| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Solve equation involving derivatives |
| Difficulty | Standard +0.3 This is a standard A-level calculus question requiring product rule differentiation, solving a quadratic equation, and using the second derivative test. While it involves multiple steps and algebraic manipulation with surds, all techniques are routine P2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07o Increasing/decreasing: functions using sign of dy/dx1.07p Points of inflection: using second derivative1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks |
|---|---|
| \(H = t^{\frac{3}{2}} + \frac{1}{10}t^{\frac{5}{2}} - \frac{3}{20}t + 17\) | |
| \(\frac{dH}{dt} = \frac{1}{2}t^{-\frac{1}{2}} + \frac{1}{20}t^{\frac{3}{2}} - \frac{9}{8}t^{\frac{1}{2}}\) | M1, A1 |
| e.g. \(\frac{1}{2}t^{-\frac{1}{2}} + \frac{1}{20}t^{\frac{3}{2}} - \frac{9}{8}t^{\frac{1}{2}} = 0\) | |
| \(\frac{1}{2} + \frac{1}{20}t - \frac{9}{8}t^2 = 0\) | A1 |
| \(5t^2 - 18t - 20 = 0\) * | dM1, A1* |
| Answer | Marks |
|---|---|
| \(t = \frac{9 + \sqrt{181}}{5} = \text{awrt } 4.491\) | B1 |
| Answer | Marks |
|---|---|
| \(\frac{d^2H}{dt^2} = -\frac{1}{4}t^{-\frac{3}{2}} + \frac{9}{40}t^{\frac{1}{2}} - \frac{3}{16}t^{-\frac{1}{2}}\) | M1 |
| Substitute \(-\left("4.49"\right)^{-\frac{3}{2}} + \frac{9}{40}\left("4.49"\right)^{\frac{1}{2}} - \frac{3}{16}\left("4.49"\right)^{-\frac{1}{2}}\) | |
| \(\frac{d^2H}{dt^2} = (-0.3\ldots) < 0 \text{ max}\) * | A1* |
**7(a)**
$H = t^{\frac{3}{2}} + \frac{1}{10}t^{\frac{5}{2}} - \frac{3}{20}t + 17$ |
$\frac{dH}{dt} = \frac{1}{2}t^{-\frac{1}{2}} + \frac{1}{20}t^{\frac{3}{2}} - \frac{9}{8}t^{\frac{1}{2}}$ | M1, A1
e.g. $\frac{1}{2}t^{-\frac{1}{2}} + \frac{1}{20}t^{\frac{3}{2}} - \frac{9}{8}t^{\frac{1}{2}} = 0$ |
$\frac{1}{2} + \frac{1}{20}t - \frac{9}{8}t^2 = 0$ | A1
$5t^2 - 18t - 20 = 0$ * | dM1, A1*
(5 marks)
**7(b)**
$t = \frac{9 + \sqrt{181}}{5} = \text{awrt } 4.491$ | B1
(1 mark)
**7(c)**
$\frac{d^2H}{dt^2} = -\frac{1}{4}t^{-\frac{3}{2}} + \frac{9}{40}t^{\frac{1}{2}} - \frac{3}{16}t^{-\frac{1}{2}}$ | M1
Substitute $-\left("4.49"\right)^{-\frac{3}{2}} + \frac{9}{40}\left("4.49"\right)^{\frac{1}{2}} - \frac{3}{16}\left("4.49"\right)^{-\frac{1}{2}}$ |
$\frac{d^2H}{dt^2} = (-0.3\ldots) < 0 \text{ max}$ * | A1*
(2 marks)
**Guidance for 7(a):**
Derivatives which are already factorised before being set $= 0$ and have the quadratic factor $20 + 18t - 5t^2$ can score this mark for e.g. $\frac{dH}{dt} = \frac{1}{40}t^{-\frac{1}{2}}\left("20 + 18t - 5t^2"\right) = 0 \Rightarrow "5t^2 - 18t - 20" =\begin{enumerate}
\item The height of a river above a fixed point on the riverbed was monitored over a 7-day period.
\end{enumerate}
The height of the river, $H$ metres, $t$ days after monitoring began, was given by
$$H = \frac { \sqrt { t } } { 20 } \left( 20 + 6 t - t ^ { 2 } \right) + 17 \quad 0 \leqslant t \leqslant 7$$
Given that $H$ has a stationary value at $t = \alpha$\\
(a) use calculus to show that $\alpha$ satisfies the equation
$$5 \alpha ^ { 2 } - 18 \alpha - 20 = 0$$
(b) Hence find the value of $\alpha$, giving your answer to 3 decimal places.\\
(c) Use further calculus to prove that $H$ is a maximum at this value of $\alpha$.
\hfill \mbox{\textit{Edexcel P2 2023 Q7 [8]}}