| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2023 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.8 This question requires knowledge of polar area formula, hyperbolic function identities (sinh θ + cosh θ = e^θ), and integration of exponentials. While it involves Further Maths content (hyperbolic functions and polar coordinates), the solution path is relatively straightforward once the key identity is recognized: simplify r² using the identity, then integrate. The algebraic manipulation is non-trivial but follows standard techniques, making this harder than average but not exceptionally difficult for Further Maths students. |
| Spec | 4.07a Hyperbolic definitions: sinh, cosh, tanh as exponentials4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Area} = \frac{1}{2}\int_0^{\pi} r^2 \, d\theta = \frac{1}{2}\int_0^{\pi}\left[2\sqrt{\sinh\theta + \cosh\theta}\right]^2 \{d\theta\}\) or \(\frac{1}{2}\int_0^{\pi} 4(\sinh\theta + \cosh\theta)\{d\theta\}\) | B1 | Correct area formula applied, including the \(\frac{1}{2}\) and correct limits; may be seen later, \(d\theta\) may be implied |
| \(= 2\left[\cosh\theta + \sinh\theta\right]_0^{\pi}\) or \(2\int_0^{\pi}\left(\frac{e^\theta - e^{-\theta}}{2} + \frac{e^\theta + e^{-\theta}}{2}\right)d\theta = 2\int_0^{\pi} e^\theta \, d\theta = 2\left[e^\theta\right]_0^{\pi}\) | M1 | Attempts the integration \(\sinh\theta \to \pm\cosh\theta\) and \(\cosh\theta \to \pm\sinh\theta\), or in terms of exponentials \(\int e^{\lambda\theta}\,d\theta = \frac{1}{\lambda}e^{\lambda\theta}\) |
| \(= 2\Big((\cosh\pi + \sinh\pi) - (\cosh 0 + \sinh 0)\Big)\) \(= 2\!\left(\!\left(\frac{e^\pi+e^{-\pi}}{2}+\frac{e^\pi-e^{-\pi}}{2}\right)-(1+0)\right)\) or \(= 2(e^\pi - e^0)\) | M1 | Applies limits to the integral, subtracts (must be an attempt to integrate) and uses exponential definitions to achieve answer in suitable form. Condone inclusion of \(i\) or a missing \(\frac{1}{2}\) from definitions. May be implied |
| \(= 2e^\pi - 2\) or \(2(e^\pi - 1)\) | A1 | Correct exact answer, no \(i\) |
| (4) | (4 marks) |
## Question 1:
**Area using polar formula, integration, and evaluation**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Area} = \frac{1}{2}\int_0^{\pi} r^2 \, d\theta = \frac{1}{2}\int_0^{\pi}\left[2\sqrt{\sinh\theta + \cosh\theta}\right]^2 \{d\theta\}$ or $\frac{1}{2}\int_0^{\pi} 4(\sinh\theta + \cosh\theta)\{d\theta\}$ | **B1** | Correct area formula applied, including the $\frac{1}{2}$ and correct limits; may be seen later, $d\theta$ may be implied |
| $= 2\left[\cosh\theta + \sinh\theta\right]_0^{\pi}$ or $2\int_0^{\pi}\left(\frac{e^\theta - e^{-\theta}}{2} + \frac{e^\theta + e^{-\theta}}{2}\right)d\theta = 2\int_0^{\pi} e^\theta \, d\theta = 2\left[e^\theta\right]_0^{\pi}$ | **M1** | Attempts the integration $\sinh\theta \to \pm\cosh\theta$ and $\cosh\theta \to \pm\sinh\theta$, or in terms of exponentials $\int e^{\lambda\theta}\,d\theta = \frac{1}{\lambda}e^{\lambda\theta}$ |
| $= 2\Big((\cosh\pi + \sinh\pi) - (\cosh 0 + \sinh 0)\Big)$ $= 2\!\left(\!\left(\frac{e^\pi+e^{-\pi}}{2}+\frac{e^\pi-e^{-\pi}}{2}\right)-(1+0)\right)$ or $= 2(e^\pi - e^0)$ | **M1** | Applies limits to the integral, subtracts (must be an attempt to integrate) and uses exponential definitions to achieve answer in suitable form. Condone inclusion of $i$ or a missing $\frac{1}{2}$ from definitions. May be implied |
| $= 2e^\pi - 2$ or $2(e^\pi - 1)$ | **A1** | Correct exact answer, no $i$ |
| | **(4)** | **(4 marks)** |1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{59a57888-8aa8-4ed8-b704-ebf3980c0344-02_300_1006_242_532}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curve with polar equation
$$r = 2 \sqrt { \sinh \theta + \cosh \theta } \quad 0 \leqslant \theta \leqslant \pi$$
The region $R$, shown shaded in Figure 1, is bounded by the initial line, the curve and the line with equation $\theta = \pi$
Use algebraic integration to determine the exact area of $R$ giving your answer in the form $p \mathrm { e } ^ { q } - r$ where $p , q$ and $r$ are real numbers to be found.
\hfill \mbox{\textit{Edexcel CP2 2023 Q1 [4]}}