| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/1 (Pre-U Further Mathematics Paper 1) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| Topic | Polar coordinates |
| Type | Polar curves with trigonometric identities |
| Difficulty | Challenging +1.8 This Pre-U Further Maths question requires multiple sophisticated techniques: proving a trigonometric identity with substitution, differentiating an inverse tan function (chain rule with implicit understanding), converting Cartesian to polar coordinates, optimizing in polar form, and computing polar area integrals. While each individual step follows established methods, the combination of diverse topics (trigonometry, calculus, coordinate geometry) and the need to connect results across parts (using part (a) in (b), deducing symmetry for total area) makes this substantially harder than typical A-level questions but not exceptionally difficult for Further Maths students who have practiced these techniques. |
| Spec | 4.08g Derivatives: inverse trig and hyperbolic functions4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| (i)(a) Use of \(\sin 2x = \frac{2t}{1+t^2}\) in \(\frac{2}{2 - \sin 2x}\) where \(t = \tan x\) | M1 | |
| \(\frac{2}{2-\sin 2x} = \frac{2}{2-\left(\frac{2t}{1+t^2}\right)} = \frac{1+t^2}{1-t+t^2}\) ANSWER GIVEN | A1 | |
| [2] | ||
| (i)(b) \(y = \tan^{-1}\left(1 - \frac{2\tan x}{\sqrt{3}}\right) \Rightarrow \frac{dy}{dx} = \frac{1}{1+\left(\frac{1-2t}{\sqrt{3}}\right)^2} \times \frac{-2}{\sqrt{3}}\sec^2 x\) | M1, A1 | |
| or \(\tan y = \ldots\) and use of implicit diffn. | ||
| \(= \frac{-2}{\sqrt{3}}(1+t^2) \times \frac{3}{4-4t+4t^2}\) | M1 | |
| \(= \frac{-\sqrt{3}}{2} \cdot \frac{1+t^2}{1-t+t^2} = \frac{-\sqrt{3}}{2} \cdot \frac{2-\sin 2x}{\text{denom}}\) | A1 | |
| so that \(k = -\sqrt{3}\) | A1 | |
| [4] | ||
| (ii)(a) Use of \(x = r\cos\theta, y = r\sin\theta\) (and \(x^2 + y^2 = r^2\)) | M1 | |
| \(\Rightarrow r^2 = 72 + r^2\sin\theta\cos\theta\) i.e. \(x, y\) completely (and correctly) eliminated | M1, A1 | |
| \(\Rightarrow r^2 = \frac{144}{2-\sin 2\theta}\) or equivalent form | M1 | |
| \(2 - \sin 2\theta \geq 1 \Rightarrow r^2 \leq 144\) so that \(r_{\max} = 12\) | M1, A1 | |
| Calculus approach fine as an alternative | ||
| Then \(\sin 2\theta = 1 \Rightarrow \theta = \frac{1}{4}\pi\) or \(\frac{3}{4}\pi\) | M1, A1 | |
| No M ft from incorrect differentiation | ||
| [7] | ||
| (ii)(b) \(A = \frac{1}{2}\int r^2 \, d\theta = \int \frac{72}{2 - \sin 2\theta} \, d\theta\) | M1 | |
| \(= 72\left[\frac{-1}{\sqrt{3}}\tan^{-1}\left(1 - \frac{2\tan x}{\sqrt{3}}\right)\right]_0^{\pi/4}\) | M1 | Use of previous result |
| \(= \frac{72}{\sqrt{3}}\left(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)\right)\) or \(\frac{72}{\sqrt{3}}\left(\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)\right)\) | A1, A1 | |
| \(= 8\pi\sqrt{3}\) CAO ft their \(k\) above | A1 | |
| Area inside \(C\) in 1st quad. = \(2A = 16\pi\sqrt{3}\) since \(C\) is symmetric in \(y = x\) ft | B1 | |
| [5] |
**(i)(a)** Use of $\sin 2x = \frac{2t}{1+t^2}$ in $\frac{2}{2 - \sin 2x}$ where $t = \tan x$ | M1 |
$\frac{2}{2-\sin 2x} = \frac{2}{2-\left(\frac{2t}{1+t^2}\right)} = \frac{1+t^2}{1-t+t^2}$ **ANSWER GIVEN** | A1 |
| [2] |
**(i)(b)** $y = \tan^{-1}\left(1 - \frac{2\tan x}{\sqrt{3}}\right) \Rightarrow \frac{dy}{dx} = \frac{1}{1+\left(\frac{1-2t}{\sqrt{3}}\right)^2} \times \frac{-2}{\sqrt{3}}\sec^2 x$ | M1, A1 |
or $\tan y = \ldots$ and use of implicit diffn. | |
$= \frac{-2}{\sqrt{3}}(1+t^2) \times \frac{3}{4-4t+4t^2}$ | M1 |
$= \frac{-\sqrt{3}}{2} \cdot \frac{1+t^2}{1-t+t^2} = \frac{-\sqrt{3}}{2} \cdot \frac{2-\sin 2x}{\text{denom}}$ | A1 |
so that $k = -\sqrt{3}$ | A1 |
| [4] |
**(ii)(a)** Use of $x = r\cos\theta, y = r\sin\theta$ (and $x^2 + y^2 = r^2$) | M1 |
$\Rightarrow r^2 = 72 + r^2\sin\theta\cos\theta$ i.e. $x, y$ completely (and correctly) eliminated | M1, A1 |
$\Rightarrow r^2 = \frac{144}{2-\sin 2\theta}$ or equivalent form | M1 |
$2 - \sin 2\theta \geq 1 \Rightarrow r^2 \leq 144$ so that $r_{\max} = 12$ | M1, A1 |
Calculus approach fine as an alternative | |
Then $\sin 2\theta = 1 \Rightarrow \theta = \frac{1}{4}\pi$ or $\frac{3}{4}\pi$ | M1, A1 |
No M ft from incorrect differentiation | |
| [7] |
**(ii)(b)** $A = \frac{1}{2}\int r^2 \, d\theta = \int \frac{72}{2 - \sin 2\theta} \, d\theta$ | M1 |
$= 72\left[\frac{-1}{\sqrt{3}}\tan^{-1}\left(1 - \frac{2\tan x}{\sqrt{3}}\right)\right]_0^{\pi/4}$ | M1 | Use of previous result
$= \frac{72}{\sqrt{3}}\left(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) - \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)\right)$ or $\frac{72}{\sqrt{3}}\left(\frac{\pi}{6} - \left(-\frac{\pi}{6}\right)\right)$ | A1, A1 |
$= 8\pi\sqrt{3}$ **CAO ft their $k$ above** | A1 |
Area inside $C$ in 1st quad. = $2A = 16\pi\sqrt{3}$ since $C$ is symmetric in $y = x$ ft | B1 |
| [5] |\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Given that $t = \tan x$, prove that $\frac{2}{2 - \sin 2x} = \frac{1 + t^2}{1 - t + t^2}$. [2]
\item Hence determine the value of the constant $k$ for which
$$\frac{d}{dx}\left\{\tan^{-1}\left(\frac{1 - 2\tan x}{\sqrt{3}}\right)\right\} = \frac{k}{2 - \sin 2x}.$$ [4]
\end{enumerate}
\item The curve $C$ has cartesian equation $x^2 - xy + y^2 = 72$.
\begin{enumerate}[label=(\alph*)]
\item Determine a polar equation for $C$ in the form $r^2 = f(\theta)$, and deduce the polar coordinates $(r, \theta)$, where $0 \leqslant \theta < 2\pi$, of the points on $C$ which are furthest from the pole $O$. [7]
\item Find the exact area of the region of the plane in the first quadrant bounded by $C$, the $x$-axis and the line $y = x$. Deduce the total area of the region of the plane which lies inside $C$ and within the first quadrant. [5]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2011 Q13 [18]}}