(a) At the stationary point, \(\frac{dy}{dx} = 0\). So, \(6(3 \times 2 - 5)^3 - k \times 2^2 = 0\).

\(6(1)^3 - 4k = 0\)

\(6 - 4k = 0\)

\(4k = 6\)

\(k = \frac{3}{2}\)

(c) \(\frac{d^2y}{dx^2} = [3x][18(3x-5)^2][-2kx]\)

Alternative method: \(486x^2 - 1623x + 1350\) or \(-1620x - 2kx\)

(d) At \(x = 2\), \(\frac{d^2y}{dx^2} = 54(3 \times 2 - 5)^2 - 4k\) or \(48\)

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.

(b) To find the stationary point, we first find the derivative \(\frac{dy}{dx}\) and set it to zero. The derivative is:

\(\frac{dy}{dx} = 3(3x+4)^{-0.5} - 1\)

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

\(3(3x+4)^{-0.5} - 1 = 0\)

Solving for \(x\):

\(x = \frac{5}{3}\)

Substitute \(x = \frac{5}{3}\) back into the original equation to find \(y\):

\(y = 2\left(3\left(\frac{5}{3}\right) + 4\right)^{0.5} - \frac{5}{3}\)

\(y = 2\left(5 + 4\right)^{0.5} - \frac{5}{3}\)

\(y = 2 \times 3 - \frac{5}{3}\)

\(y = 6 - \frac{5}{3}\)

\(y = \frac{18}{3} - \frac{5}{3}\)

\(y = \frac{13}{3}\)

Thus, the coordinates of the stationary point are \(\left( \frac{5}{3}, \frac{13}{3} \right)\).

(c) To determine the nature of the stationary point, we find the second derivative \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -\frac{9}{2}(3x+4)^{-1.5}\)

At \(x = \frac{5}{3}\), \(\frac{d^2y}{dx^2}\) is negative, indicating that the point is a maximum.

To find the gradient of the curve, we need to differentiate the given equation:

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

The derivative \(\frac{dy}{dx}\) is:

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

Substitute \(x = \frac{1}{4}\) into the derivative:

\(\frac{dy}{dx} = \frac{1}{2k}(\frac{1}{4})^{-\frac{1}{2}} - \frac{1}{2}(\frac{1}{4})^{-\frac{3}{2}}\).

Simplify \((\frac{1}{4})^{-\frac{1}{2}} = 2\) and \((\frac{1}{4})^{-\frac{3}{2}} = 8\):

\(\frac{dy}{dx} = \frac{1}{2k} \cdot 2 - \frac{1}{2} \cdot 8\).

\(\frac{dy}{dx} = \frac{1}{k} - 4\).

It is given that \(\frac{dy}{dx} = 3\) when \(x = \frac{1}{4}\):

\(3 = \frac{1}{k} - 4\).

Solving for \(k\):

\(3 + 4 = \frac{1}{k}\).

\(7 = \frac{1}{k}\).

\(k = \frac{1}{7}\).

(a) To find \(\frac{dy}{dx}\), differentiate \(y = 2x + 1 + \frac{1}{2x+1}\).

The derivative of \(2x + 1\) is 2.

For \(\frac{1}{2x+1}\), use the chain rule: \(\frac{d}{dx} \left( (2x+1)^{-1} \right) = -1(2x+1)^{-2} \cdot 2 = -2(2x+1)^{-2}\).

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

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx} = 2 - 2(2x+1)^{-2}\).

The derivative of \(-2(2x+1)^{-2}\) is \(4(2x+1)^{-3} \cdot 2 = 8(2x+1)^{-3}\).

Thus, \(\frac{d^2y}{dx^2} = 8(2x+1)^{-3}\).

(b) To find the stationary point, set \(\frac{dy}{dx} = 0\).

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

\((2x+1)^{-2} = 1\) implies \(2x+1 = \pm 1\).

Solving \(2x+1 = 1\), we get \(x = 0\).

Substitute \(x = 0\) into \(y = 2x + 1 + \frac{1}{2x+1}\) to find \(y = 2(0) + 1 + \frac{1}{1} = 2\).

Thus, the stationary point is \((0, 2)\).

Since \(\frac{d^2y}{dx^2} = 8(2x+1)^{-3} > 0\) for \(x > -\frac{1}{2}\), the stationary point is a minimum.