To find the stationary points, we first differentiate the equation \(y = \frac{k^2}{x+2} + x\) with respect to \(x\):

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

Set \(\frac{dy}{dx} = 0\) to find the stationary points:

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

Solving for \(x\), we get:

\((x+2)^2 = k^2\).

\(x+2 = \pm k\).

Thus, \(x = -2 + k\) or \(x = -2 - k\).

Next, we find the second derivative to determine the nature of the stationary points:

\(\frac{d^2y}{dx^2} = 2k^2(x+2)^{-3}\).

Substitute \(x = -2 + k\) into \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{2}{k}\), which is positive, indicating a minimum.

Substitute \(x = -2 - k\) into \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = \frac{-2}{k}\), which is negative, indicating a maximum.

To find the gradient of the curve, we need to differentiate \(y = \frac{2}{\sqrt{5x - 6}}\) with respect to \(x\).

First, rewrite the equation as \(y = 2(5x - 6)^{-\frac{1}{2}}\).

Using the chain rule, the derivative \(\frac{dy}{dx}\) is:

\(\frac{dy}{dx} = 2 \times -\frac{1}{2} \times (5x - 6)^{-\frac{3}{2}} \times 5\).

Simplifying, \(\frac{dy}{dx} = -\frac{5}{(5x - 6)^{\frac{3}{2}}}\).

Substitute \(x = 2\) into the derivative:

\(\frac{dy}{dx} = -\frac{5}{(5(2) - 6)^{\frac{3}{2}}} = -\frac{5}{(10 - 6)^{\frac{3}{2}}} = -\frac{5}{4^{\frac{3}{2}}}\).

Since \(4^{\frac{3}{2}} = 8\), the gradient is \(-\frac{5}{8}\).

To find the stationary point, we first differentiate the equation \(y = ax^{\frac{1}{2}} - 2x\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{1}{2}ax^{-\frac{1}{2}} - 2\).

At a stationary point, \(\frac{dy}{dx} = 0\). Substituting \(x = 9\) into the derivative:

\(0 = \frac{1}{2}a(9)^{-\frac{1}{2}} - 2\).

This simplifies to:

\(0 = \frac{a}{6} - 2\).

Solving for \(a\):

\(\frac{a}{6} = 2\)

\(a = 12\).

Now, substitute \(a = 12\) back into the original equation to find the \(y\)-coordinate of \(P\):

\(y = 12 \times 9^{\frac{1}{2}} - 2 \times 9\).

\(y = 12 \times 3 - 18\).

\(y = 36 - 18\).

\(y = 18\).

We start with the equations:

1. \(u = x^2 y\) 2. \(y + 3x = 9\)

Express \(y\) in terms of \(x\):

\(y = 9 - 3x\)

Substitute \(y\) into the equation for \(u\):

\(u = x^2 (9 - 3x) = 9x^2 - 3x^3\)

Differentiate \(u\) with respect to \(x\):

\(\frac{du}{dx} = 18x - 9x^2\)

Set \(\frac{du}{dx} = 0\) to find the stationary points:

\(18x - 9x^2 = 0\)

\(9x(2 - x) = 0\)

\(x = 0\) or \(x = 2\)

Since \(x\) cannot be zero (as \(x\) is non-zero), we have \(x = 2\).

Substitute \(x = 2\) back into \(y = 9 - 3x\):

\(y = 9 - 3(2) = 3\)

Calculate \(u\):

\(u = (2)^2 (3) = 12\)

To determine if this is a maximum or minimum, find the second derivative:

\(\frac{d^2u}{dx^2} = 18 - 18x\)

Evaluate at \(x = 2\):

\(\frac{d^2u}{dx^2} = 18 - 18(2) = -18\)

Since \(\frac{d^2u}{dx^2} < 0\), the stationary point is a maximum.

(i) Substitute \(u = x^{\frac{1}{2}}\), then \(f'(x) = 3u + \frac{3}{u} - 10 = 0\).

Multiply through by \(u\): \(3u^2 - 10u + 3 = 0\).

Factorize: \((3u - 1)(u - 3) = 0\).

So, \(u = \frac{1}{3}\) or \(u = 3\).

Since \(u = x^{\frac{1}{2}}\), \(x = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\) or \(x = 3^2 = 9\).

(ii) Differentiate \(f'(x)\) to find \(f''(x) = \frac{3}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{-\frac{3}{2}}\).

At \(x = \frac{1}{9}\), \(f''(x) = \frac{3}{2}(3) - \frac{3}{2}(27) = -36 < 0\), indicating a maximum.

At \(x = 9\), \(f''(x) = \frac{3}{2} \times \frac{1}{3} - \frac{3}{2} \times \frac{1}{27} = \frac{4}{9} > 0\), indicating a minimum.