(a) To find the stationary point, we need to differentiate \(y = \frac{x^3}{e^x - 1}\) and set the derivative to zero.

Using the quotient rule, \(\frac{d}{dx} \left( \frac{x^3}{e^x - 1} \right) = \frac{(e^x - 1)(3x^2) - x^3(e^x)}{(e^x - 1)^2}\).

Setting the numerator to zero for a stationary point: \((e^x - 1)(3x^2) - x^3(e^x) = 0\).

Simplifying gives \(3x^2 e^x - 3x^2 - x^3 e^x = 0\).

Rearranging, \(x^2 e^x (3 - x) = 3x^2\).

Dividing by \(x^2\) (assuming \(x \neq 0\)), \(e^x (3 - x) = 3\).

Thus, \(3 - x = 3e^{-x}\) or \(x = 3(1 - e^{-x})\).

Therefore, \(p = 3(1 - e^{-p})\).

(b) To verify \(p\) lies between 2.5 and 3, calculate \(f(p) = 3(1 - e^{-p}) - p\).

For \(p = 2.5\), \(f(2.5) = 3(1 - e^{-2.5}) - 2.5 \approx 0.351\).

For \(p = 3\), \(f(3) = 3(1 - e^{-3}) - 3 \approx -0.950\).

Since \(f(2.5) > 0\) and \(f(3) < 0\), by the Intermediate Value Theorem, \(p\) lies between 2.5 and 3.

(c) Using the iterative formula \(p_{n+1} = 3(1 - e^{-p_n})\), start with an initial guess, say \(p_0 = 2.5\).

Calculate successive iterations:

\(p_1 = 3(1 - e^{-2.5}) \approx 2.7639\)

\(p_2 = 3(1 - e^{-2.7639}) \approx 2.8113\)

\(p_3 = 3(1 - e^{-2.8113}) \approx 2.8212\)

\(p_4 = 3(1 - e^{-2.8212}) \approx 2.8231\)

\(p_5 = 3(1 - e^{-2.8231}) \approx 2.8235\)

The iterations converge to \(p = 2.82\) to 2 decimal places.

(a) Differentiate \(y = \frac{x}{\cos^2 x}\) using the quotient rule:

\(\frac{dy}{dx} = \frac{\cos^2 x + 2x \sin x \cos x}{\cos^4 x}\).

At \(x = a\), set \(\frac{dy}{dx} = 12\):

\(12 = \frac{\cos^2 a + 2a \sin a \cos a}{\cos^4 a}\).

Simplify to obtain:

\(12 \cos^3 a = \cos a + 2a \sin a\).

\(\cos^3 a = \frac{\cos a + 2a \sin a}{12}\).

\(a = \cos^{-1} \left( \sqrt[3]{\frac{\cos a + 2a \sin a}{12}} \right)\).

(b) Calculate \(\cos 0.9 \approx 0.622\) and \(\cos 1 \approx 0.540\).

\(\cos 0.9 > 0.553\) and \(\cos 1 < 0.570\), so \(0.9 < a < 1\).

(c) Use the iterative formula:

\(a_{n+1} = \cos^{-1} \left( \sqrt[3]{\frac{\cos a_n + 2a_n \sin a_n}{12}} \right)\).

Starting with \(a_0 = 0.9\), iterate to find:

\(a_1 \approx 0.95\), \(a_2 \approx 0.9743\), \(a_3 \approx 0.9694\), \(a_4 \approx 0.9704\).

Final answer: \(a \approx 0.97\).

(a) To find the stationary point, we need to differentiate \(y = x \sqrt{\sin x}\) and set the derivative to zero. Using the product rule:

\(\frac{dy}{dx} = \sqrt{\sin x} + x \cdot \frac{1}{2\sqrt{\sin x}} \cdot \cos x\)

\(= \sqrt{\sin x} + \frac{x \cos x}{2\sqrt{\sin x}}\)

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

\(\sqrt{\sin x} + \frac{x \cos x}{2\sqrt{\sin x}} = 0\)

\(2\sin x + x \cos x = 0\)

\(\tan x = -\frac{1}{2}x\)

(b) Calculate \(\tan 2\) and \(\tan 2.5\):

\(\tan 2 \approx -2.18\) and \(\tan 2.5 \approx -0.747\)

Since \(-2.18 < -1 < -0.747\), \(a\) lies between 2 and 2.5.

(c) The iterative formula is \(x_{n+1} = \pi - \arctan\left(\frac{1}{2}x_n\right)\). If it converges, it converges to \(a\) where \(\tan a = -\frac{1}{2}a\).

(d) Using the iterative formula:

Start with \(x_0 = 2.25\):

\(x_1 = \pi - \arctan\left(\frac{1}{2} \times 2.25\right) \approx 2.2974\)

\(x_2 = \pi - \arctan\left(\frac{1}{2} \times 2.2974\right) \approx 2.2871\)

\(x_3 = \pi - \arctan\left(\frac{1}{2} \times 2.2871\right) \approx 2.2893\)

\(x_4 = \pi - \arctan\left(\frac{1}{2} \times 2.2893\right) \approx 2.2888\)

Thus, \(a \approx 2.29\) to 2 decimal places.

(a) To find \(\frac{dy}{dx}\), use the chain rule. Let \(u = \tan x\), then \(y = \sqrt{u}\). The derivative \(\frac{dy}{du} = \frac{1}{2\sqrt{u}}\) and \(\frac{du}{dx} = \sec^2 x\). Therefore, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{\tan x}} \cdot \sec^2 x = \frac{1 + \tan^2 x}{2\sqrt{\tan x}}\).

Verify at \(x = \frac{1}{4}\pi\): \(\tan \frac{1}{4}\pi = 1\), so \(\frac{dy}{dx} = \frac{1 + 1^2}{2\sqrt{1}} = 1\).

(b) Substitute \(t = \tan a\) into the equation from part (a): \(\frac{1 + t^2}{2\sqrt{t}} = 1\). Rearrange to get \((1 + t^2)^2 = 4t\), leading to \(t^4 + 2t^2 - 4t + 1 = 0\). Divide by \(t - 1\) to obtain \(t^3 + t^2 + 3t - 1 = 0\).

(c) Use the iterative formula \(a_{n+1} = \arctan \left( \frac{1}{3} (1 - \tan^2 a_n - \tan^3 a_n) \right)\). Start with an initial guess, e.g., \(a_0 = 0.3\), and iterate: \(a_1 = 0.2854\), \(a_2 = 0.2894\), \(a_3 = 0.2883\), etc. Continue until \(a\) stabilizes to 2 decimal places: \(a = 0.29\).

(a) To find the maximum point, we differentiate \(y = \frac{\arctan x}{\sqrt{x}}\) using the quotient rule:

\(\frac{dy}{dx} = \frac{\sqrt{x} \cdot \frac{1}{1+x^2} - \arctan x \cdot \frac{1}{2\sqrt{x}}}{x}\).

Setting \(\frac{dy}{dx} = 0\) gives:

\(\sqrt{x} \cdot \frac{1}{1+x^2} = \arctan x \cdot \frac{1}{2\sqrt{x}}\).

Rearranging and simplifying leads to:

\(a = \tan \left( \frac{2a}{1 + a^2} \right)\).

(b) Calculate \(a = \tan \left( \frac{2a}{1 + a^2} \right)\) for \(a = 1.3\) and \(a = 1.5\):

For \(a = 1.3\), \(\tan \left( \frac{2 \times 1.3}{1 + 1.3^2} \right) \approx 1.448\).

For \(a = 1.5\), \(\tan \left( \frac{2 \times 1.5}{1 + 1.5^2} \right) \approx 1.322\).

Since \(1.3 < 1.448\) and \(1.5 > 1.322\), \(a\) lies between 1.3 and 1.5.

(c) Using the iterative formula \(a_{n+1} = \tan \left( \frac{2a_n}{1 + a_n^2} \right)\):

Start with \(a_0 = 1.4\).

\(a_1 = \tan \left( \frac{2 \times 1.4}{1 + 1.4^2} \right) \approx 1.3914\).

\(a_2 = \tan \left( \frac{2 \times 1.3914}{1 + 1.3914^2} \right) \approx 1.3900\).

\(a_3 = \tan \left( \frac{2 \times 1.3900}{1 + 1.3900^2} \right) \approx 1.3900\).

Thus, \(a \approx 1.39\) correct to 2 decimal places.