| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area enclosed by polar curve |
| Difficulty | Challenging +1.8 This is a Further Maths polar coordinates question requiring multiple techniques: solving a transcendental equation (ln(1+sin θ)=0), applying the polar area formula with integration of ln²(1+sin θ), and geometric reasoning about circularity. While the individual steps are standard for FM students, the combination and the non-standard integrand make this significantly harder than typical A-level questions but not exceptionally challenging for Further Maths. |
| 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 |
|---|---|---|
| 6 | (a) | ln(1 + sinθ) = 0 => 1 + sinθ = 1 => sinθ = 0 |
| so α = 0 and β = π | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| Answer | Marks |
|---|---|
| (b) | π |
| Answer | Marks |
|---|---|
| = 0.4162 (4 sf) cao | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.2 |
| 1.1 | Correct formula for area with r |
| Answer | Marks |
|---|---|
| BC | Incorrect formula = M0A0 |
| Answer | Marks |
|---|---|
| (c) | π |
| Answer | Marks |
|---|---|
| R = 0.3640 (4 sf) so the curve is not circular | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 3.2a | or radius R = 0.3466 (4 sf) |
| Answer | Marks |
|---|---|
| is not 0.4162 (4 sf) | It must be clear that the r value |
Question 6:
6 | (a) | ln(1 + sinθ) = 0 => 1 + sinθ = 1 => sinθ = 0
so α = 0 and β = π | M1
A1
[2] | 1.1a
2.2a
(b) | π
1
A= ∫(ln(1+sinθ))2dθ
2
0
= 0.4162 (4 sf) cao | M1
A1
[2] | 1.2
1.1 | Correct formula for area with r
correctly substituted and their
limits. Must be unambiguous but
can be implied by correct
answer/later work
BC | Incorrect formula = M0A0
Condone missing dθ
(c) | π
θ= ⇒r =ln2=0.6931 (4 sf) which
2
would be the diameter, D, of the circle
But A = 0.4162 (4 sf) => D = 0.7280 (4 sf) or
R = 0.3640 (4 sf) so the curve is not circular | M1
A1
[2] | 3.1a
3.2a | or radius R = 0.3466 (4 sf)
condone correct R or D without
reasoning
or R = 0.3466 (4 sf) (or D =
0.6931) => A = 0.3773 (4 sf) which
is not 0.4162 (4 sf) | It must be clear that the r value
would be the diameter of the
circle; the calculation alone is
insufficient for M1.
M1 can be implied by area
2
ln2
given as π
2
Explanation must include
comparison of R’s, D’s or A’s
and conclusion . Allow correct
working to 3 sf.6 The equation of a curve in polar coordinates is $r = \ln ( 1 + \sin \theta )$ for $\alpha \leqslant \theta \leqslant \beta$ where $\alpha$ and $\beta$ are non-negative angles. The curve consists of a single closed loop through the pole.
\begin{enumerate}[label=(\alph*)]
\item By solving the equation $r = 0$, determine the smallest possible values of $\alpha$ and $\beta$.
\item Find the area enclosed by the curve, giving your answer to 4 significant figures.
\item Hence, by considering the value of $r$ at $\theta = \frac { \alpha + \beta } { 2 }$, show that the loop is not circular.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2020 Q6 [6]}}