| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2022 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Sketch polar curve |
| Difficulty | Standard +0.8 This is a multi-part Further Maths polar coordinates question requiring: (a) sketching with specific point identification, (b) optimization using y = r sin θ, (c) polar area integration, and (d) Cartesian conversion. While the techniques are standard for FM students, the sec² function and the multi-step nature with 4 parts requiring different skills makes this moderately challenging but still within typical FM scope. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| [Sketch with correct position and domain, \(r\) strictly increasing, decreasing positive gradient] | B1 | Correct position and domain and \(r\) strictly increasing. |
| [Decreasing positive gradient shown] | B1 | Decreasing positive gradient. |
| \((a, 0)\ \left(2a, \frac{1}{4}\pi\right)\) | B1 | Can be labelled on their sketch. |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = (2a)\sin\left(\frac{1}{4}\pi\right)\) | M1 | Uses \(y = r\sin\theta\) |
| \(a\sqrt{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}a^2\int_0^{\frac{1}{4}\pi}\sec^4\theta\, d\theta\) | M1 | Forms \(\frac{1}{2}\int r^2\, d\theta\) with correct limits. |
| \(= \frac{1}{2}a^2\int_0^{\frac{1}{4}\pi}\sec^2\theta(\tan^2\theta + 1)\,d\theta\) | M1 | Applies \(\sec^2\theta = \tan^2\theta + 1\) to obtain an integrable form. |
| \(= \frac{1}{2}a^2\left[\frac{1}{3}\tan^3\theta + \tan\theta\right]_0^{\frac{1}{4}\pi}\) | A1 | |
| \(= \frac{2}{3}a^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x^2 + y^2 = r^2,\quad x = r\cos\theta\) | M1 | Substitutes to eliminate either \(r\) or \(\theta\). |
| \(r^2\cos^2\theta = ar\) leading to \(x^2 = a\sqrt{x^2+y^2}\) leading to \(x^4 = a^2(x^2+y^2)\) | M1 | Completes substitution to eliminate both \(r\) and \(\theta\) and makes \(y\) or \(y^2\) the subject. |
| \(y = \sqrt{a^{-2}x^4 - x^2} = x\sqrt{a^{-2}x^2 - 1}\) | A1 | AEF |
## Question 5(a):
[Sketch with correct position and domain, $r$ strictly increasing, decreasing positive gradient] | B1 | Correct position and domain and $r$ strictly increasing.
[Decreasing positive gradient shown] | B1 | Decreasing positive gradient.
$(a, 0)\ \left(2a, \frac{1}{4}\pi\right)$ | B1 | Can be labelled on their sketch.
**Total: 3**
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## Question 5(b):
$y = (2a)\sin\left(\frac{1}{4}\pi\right)$ | M1 | Uses $y = r\sin\theta$
$a\sqrt{2}$ | A1 |
**Total: 2**
---
## Question 5(c):
$\frac{1}{2}a^2\int_0^{\frac{1}{4}\pi}\sec^4\theta\, d\theta$ | M1 | Forms $\frac{1}{2}\int r^2\, d\theta$ with correct limits.
$= \frac{1}{2}a^2\int_0^{\frac{1}{4}\pi}\sec^2\theta(\tan^2\theta + 1)\,d\theta$ | M1 | Applies $\sec^2\theta = \tan^2\theta + 1$ to obtain an integrable form.
$= \frac{1}{2}a^2\left[\frac{1}{3}\tan^3\theta + \tan\theta\right]_0^{\frac{1}{4}\pi}$ | A1 |
$= \frac{2}{3}a^2$ | A1 |
**Total: 4**
---
## Question 5(d):
$x^2 + y^2 = r^2,\quad x = r\cos\theta$ | M1 | Substitutes to eliminate either $r$ or $\theta$.
$r^2\cos^2\theta = ar$ leading to $x^2 = a\sqrt{x^2+y^2}$ leading to $x^4 = a^2(x^2+y^2)$ | M1 | Completes substitution to eliminate both $r$ and $\theta$ and makes $y$ or $y^2$ the subject.
$y = \sqrt{a^{-2}x^4 - x^2} = x\sqrt{a^{-2}x^2 - 1}$ | A1 | AEF
**Total: 3**5 The curve $C$ has polar equation $r = \operatorname { asec } ^ { 2 } \theta$, where $a$ is a positive constant and $0 \leqslant \theta \leqslant \frac { 1 } { 4 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$, stating the polar coordinates of the point of intersection of $C$ with the initial line and also with the half-line $\theta = \frac { 1 } { 4 } \pi$.
\item Find the maximum distance of a point of $C$ from the initial line.
\item Find the area of the region enclosed by $C$, the initial line and the half-line $\theta = \frac { 1 } { 4 } \pi$.
\item Find, in the form $y = f ( x )$, the Cartesian equation of $C$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2022 Q5 [12]}}