| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Maximum/minimum distance from pole or line |
| Difficulty | Challenging +1.3 This is a multi-part Further Maths polar coordinates question requiring differentiation to find extrema, solving a transcendental equation, sketching, and area integration. While it involves several techniques, each step follows standard procedures: part (i) is routine differentiation, part (ii) uses the standard perpendicular distance formula leading to a straightforward transcendental equation with verification (not full solution), part (iii) is a sketch, and part (iv) is a standard polar area integral. The question is more demanding than typical A-level due to the Further Maths content and multi-step nature, but doesn't require exceptional insight. |
| Spec | 1.09a Sign change methods: locate roots4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dr}{d\theta} = -\frac{a}{(1+\theta)^2}\) | M1 | |
| Since \(a > 0\), then \(\frac{dr}{d\theta} < 0\), \(\forall\) points of \(C\) | A1 | [2] |
| OR \(a(> 0)\) constant and as \(\theta(> 0)\) increases, \(1 + \theta\) increases | M1 | |
| \(\therefore \frac{a}{1+\theta}\) decreases | A1 | |
| (ii) \(y = a\sin \theta/(1+\theta)\) | M1 | |
| \(\frac{dy}{d\theta} = 0 \Rightarrow \ldots \Rightarrow (1+\theta)\cos \theta - \sin \theta = 0\) | A1 | |
| \(\Rightarrow \tan \theta = 1 + \theta\) (AG) | A1 | |
| \(\tan \theta - 1 - \theta = -0.135\) when \(\theta = 1.1\): \(\tan \theta - 1 - \theta = +0.372\) when \(\theta = 1.2\) | B1 | [4] |
| Answer | Marks |
|---|---|
| \(\tan(1.1) \approx 1.96 < 2.1\), \(\tan(1.2) \approx 2.37 > 2.2\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Approximately correct shape and placement for \(0 \leq \theta \lesssim \pi/2\) and passing through \((a, 0)\) and \((0.4a, \pi/2)\) approximately, indicated in some way | B1B1 | |
| Maximum in interval \((\pi/4, \pi/2)\) | B1 | [3] |
| (iv) \(A = (a^2/2) \int_0^{\pi/2} (1+\theta)^{-2} d\theta\) | M1 | |
| \(= (a^2/2) \left[-(1+\theta)^{-1}\right]_0^{\pi/2}\) | A1 | |
| \(= \ldots = \pi a^2/2(\pi + 2)\) | A1 | |
| Do not accept double minus signs or fractions in the denominator for the final mark. | [3] |
**(i)** $\frac{dr}{d\theta} = -\frac{a}{(1+\theta)^2}$ | M1 |
Since $a > 0$, then $\frac{dr}{d\theta} < 0$, $\forall$ points of $C$ | A1 | [2] |
**OR** $a(> 0)$ constant and as $\theta(> 0)$ increases, $1 + \theta$ increases | M1 |
$\therefore \frac{a}{1+\theta}$ decreases | A1 |
**(ii)** $y = a\sin \theta/(1+\theta)$ | M1 |
$\frac{dy}{d\theta} = 0 \Rightarrow \ldots \Rightarrow (1+\theta)\cos \theta - \sin \theta = 0$ | A1 |
$\Rightarrow \tan \theta = 1 + \theta$ (AG) | A1 |
$\tan \theta - 1 - \theta = -0.135$ when $\theta = 1.1$: $\tan \theta - 1 - \theta = +0.372$ when $\theta = 1.2$ | B1 | [4] |
**OR** equivalent argument for B1 such as:
$\tan(1.1) \approx 1.96 < 2.1$, $\tan(1.2) \approx 2.37 > 2.2$ | B1 |
**(iii)** Sketch:
Approximately correct shape and placement for $0 \leq \theta \lesssim \pi/2$ and passing through $(a, 0)$ and $(0.4a, \pi/2)$ approximately, indicated in some way | B1B1 |
Maximum in interval $(\pi/4, \pi/2)$ | B1 | [3] |
**(iv)** $A = (a^2/2) \int_0^{\pi/2} (1+\theta)^{-2} d\theta$ | M1 |
$= (a^2/2) \left[-(1+\theta)^{-1}\right]_0^{\pi/2}$ | A1 |
$= \ldots = \pi a^2/2(\pi + 2)$ | A1 |
Do not accept double minus signs or fractions in the denominator for the final mark. | [3] |
---11 The curve $C$ has polar equation
$$r = \frac { a } { 1 + \theta }$$
where $a$ is a positive constant and $0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi$.\\
(i) Show that $r$ decreases as $\theta$ increases.\\
(ii) The point $P$ of $C$ is further from the initial line than any other point of $C$. Show that, at $P$,
$$\tan \theta = 1 + \theta$$
and verify that this equation has a root between 1.1 and 1.2.\\
(iii) Draw a sketch of $C$.\\
(iv) Find the area of the region bounded by the initial line, the line $\theta = \frac { 1 } { 2 } \pi$ and $C$, leaving your answer in terms of $\pi$ and $a$.
\hfill \mbox{\textit{CAIE FP1 2010 Q11 [12]}}