| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Standard +0.8 This is a Further Maths polar coordinates question requiring curve sketching and area calculation with a non-standard substitution. While the substitution is suggested, students must recognize that r² = ln(1+θ) means r = √(ln(1+θ)), handle the domain carefully (ln(1+θ) ≥ 0 requires θ ≥ 0), apply the polar area formula ½∫r²dθ which simplifies nicely here, and work through integration by parts after substitution. The conceptual demand of polar curves plus multi-step integration places this moderately above average difficulty. |
| Spec | 1.08h Integration by substitution4.09a 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 |
| Correct shape, domain and orientation (closed loop curve) | B1 | Correct shape, domain and orientation |
| Initial line is tangential to \(C\) at the pole | B1 | The initial line is tangential to \(C\) at the pole |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \dfrac{1}{2}\displaystyle\int_0^{2\pi} \ln(1+\theta)\, \mathrm{d}\theta\) | M1 | States \(\frac{1}{2}\int r^2\, \mathrm{d}\theta\) with correct expression and limits |
| \(\displaystyle\int_0^{2\pi} \ln(1+\theta)\,\mathrm{d}\theta = \int_1^{2\pi+1} \ln u\, \mathrm{d}u\) | M1 | Applies given substitution correctly, changes their limits |
| \(\displaystyle\int_1^{2\pi+1} \ln u\, \mathrm{d}u = \bigl[u \ln u\bigr]_1^{2\pi+1} - \bigl[u\bigr]_1^{2\pi+1}\) | M1 A1 | Integrates \(\ln u\) (by parts or otherwise) |
| \(A = \left(\pi + \dfrac{1}{2}\right)\ln(2\pi+1) - \pi\) | A1 | AEF, must be exact |
| Total: 5 |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct shape, domain and orientation (closed loop curve) | **B1** | Correct shape, domain and orientation |
| Initial line is tangential to $C$ at the pole | **B1** | The initial line is tangential to $C$ at the pole |
| **Total: 2** | | |
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## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \dfrac{1}{2}\displaystyle\int_0^{2\pi} \ln(1+\theta)\, \mathrm{d}\theta$ | **M1** | States $\frac{1}{2}\int r^2\, \mathrm{d}\theta$ with correct expression and limits |
| $\displaystyle\int_0^{2\pi} \ln(1+\theta)\,\mathrm{d}\theta = \int_1^{2\pi+1} \ln u\, \mathrm{d}u$ | **M1** | Applies given substitution correctly, changes their limits |
| $\displaystyle\int_1^{2\pi+1} \ln u\, \mathrm{d}u = \bigl[u \ln u\bigr]_1^{2\pi+1} - \bigl[u\bigr]_1^{2\pi+1}$ | **M1 A1** | Integrates $\ln u$ (by parts or otherwise) |
| $A = \left(\pi + \dfrac{1}{2}\right)\ln(2\pi+1) - \pi$ | **A1** | AEF, must be exact |
| **Total: 5** | | |2 The curve $C$ has polar equation $r ^ { 2 } = \ln ( 1 + \theta )$, for $0 \leqslant \theta \leqslant 2 \pi$.\\
(i) Sketch $C$.\\
(ii) Using the substitution $u = 1 + \theta$, or otherwise, find the area of the region bounded by $C$ and the initial line, leaving your answer in an exact form.\\
\hfill \mbox{\textit{CAIE FP1 2019 Q2 [7]}}