| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Deduce integral value from area |
| Difficulty | Challenging +1.8 This is a substantial Further Maths polar coordinates question requiring multiple techniques: sketching polar curves, computing polar area (∫½r²dθ), converting to Cartesian form, and the key insight of deducing a Cartesian integral from polar area. Part (d) requires recognizing the geometric connection between the two representations. While systematic, it demands strong technical facility and conceptual understanding across coordinate systems. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sketch showing curve approaching \(\theta=0\) | B1 | |
| Maximum distance of \(C\) from the pole is \(a\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}a^2\int_0^{\frac{1}{4}\pi}\tan^2\theta\, \mathrm{d}\theta\) | M1 | |
| \(= \frac{1}{2}a^2\int_0^{\frac{1}{4}\pi}\sec^2\theta - 1\,\mathrm{d}\theta = \frac{1}{2}a^2\left[\tan\theta - \theta\right]_0^{\frac{1}{4}\pi}\) | M1 A1 | |
| \(= \frac{1}{2}a^2\left(1-\frac{1}{4}\pi\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\sqrt{x^2+y^2} = a\frac{y}{x}\) | B1 | |
| \(x^2(x^2+y^2) = a^2y^2 \Rightarrow y^2(a^2-x^2) = x^4\) | M1 | |
| \(y = \frac{x^2}{\sqrt{(a^2-x^2)}}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}\left(a\cos\frac{1}{4}\pi\right)\left(a\sin\frac{1}{4}\pi\right) - \frac{1}{2}a^2\left(1-\frac{1}{4}\pi\right) = \frac{1}{4}a^2\left(\frac{1}{2}\pi-1\right)\) | M1 A1 FT | A1 FT their (b) |
## Question 5:
**Part 5(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sketch showing curve approaching $\theta=0$ | B1 | |
| Maximum distance of $C$ from the pole is $a$ | B1 | |
**Part 5(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}a^2\int_0^{\frac{1}{4}\pi}\tan^2\theta\, \mathrm{d}\theta$ | M1 | |
| $= \frac{1}{2}a^2\int_0^{\frac{1}{4}\pi}\sec^2\theta - 1\,\mathrm{d}\theta = \frac{1}{2}a^2\left[\tan\theta - \theta\right]_0^{\frac{1}{4}\pi}$ | M1 A1 | |
| $= \frac{1}{2}a^2\left(1-\frac{1}{4}\pi\right)$ | A1 | |
**Part 5(c):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\sqrt{x^2+y^2} = a\frac{y}{x}$ | B1 | |
| $x^2(x^2+y^2) = a^2y^2 \Rightarrow y^2(a^2-x^2) = x^4$ | M1 | |
| $y = \frac{x^2}{\sqrt{(a^2-x^2)}}$ | A1 | AG |
**Part 5(d):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}\left(a\cos\frac{1}{4}\pi\right)\left(a\sin\frac{1}{4}\pi\right) - \frac{1}{2}a^2\left(1-\frac{1}{4}\pi\right) = \frac{1}{4}a^2\left(\frac{1}{2}\pi-1\right)$ | M1 A1 FT | A1 FT their (b) |
---5 The curve $C$ has polar equation $r = \operatorname { atan } \theta$, where $a$ is a positive constant and $0 \leqslant \theta \leqslant \frac { 1 } { 4 } \pi$.
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$ and state the greatest distance of a point on $C$ from the pole.
\item Find the exact value of the area of the region bounded by $C$ and the half-line $\theta = \frac { 1 } { 4 } \pi$.
\item Show that $C$ has Cartesian equation $\mathrm { y } = \frac { \mathrm { x } ^ { 2 } } { \sqrt { \mathrm { a } ^ { 2 } - \mathrm { x } ^ { 2 } } }$.
\item Using your answer to part (b), deduce the exact value of $\int _ { 0 } ^ { \frac { 1 } { 2 } a \sqrt { 2 } } \frac { x ^ { 2 } } { \sqrt { a ^ { 2 } - x ^ { 2 } } } d x$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q5 [11]}}