(i) Sketch the graphs of \(y = \cot x\) and \(y = 1 + x^2\) over the interval \(0 < x < \frac{1}{2}\pi\). The graph of \(\cot x\) decreases from infinity to 0, while \(y = 1 + x^2\) is a parabola opening upwards starting at 1. They intersect at only one point in this interval, indicating one root.

(ii) Calculate \(\cot 0.5 - (1 + 0.5^2)\) and \(\cot 0.8 - (1 + 0.8^2)\). The sign changes between these values, confirming a root exists between 0.5 and 0.8.

(iii) Use the iterative formula \(x_{n+1} = \arctan\left( \frac{1}{1 + x_n^2} \right)\) starting with \(x_0 = 0.5\). Iterate until the result stabilizes to 2 decimal places:

1. \(x_1 = \arctan\left( \frac{1}{1 + 0.5^2} \right) = 0.5880\)

2. \(x_2 = \arctan\left( \frac{1}{1 + 0.5880^2} \right) = 0.6176\)

3. \(x_3 = \arctan\left( \frac{1}{1 + 0.6176^2} \right) = 0.6233\)

4. \(x_4 = \arctan\left( \frac{1}{1 + 0.6233^2} \right) = 0.6216\)

5. \(x_5 = \arctan\left( \frac{1}{1 + 0.6216^2} \right) = 0.6219\)

The root correct to 2 decimal places is 0.62.

(a) Sketch the graphs of \(y = e^x - 3\) and \(y = \sqrt{x}\). The graph of \(y = e^x - 3\) cuts the vertical axis at (0, -2) and has an increasing gradient. The graph of \(y = \sqrt{x}\) starts at (0, 0) and has a reducing gradient. The intersection of these graphs indicates the root.

(b) Calculate \(e^1 - 3 = -0.28\) and \(e^2 - 3 = 4.39\). Since \(\sqrt{1} = 1\) and \(\sqrt{2} \approx 1.41\), we have \(1 > -0.28\) and \(1.41 < 4.39\). Therefore, the root lies between 1 and 2.

(c) Rearrange \(\sqrt{x} = e^x - 3\) to \(x = \ln(3 + \sqrt{x})\). The iterative formula \(x_{n+1} = \ln(3 + \sqrt{x_n})\) converges to the root of the equation.

(d) Using the iterative formula, start with an initial guess, e.g., \(x_1 = 1\):

\(x_2 = \ln(3 + \sqrt{1}) = 1.3864\)

\(x_3 = \ln(3 + \sqrt{1.3864}) = 1.4297\)

\(x_4 = \ln(3 + \sqrt{1.4297}) = 1.4341\)

\(x_5 = \ln(3 + \sqrt{1.4341}) = 1.4344\)

\(x_6 = \ln(3 + \sqrt{1.4344}) = 1.4345\)

The root correct to 2 decimal places is 1.43.

(i) To show that \(4x^2 - 1 = \cot x\) has only one root in the interval \(0 < x < \frac{1}{2}\pi\), sketch the graphs of \(y = 4x^2 - 1\) and \(y = \cot x\). The graph of \(y = 4x^2 - 1\) is a parabola opening upwards with vertex at \((0, -1)\). The graph of \(y = \cot x\) has vertical asymptotes at \(x = 0\) and \(x = \frac{1}{2}\pi\), and decreases from positive infinity to negative infinity. The two graphs intersect only once in the interval \(0 < x < \frac{1}{2}\pi\).

(ii) To verify the root lies between 0.6 and 1, calculate \(4x^2 - 1 - \cot x\) at \(x = 0.6\) and \(x = 1\). At \(x = 0.6\), \(4(0.6)^2 - 1 - \cot(0.6)\) is negative, and at \(x = 1\), \(4(1)^2 - 1 - \cot(1)\) is positive. This indicates a sign change, confirming a root exists between 0.6 and 1.

(iii) Using the iterative formula \(x_{n+1} = \frac{1}{2}\sqrt{1 + \cot x_n}\), start with an initial guess, say \(x_0 = 0.7\). Calculate successive iterations:

\(x_1 = \frac{1}{2}\sqrt{1 + \cot(0.7)}\)

\(x_2 = \frac{1}{2}\sqrt{1 + \cot(x_1)}\)

Continue until the value stabilizes to 2 decimal places. The iterations converge to approximately 0.73.

(i) Integrate by parts: let \(u = x\) and \(dv = e^{\frac{1}{2}x} \, dx\). Then \(du = dx\) and \(v = 2e^{\frac{1}{2}x}\).

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

\(= 2xe^{\frac{1}{2}x} - 4e^{\frac{1}{2}x} + C\)

Substitute limits and equate to 6: \(2ae^{\frac{1}{2}a} - 4e^{\frac{1}{2}a} = 6\)

Rearrange: \(a = e^{-\frac{1}{2}a} + 2\)

(ii) Sketch the graphs of \(y = x\) and \(y = 2 + e^{-\frac{1}{2}x}\). They intersect at one point, indicating one root.

(iii) Check the sign of \(x - e^{-\frac{1}{2}x} - 2\) at \(x = 2\) and \(x = 2.5\). The sign changes, confirming a root between these values.

(iv) Use the iterative formula \(x_{n+1} = 2 + e^{-\frac{1}{2}x_n}\) with initial guess \(x_0 = 2\).

Calculate iterations: \(x_1 = 2.3033\), \(x_2 = 2.3113\), \(x_3 = 2.3105\), \(x_4 = 2.3107\).

The value of a is 2.31 to 2 decimal places.

(i) Sketch the graphs of \(y = 2 - x\) and \(y = \ln x\). The intersection point of these graphs represents the root of the equation \(2 - x = \ln x\). Since \(y = 2 - x\) is a straight line with a negative slope and \(y = \ln x\) is a logarithmic curve, they intersect at only one point, indicating one root.

(ii) Evaluate \(2 - x - \ln x\) at \(x = 1.4\) and \(x = 1.7\):

For \(x = 1.4\), \(2 - 1.4 - \ln(1.4) \approx 0.246\).

For \(x = 1.7\), \(2 - 1.7 - \ln(1.7) \approx -0.036\).

Since the sign changes between \(x = 1.4\) and \(x = 1.7\), the root lies in this interval.

(iii) Rearrange the equation \(x = \frac{1}{3}(4 + x - 2 \ln x)\) to \(2 - x = \ln x\), confirming that the root satisfies both equations.

(iv) Use the iterative formula:

\(x_1 = 1.5\)

\(x_2 = \frac{1}{3}(4 + 1.5 - 2 \ln(1.5)) \approx 1.5581\)

\(x_3 = \frac{1}{3}(4 + 1.5581 - 2 \ln(1.5581)) \approx 1.5560\)

\(x_4 = \frac{1}{3}(4 + 1.5560 - 2 \ln(1.5560)) \approx 1.5563\)

\(x_5 = \frac{1}{3}(4 + 1.5563 - 2 \ln(1.5563)) \approx 1.5562\)

The root correct to 2 decimal places is 1.56.