| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2020 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Polar curve with substitution integral |
| Difficulty | Challenging +1.8 This is a challenging Further Maths polar coordinates question requiring curve sketching, integration by parts with substitution (non-trivial algebra), and implicit differentiation of polar curves. While the techniques are standard for FM students, the multi-step integration proof and the implicit equation derivation require careful execution and algebraic manipulation beyond routine exercises. |
| 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 |
|---|---|---|
| \(\theta = 0\) [curve sketch with correct shape, \(r\) decreasing] | B1 | Correct shape, \(r\) decreasing. |
| [Section \(\frac{1}{2}\pi \leq \theta \leq \pi\) correct, becoming tangential at \(\theta = \pi\)] | B1 | Section \(\frac{1}{2}\pi \leq \theta \leq \pi\) correct, becoming tangential at \(\theta = \pi\). |
| \((\ln(1+\pi), 0)\) | B1 | May be seen on their diagram. Allow \((1.42, 0)\) |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}\int_0^{\pi} \ln^2(1+\pi-\theta)\, d\theta = \frac{1}{2}\int_1^{1+\pi} \ln^2 u\, du\) | M1 | Uses correct formula with correct limits. |
| \(= \left[\frac{1}{2}u\ln^2 u\right]_1^{1+\pi} - \int_1^{1+\pi} \ln u\, du\) | M1 A1 | Integrates by parts once. |
| \(= \left[\frac{1}{2}u\ln^2 u\right]_1^{1+\pi} - \left([u\ln u]_1^{1+\pi} - \int_1^{1+\pi} 1\, du\right)\) | M1 | Integrates by parts again (or uses known result for integral of \(\ln u\)). |
| \(= \left[\frac{1}{2}u\ln^2 u - u\ln u + u\right]_1^{1+\pi}\) | A1 | |
| \(\frac{1}{2}(1+\pi)\ln^2(1+\pi) - (1+\pi)\ln(1+\pi) + \pi = \frac{1}{2}(1+\pi)\ln(1+\pi)(\ln(1+\pi)-2)+\pi\) | A1 | AG |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \ln(1 + \pi - \theta)\sin\theta\) | B1 | Uses \(y = r\sin\theta\) |
| \(\frac{dy}{d\theta} = \ln(1 + \pi - \theta)\cos\theta - \frac{\sin\theta}{1 + \pi - \theta} = 0\) | M1 A1 | Sets derivative equal to zero |
| \(\cos\theta \neq 0 \Rightarrow (1 + \pi - \theta)\ln(1 + \pi - \theta) - \tan\theta = 0\) | A1 | AG |
| \((1 + \pi - 1.2)\ln(1 + \pi - 1.2) - \tan 1.2 = 0.602\) and \((1 + \pi - 1.3)\ln(1 + \pi - 1.3) - \tan 1.3 = -0.634\) | B1 | Shows sign change. Values correct to 2sf should be shown |
| Total: 5 |
## Question 5:
### Part (a):
| $\theta = 0$ [curve sketch with correct shape, $r$ decreasing] | B1 | Correct shape, $r$ decreasing. |
| [Section $\frac{1}{2}\pi \leq \theta \leq \pi$ correct, becoming tangential at $\theta = \pi$] | B1 | Section $\frac{1}{2}\pi \leq \theta \leq \pi$ correct, becoming tangential at $\theta = \pi$. |
| $(\ln(1+\pi), 0)$ | B1 | May be seen on their diagram. Allow $(1.42, 0)$ |
| **Total: 3** | | |
### Part (b):
| $\frac{1}{2}\int_0^{\pi} \ln^2(1+\pi-\theta)\, d\theta = \frac{1}{2}\int_1^{1+\pi} \ln^2 u\, du$ | M1 | Uses correct formula with correct limits. |
| $= \left[\frac{1}{2}u\ln^2 u\right]_1^{1+\pi} - \int_1^{1+\pi} \ln u\, du$ | M1 A1 | Integrates by parts once. |
| $= \left[\frac{1}{2}u\ln^2 u\right]_1^{1+\pi} - \left([u\ln u]_1^{1+\pi} - \int_1^{1+\pi} 1\, du\right)$ | M1 | Integrates by parts again (or uses known result for integral of $\ln u$). |
| $= \left[\frac{1}{2}u\ln^2 u - u\ln u + u\right]_1^{1+\pi}$ | A1 | |
| $\frac{1}{2}(1+\pi)\ln^2(1+\pi) - (1+\pi)\ln(1+\pi) + \pi = \frac{1}{2}(1+\pi)\ln(1+\pi)(\ln(1+\pi)-2)+\pi$ | A1 | AG |
| **Total: 6** | | |
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \ln(1 + \pi - \theta)\sin\theta$ | **B1** | Uses $y = r\sin\theta$ |
| $\frac{dy}{d\theta} = \ln(1 + \pi - \theta)\cos\theta - \frac{\sin\theta}{1 + \pi - \theta} = 0$ | **M1 A1** | Sets derivative equal to zero |
| $\cos\theta \neq 0 \Rightarrow (1 + \pi - \theta)\ln(1 + \pi - \theta) - \tan\theta = 0$ | **A1** | AG |
| $(1 + \pi - 1.2)\ln(1 + \pi - 1.2) - \tan 1.2 = 0.602$ and $(1 + \pi - 1.3)\ln(1 + \pi - 1.3) - \tan 1.3 = -0.634$ | **B1** | Shows sign change. Values correct to 2sf should be shown |
| **Total: 5** | | |
---5 The curve $C$ has polar equation $r = \ln ( 1 + \pi - \theta )$, for $0 \leqslant \theta \leqslant \pi$.
\begin{enumerate}[label=(\alph*)]
\item Sketch $C$ and state the polar coordinates of the point of $C$ furthest from the pole.
\item Using the substitution $u = 1 + \pi - \theta$, or otherwise, show that the area of the region enclosed by $C$ and the initial line is
$$\frac { 1 } { 2 } ( 1 + \pi ) \ln ( 1 + \pi ) ( \ln ( 1 + \pi ) - 2 ) + \pi$$
\item Show that, at the point of $C$ furthest from the initial line,
$$( 1 + \pi - \theta ) \ln ( 1 + \pi - \theta ) - \tan \theta = 0$$
and verify that this equation has a root between 1.2 and 1.3.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2020 Q5 [14]}}