(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.