| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Tangent parallel/perpendicular to initial line |
| Difficulty | Challenging +1.2 This is a Further Maths FP2 polar coordinates question requiring the tangent condition formula (dy/dx or perpendicularity) and area integration. Part (a) needs solving dr/dθ and the perpendicularity condition, while part (b) is a standard polar area integral with trigonometric substitution. The techniques are bookwork for FP2 students, though the algebra requires care. Moderately above average difficulty due to being Further Maths content with multi-step calculus. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x = r\cos\theta \Rightarrow x = (1+\tan\theta)\cos\theta\) | M1 | States or implies \(x = r\cos\theta\) |
| \(x = \cos\theta + \sin\theta\), \(\frac{dx}{d\theta} = \cos\theta - \sin\theta\) | M1A1 | M1: Attempt to differentiate \(x = r\cos\theta\) or \(x = r\sin\theta\); A1: Correct derivative |
| Alt: \(\frac{dx}{d\theta} = \sec^2\theta\cos\theta + (1+\tan\theta)(-\sin\theta)\) | M1A1 | M1: Attempt to differentiate using product rule (dep on first M1); A1: correct (unsimplified) differentiation |
| \(\frac{dx}{d\theta} = 0 \Rightarrow \tan\theta = 1 \Rightarrow \theta = \ldots\) | dM1 | Set derivative \(= 0\) and attempt to solve for \(\theta\) (dependent on second M mark) |
| \(\theta = \frac{\pi}{4}\), \(r = 2\) | A1 | Both |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int r^2\,d\theta = \int (1+\tan\theta)^2\,d\theta\) | M1 | Use of \(\int r^2\,d\theta\) and \(r = 1+\tan\theta\); no limits needed |
| \((1+\tan\theta)^2 = 1 + 2\tan\theta + \tan^2\theta = 1 + 2\tan\theta + \sec^2\theta - 1\) | M1 | Expands and uses the correct identity |
| \(\int (2\tan\theta + \sec^2\theta)\,d\theta\) | A1 | Correct expression; need not be simplified, no limits needed |
| \(\left[2\ln\sec\theta + \tan\theta\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}\) | dM1A1 | dM1: Attempt to integrate — at least one trig term integrated (dependent on second M mark); A1: Correct integration, need not be simplified or include limits |
| \(R = \frac{1}{2}\left\{\left(2\ln\sec\frac{\pi}{3}+\tan\frac{\pi}{3}\right) - \left(2\ln\sec\frac{\pi}{4}+\tan\frac{\pi}{4}\right)\right\}\) | dM1 | Substitutes \(\frac{\pi}{3}\) and their \(\frac{\pi}{4}\) and subtracts (dependent on 2 previous method marks in (b)) |
| \(R = \frac{1}{2}\left\{\ln 2 + \sqrt{3} - 1\right\}\) | A1 | cao and cso |
## Question 8(a):
| Working | Mark | Guidance |
|---------|------|----------|
| $x = r\cos\theta \Rightarrow x = (1+\tan\theta)\cos\theta$ | M1 | States or implies $x = r\cos\theta$ |
| $x = \cos\theta + \sin\theta$, $\frac{dx}{d\theta} = \cos\theta - \sin\theta$ | M1A1 | M1: Attempt to differentiate $x = r\cos\theta$ or $x = r\sin\theta$; A1: Correct derivative |
| Alt: $\frac{dx}{d\theta} = \sec^2\theta\cos\theta + (1+\tan\theta)(-\sin\theta)$ | M1A1 | M1: Attempt to differentiate using product rule (dep on first M1); A1: correct (unsimplified) differentiation |
| $\frac{dx}{d\theta} = 0 \Rightarrow \tan\theta = 1 \Rightarrow \theta = \ldots$ | dM1 | Set derivative $= 0$ and attempt to solve for $\theta$ (dependent on second M mark) |
| $\theta = \frac{\pi}{4}$, $r = 2$ | A1 | Both |
**NB:** Use of $x = r\sin\theta$ can score M0M1A0M1A0 max
---
## Question 8(b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int r^2\,d\theta = \int (1+\tan\theta)^2\,d\theta$ | M1 | Use of $\int r^2\,d\theta$ and $r = 1+\tan\theta$; no limits needed |
| $(1+\tan\theta)^2 = 1 + 2\tan\theta + \tan^2\theta = 1 + 2\tan\theta + \sec^2\theta - 1$ | M1 | Expands and uses the correct identity |
| $\int (2\tan\theta + \sec^2\theta)\,d\theta$ | A1 | Correct expression; need not be simplified, no limits needed |
| $\left[2\ln\sec\theta + \tan\theta\right]_{\frac{\pi}{4}}^{\frac{\pi}{3}}$ | dM1A1 | dM1: Attempt to integrate — at least one trig term integrated (dependent on second M mark); A1: Correct integration, need not be simplified or include limits |
| $R = \frac{1}{2}\left\{\left(2\ln\sec\frac{\pi}{3}+\tan\frac{\pi}{3}\right) - \left(2\ln\sec\frac{\pi}{4}+\tan\frac{\pi}{4}\right)\right\}$ | dM1 | Substitutes $\frac{\pi}{3}$ and their $\frac{\pi}{4}$ and subtracts (dependent on 2 previous method marks in (b)) |
| $R = \frac{1}{2}\left\{\ln 2 + \sqrt{3} - 1\right\}$ | A1 | cao and cso |
**Total: 12 marks**
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Could you please share the correct pages that contain the actual mark scheme content? I'd be happy to format it as requested once I can see the questions, answers, mark allocations, and guidance notes.8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c21767d7-7331-47f7-8e59-06a0727c67c5-13_771_1036_260_593}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of part of the curve $C$ with polar equation
$$r = 1 + \tan \theta , \quad 0 \leqslant \theta < \frac { \pi } { 2 }$$
The tangent to the curve $C$ at the point $P$ is perpendicular to the initial line.
\begin{enumerate}[label=(\alph*)]
\item Find the polar coordinates of the point $P$.
The point $Q$ lies on the curve $C$, where $\theta = \frac { \pi } { 3 }$\\
The shaded region $R$ is bounded by $O P , O Q$ and the curve $C$, as shown in Figure 1
\item Find the exact area of $R$, giving your answer in the form
$$\frac { 1 } { 2 } ( \ln p + \sqrt { q } + r )$$
where $p , q$ and $r$ are integers to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2014 Q8 [12]}}