| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Numerical integration |
| Type | Trapezium rule applied to real-world data |
| Difficulty | Standard +0.3 This is a standard P3 applied calculus question with routine techniques: solving f(x)=0 using sign change (part a), finding maximum via differentiation and rearranging (part b), and applying a given iteration formula (part c). All steps are algorithmic with no novel insight required, making it slightly easier than average. |
| Spec | 1.06b Gradient of e^(kx): derivative and exponential model1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(h=0\), cancels by \(d\): \(0 = 1.5d - 0.5d\,e^{0.02d} \Rightarrow e^{0.02d} = 3\) | M1 | Sets \(h=0\), cancels/factors \(d\), proceeds to form \(ae^{\pm 0.02d}=b\) where \(ab>0\) |
| \(0.02d = \ln 3 \Rightarrow d = 54.93\) | dM1 A1 | Solves using logs; dependent on previous M; 54.93 (m) cao; at least one intermediate stage required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dh}{dx} = 1.5 - \left(0.5e^{0.02x} + 0.5x \times 0.02\,e^{0.02x}\right)\) | M1A1 | Product rule on \(xe^{0.02x}\) achieving \(\pm Axe^{0.02x} \pm Be^{0.02x}\); A can be 1 |
| Sets \(0 = 1.5 - 0.5e^{0.02x} - 0.5x \times 0.02\,e^{0.02x} \Rightarrow e^{0.02x}(0.5 + 0.5x \times 0.02) = 1.5\) | dM1 | Attempts to make \(e^{\pm 0.02x}\) the subject; condone slips in rearrangement |
| \(e^{0.02x} = \frac{1.5}{(0.5+0.5x \times 0.02)} = \frac{150}{50+x} \Rightarrow x = 50\ln\!\left(\frac{150}{50+x}\right)\) * | A1* | Achieves given answer with no errors; must see unsimplified expression or intermediate step before final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| awrt \(31.43\) | M1 A1 | Attempts iteration formula correctly seen once; e.g. award for \(x_2 = 50\ln\!\left(\frac{150}{80}\right)\) or awrt 31.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| awrt \(30.88\) metres | A1 | Units required; if contradiction between answers, main body of work takes precedence |
# Question 8:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $h=0$, cancels by $d$: $0 = 1.5d - 0.5d\,e^{0.02d} \Rightarrow e^{0.02d} = 3$ | M1 | Sets $h=0$, cancels/factors $d$, proceeds to form $ae^{\pm 0.02d}=b$ where $ab>0$ |
| $0.02d = \ln 3 \Rightarrow d = 54.93$ | dM1 A1 | Solves using logs; dependent on previous M; 54.93 (m) cao; at least one intermediate stage required |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dh}{dx} = 1.5 - \left(0.5e^{0.02x} + 0.5x \times 0.02\,e^{0.02x}\right)$ | M1A1 | Product rule on $xe^{0.02x}$ achieving $\pm Axe^{0.02x} \pm Be^{0.02x}$; A can be 1 |
| Sets $0 = 1.5 - 0.5e^{0.02x} - 0.5x \times 0.02\,e^{0.02x} \Rightarrow e^{0.02x}(0.5 + 0.5x \times 0.02) = 1.5$ | dM1 | Attempts to make $e^{\pm 0.02x}$ the subject; condone slips in rearrangement |
| $e^{0.02x} = \frac{1.5}{(0.5+0.5x \times 0.02)} = \frac{150}{50+x} \Rightarrow x = 50\ln\!\left(\frac{150}{50+x}\right)$ * | A1* | Achieves given answer with no errors; must see unsimplified expression or intermediate step before final answer |
## Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| awrt $31.43$ | M1 A1 | Attempts iteration formula correctly seen once; e.g. award for $x_2 = 50\ln\!\left(\frac{150}{80}\right)$ or awrt 31.4 |
## Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| awrt $30.88$ metres | A1 | Units required; if contradiction between answers, main body of work takes precedence |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5a695b86-1660-4c06-ac96-4cdb07af9a2e-26_499_551_246_758}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 is a graph showing the path of a golf ball after the ball has been hit until it first hits the ground.
The vertical height, $h$ metres, of the ball above the ground has been plotted against the horizontal distance travelled, $x$ metres, measured from where the ball was hit.
The ball travels a horizontal distance of $d$ metres before it first hits the ground.\\
The ball is modelled as a particle travelling in a vertical plane above horizontal ground.\\
The path of the ball is modelled by the equation
$$h = 1.5 x - 0.5 x \mathrm { e } ^ { 0.02 x } \quad 0 \leqslant x \leqslant d$$
\section*{Use the model to answer parts (a), (b) and (c).}
\begin{enumerate}[label=(\alph*)]
\item Find the value of $d$, giving your answer to 2 decimal places.\\
(Solutions relying entirely on calculator technology are not acceptable.)
\item Show that the maximum value of $h$ occurs when
$$x = 50 \ln \left( \frac { 150 } { x + 50 } \right)$$
Using the iteration formula
$$x _ { n + 1 } = 50 \ln \left( \frac { 150 } { x _ { n } + 50 } \right) \quad \text { with } x _ { 1 } = 30$$
\item \begin{enumerate}[label=(\roman*)]
\item find the value of $x _ { 2 }$ to 2 decimal places,
\item find, by repeated iteration, the horizontal distance travelled by the golf ball before it reaches its maximum height. Give your answer to 2 decimal places.\\
\includegraphics[max width=\textwidth, alt={}, center]{5a695b86-1660-4c06-ac96-4cdb07af9a2e-26_2270_56_309_1981}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel P3 2024 Q8 [10]}}