(a) Sketch the graph of \(y = 4 - x^2\) and \(y = \sec \frac{1}{2}x\). The intersection of these graphs in the interval \(0 \leq x < \pi\) shows the root. The graph of \(y = 4 - x^2\) is a downward-opening parabola with vertex at (0, 4) and intersects the x-axis at \(x = 2\) and \(x = -2\). The graph of \(y = \sec \frac{1}{2}x\) has a vertical asymptote at \(x = \pi\) and intersects the y-axis at (0, 1). The graphs intersect exactly once in the interval \(0 \leq x < \pi\).

(b) Calculate \(4 - x^2\) and \(\sec \frac{1}{2}x\) at \(x = 1\) and \(x = 2\):

At \(x = 1\), \(4 - 1^2 = 3\) and \(\sec \frac{1}{2}(1) = \sec 0.5 \approx 1.139\).

At \(x = 2\), \(4 - 2^2 = 0\) and \(\sec \frac{1}{2}(2) = \sec 1 \approx 1.850\).

Since \(3 > 1.139\) and \(0 < 1.850\), the root lies between 1 and 2.

(c) Use the iterative formula \(x_{n+1} = \sqrt{4 - \sec \frac{1}{2}x_n}\):

Start with \(x_0 = 1.5\).

\(x_1 = \sqrt{4 - \sec \frac{1}{2}(1.5)} \approx 1.6161\)

\(x_2 = \sqrt{4 - \sec \frac{1}{2}(1.6161)} \approx 1.6025\)

\(x_3 = \sqrt{4 - \sec \frac{1}{2}(1.6025)} \approx 1.6003\)

\(x_4 = \sqrt{4 - \sec \frac{1}{2}(1.6003)} \approx 1.6000\)

The root correct to 2 decimal places is 1.60.

(a) Sketch the graphs of \(y = \cot \frac{1}{2}x\) and \(y = 1 + e^{-x}\). The intersection of these graphs in the interval \(0 < x \leq \pi\) shows there is exactly one root.

(b) Calculate the values at \(x = 1\) and \(x = 1.5\):

For \(x = 1\), \(\cot \frac{1}{2} \times 1 = \cot 0.5 \approx 1.8305\) and \(1 + e^{-1} \approx 1.3679\).

For \(x = 1.5\), \(\cot \frac{1}{2} \times 1.5 = \cot 0.75 \approx 1.3764\) and \(1 + e^{-1.5} \approx 1.2231\).

The change in sign between these values confirms a root between 1 and 1.5.

(c) Using the iterative formula:

Start with an initial guess, say \(x_0 = 1\).

\(x_1 = 2 \arctan \left( \frac{1}{1 + e^{-1}} \right) \approx 1.3524\)

\(x_2 = 2 \arctan \left( \frac{1}{1 + e^{-1.3524}} \right) \approx 1.3402\)

\(x_3 = 2 \arctan \left( \frac{1}{1 + e^{-1.3402}} \right) \approx 1.3398\)

\(x_4 = 2 \arctan \left( \frac{1}{1 + e^{-1.3398}} \right) \approx 1.3398\)

The root correct to 2 decimal places is 1.34.

(a) To show that the equation \(\csc x = 1 + e^{-\frac{1}{2}x}\) has exactly two roots in the interval \(0 < x < \pi\), sketch the graphs of \(y = \csc x\) and \(y = 1 + e^{-\frac{1}{2}x}\).

The graph of \(y = \csc x\) is U-shaped, roughly symmetrical about \(x = \frac{\pi}{2}\), with \(y\left(\frac{\pi}{2}\right) = 1\), and the domain is at least \(\left(\frac{\pi}{6}, \frac{5\pi}{6}\right)\).

The graph of \(y = 1 + e^{-\frac{1}{2}x}\) is an exponential curve with \(y(0) = 2\), a negative gradient, always increasing, and \(y(\pi) > 1\). Mark the intersections with dots or crosses to show the roots.

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

Calculate the iterations:

\(x_2 = 2.3217\)

\(x_3 = 2.2760\)

\(x_4 = 2.2824\)

The sequence converges to 2.28 when rounded to 2 decimal places.

(a) To show that the equation \(x^5 = 2 + x\) has exactly one real root, sketch the graphs of \(y = x^5\) and \(y = x + 2\). The intersection of these graphs represents the solution to the equation. Since \(x^5\) is a strictly increasing function and \(x + 2\) is a linear function, they intersect at exactly one point, confirming one real root.

(b) Consider the iterative formula \(x_{n+1} = \frac{4x_n^5 + 2}{5x_n^4 - 1}\). If this sequence converges, it must satisfy the equation \(x = \frac{4x^5 + 2}{5x^4 - 1}\). Rearranging gives \(x^5 = 2 + x\), which is the equation from part (a). Therefore, if the sequence converges, it converges to the root of the equation.

(c) Using the iterative formula with \(x_1 = 1.5\):

1. Calculate \(x_2 = \frac{4(1.5)^5 + 2}{5(1.5)^4 - 1} = 1.27103\).

2. Calculate \(x_3 = \frac{4(1.27103)^5 + 2}{5(1.27103)^4 - 1} = 1.26713\).

3. Calculate \(x_4 = \frac{4(1.26713)^5 + 2}{5(1.26713)^4 - 1} = 1.26700\).

4. Calculate \(x_5 = \frac{4(1.26700)^5 + 2}{5(1.26700)^4 - 1} = 1.26700\).

The sequence converges to 1.267, correct to 3 decimal places.

(a) Sketch the graph \(y = \sec x\) and the line \(y = 2 - \frac{1}{2}x\). The intersection point in the interval \(0 \leq x < \frac{1}{2}\pi\) indicates the root. The graphs intersect exactly once in this interval.

(b) Calculate \(\sec(0.8) \approx 1.139\) and \(2 - \frac{1}{2}(0.8) = 1.6\). Calculate \(\sec(1) \approx 1.850\) and \(2 - \frac{1}{2}(1) = 1.5\). Since \(\sec(0.8) < 1.6\) and \(\sec(1) > 1.5\), the root lies between 0.8 and 1.

(c) Using the iterative formula:

\(x_1 = \cos^{-1}\left(\frac{2}{4-0.8}\right) \approx 0.9273\)

\(x_2 = \cos^{-1}\left(\frac{2}{4-0.9273}\right) \approx 0.8834\)

\(x_3 = \cos^{-1}\left(\frac{2}{4-0.8834}\right) \approx 0.8800\)

\(x_4 = \cos^{-1}\left(\frac{2}{4-0.8800}\right) \approx 0.8798\)

The root correct to 2 decimal places is 0.88.