Solution — 2024 Question 4 Graphs| Differentiation & Applications| Integration & Applications

(a)

Differentiating, $\frac{d y}{d x} = 15 x^{2} + 2 p x + 12.$

At the stationary point where $x = 2,$ $\frac{d y}{d x} = 0$ :

\begin{matrix} 15 (4) + 4 p + 12 & = 0 \\ 72 + 4 p & = 0 \\ 4 p & = - 72 \\ p & = - 18 ∎ \end{matrix}

(b)

\begin{matrix} \int_{0}^{2} (5 x^{3} - 18 x^{2} + 12 x + q) d x = 12 \\ {[\frac{5 x^{4}}{4} - 6 x^{3} + 6 x^{2} + q x]}_{0}^{2} = 12 \\ \frac{5 {(2)}^{4}}{4} - 6 {(2)}^{3} + 6 {(2)}^{2} + q (2) - (\frac{5 {(0)}^{4}}{4} - 6 {(0)}^{3} + 6 {(0)}^{2} + q (0)) = 12 \\ - 4 + 2 q = 12 \\ 2 q = 16 \\ q = 8 ∎ \end{matrix}

(c)