 = 1.2032\)

\(x_4 = 1.2002\)

\(x_5 = 1.2000\)

Thus, \(a = 1.20\) to 2 decimal places.

(c) Use the quotient rule to find \(f'(x) = \frac{(e^{2x} - 1)(2e^{2x}) - (e^{2x} + 1)(2e^{2x})}{(e^{2x} - 1)^2}\).

Simplify to \(f'(x) = \frac{-4e^{2x}}{(e^{2x} - 1)^2}\).

Set \(f'(x) = -8\):

\(\frac{-4e^{2x}}{(e^{2x} - 1)^2} = -8\)

\(2(e^{2x})^2 - 5e^{2x} + 2 = 0\)

Solve the quadratic equation for \(e^{2x}\):

\(e^{2x} = 2\) or \(e^{2x} = \frac{1}{2}\)

Thus, \(x = \frac{1}{2} \ln 2\).

(i) Differentiate \(y = e^{-2x} \ln(x-1)\) using the product rule:

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

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

\(-2e^{-2x} \ln(x-1) + \frac{e^{-2x}}{x-1} = 0\).

Simplify to find \(x = 1 + e^{\frac{1}{2(x-1)}}\).

(ii) Calculate \(f(x) = \ln(x-1) - \frac{1}{2(x-1)}\):

\(f(2.2) = -0.234\), \(f(2.6) = 0.317\).

Calculate \(f(x) = 2e^{-2x} \ln(x-1) + \frac{e^{-2x}}{x-1}\):

\(f(2.2) = 0.005\), \(f(2.6) = -0.0017\).

Since \(f(x)\) changes sign, \(p\) is between 2.2 and 2.6.

(iii) Use the iterative formula \(p_{n+1} = 1 + \exp\left( \frac{1}{2(p_n-1)} \right)\):

Start with \(p_0 = 2.4\).

\(p_1 = 2.4152\)

\(p_2 = 2.4213\)

\(p_3 = 2.4198\)

\(p_4 = 2.4204\)

\(p_5 = 2.4201\)

\(p_6 = 2.4203\)

\(p_7 = 2.4202\)

Thus, \(p \approx 2.42\) to 2 decimal places.

(i) Differentiate the curves: \(y = 4 \cos \frac{1}{2} x\) gives \(\frac{dy}{dx} = -2 \sin \frac{1}{2} x\) and \(y = \frac{1}{4-x}\) gives \(\frac{dy}{dx} = \frac{1}{(4-x)^2}\).

For perpendicular tangents, the product of the derivatives is \(-1\):

\(-2 \sin \frac{1}{2} a \cdot \frac{1}{(4-a)^2} = -1\)

Rearrange to find \(a\):

\(2 \sin \frac{1}{2} a = (4-a)^2\)

\(a = 4 - \sqrt{2 \sin \frac{1}{2} a}\)

(ii) Calculate values at \(a = 2\) and \(a = 3\):

For \(a = 2\), \(2 < 2.7027 \ldots\)

For \(a = 3\), \(3 > 2.587 \ldots\)

Thus, \(a\) lies between 2 and 3.

(iii) Use the iterative formula \(a_{n+1} = 4 - \sqrt{2 \sin \frac{1}{2} a_n}\):

Start with \(a_1 = 2\):

\(a_2 = 2.70272\)

\(a_3 = 2.60285\)

\(a_4 = 2.61152\)

\(a_5 = 2.61070\)

\(a_6 = 2.61077\)

Final answer: \(a = 2.611\)

(a) To solve \(\int_0^a xe^{-2x} \, dx = \frac{1}{8}\), we start by integrating by parts. Let \(u = x\) and \(dv = e^{-2x} \, dx\). Then \(du = dx\) and \(v = -\frac{1}{2}e^{-2x}\).

Using integration by parts, \(\int xe^{-2x} \, dx = -\frac{1}{2}xe^{-2x} + \frac{1}{2} \int e^{-2x} \, dx\).

Integrating \(\int e^{-2x} \, dx\) gives \(-\frac{1}{2}e^{-2x}\), so \(\int xe^{-2x} \, dx = -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\).

Evaluating from 0 to a, we have:

\(\left[ -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} \right]_0^a = -\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} + \frac{1}{4}\).

Setting this equal to \(\frac{1}{8}\), we solve:

\(-\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} + \frac{1}{4} = \frac{1}{8}\).

Rearranging gives \(-\frac{1}{2}ae^{-2a} - \frac{1}{4}e^{-2a} = -\frac{1}{8}\).

Multiplying through by -1 and simplifying, we find \(a = \frac{1}{2} \ln(4a + 2)\).

(b) To verify a lies between 0.5 and 1, calculate:

For \(a = 0.5\), \(\frac{1}{2} \ln(4 \times 0.5 + 2) = \frac{1}{2} \ln(4) = 0.693...\).

For \(a = 1\), \(\frac{1}{2} \ln(4 \times 1 + 2) = \frac{1}{2} \ln(6) = 0.895...\).

Since \(0.5 < 0.693 < 1\), a is between 0.5 and 1.

(c) Using the iterative formula \(a_{n+1} = \frac{1}{2} \ln(4a_n + 2)\), start with \(a_0 = 0.75\):

\(a_1 = \frac{1}{2} \ln(4 \times 0.75 + 2) = 0.8047\)

\(a_2 = \frac{1}{2} \ln(4 \times 0.8047 + 2) = 0.8261\)

\(a_3 = \frac{1}{2} \ln(4 \times 0.8261 + 2) = 0.8343\)

\(a_4 = \frac{1}{2} \ln(4 \times 0.8343 + 2) = 0.8373\)

\(a_5 = \frac{1}{2} \ln(4 \times 0.8373 + 2) = 0.8385\)

Continuing this process, we find \(a \approx 0.84\) to 2 decimal places.

(i) To solve \(\int_{1}^{a} \ln(2x) \, dx = 1\), we first integrate:

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

Evaluating from 1 to \(a\):

\([a \ln(2a) - a] - [1 \ln(2) - 1] = 1\).

Simplifying gives:

\(a \ln(2a) - a - \ln(2) + 1 = 1\).

\(a \ln(2a) - a = \ln(2)\).

\(a(\ln 2 + \ln a) - a = \ln(2)\).

\(a \ln a + a \ln 2 - a = \ln(2)\).

\(a \ln a = a - a \ln 2 + \ln(2)\).

\(a \ln a = a(1 - \ln 2) + \ln(2)\).

Rearranging gives:

\(a = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{a} \right)\).

(ii) Using the iterative formula:

Start with an initial guess, say \(a_1 = 2\).

\(a_2 = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{2} \right) \approx 1.9358\).

\(a_3 = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{1.9358} \right) \approx 1.9406\).

\(a_4 = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{1.9406} \right) \approx 1.9398\).

\(a_5 = \frac{1}{2} \exp \left( 1 + \frac{\ln 2}{1.9398} \right) \approx 1.9400\).

The value of \(a\) correct to 2 decimal places is \(1.94\).