Question 10 - A-Level Maths

CAIE FP1 2013 November — Question 10 13 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2013
Session	November
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Polar coordinates
Type	Maximum/minimum distance from pole or line
Difficulty	Standard +0.8 This is a multi-part Further Maths polar coordinates question requiring differentiation to find a maximum, curve sketching, and area calculation using integration. While the techniques are standard for FP1, the combination of skills and the need to interpret the maximum condition (dr/dθ = 0) elevates it above routine single-method questions. The integration for area is straightforward but requires careful algebraic manipulation.
Spec	4.09a Polar coordinates: convert to/from cartesian 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve

10 The curve \(C\) has polar equation \(r = 2 \sin \theta ( 1 - \cos \theta )\), for \(0 \leqslant \theta \leqslant \pi\). Find \(\frac { \mathrm { d } r } { \mathrm {~d} \theta }\) and hence find the polar coordinates of the point of \(C\) that is furthest from the pole. Sketch \(C\). Find the exact area of the sector from \(\theta = 0\) to \(\theta = \frac { 1 } { 4 } \pi\).

Show mark scheme Show mark scheme source

Question 10:

Part (stationary points):

Answer	Marks	Guidance
\(\frac{dr}{d\theta} = 2\cos\theta - 2\cos 2\theta\)	M1
\(2c - 2(2c^2 - 1) = 0 \Rightarrow 2c^2 - c - 1 = 0\)	A1
\(\Rightarrow (2c+1)(c-1) = 0\)	M1
\(\Rightarrow c = -\frac{1}{2}\) or \(1\)	A1
Required points on \(C\): \(\left(\frac{3}{2}\sqrt{3}, \frac{2}{3}\pi\right)\)	A1	(5)

Part (sketch):

Answer	Marks	Guidance
Approximate shape and location	B1
Accurate scaling	B1	(2)

Part (area):

Answer	Marks	Guidance
Area \(= \frac{1}{2}\int_0^{\pi/4} 4\sin^2\theta(1 - 2\cos\theta + \cos^2\theta)d\theta\)	M1 M1
\(= \int_0^{\pi/4}\left(1 - \cos 2\theta - 4\cos\theta\sin^2\theta + \frac{1}{4}[1-\cos 4\theta]\right)d\theta\)	A1A1
\(= \left[\frac{5\theta}{4} - \frac{\sin 2\theta}{2} - 4\frac{\sin^3\theta}{3} - \frac{\sin 4\theta}{16}\right]_0^{\pi/4}\)	M1
\(= \frac{5}{16}\pi - \frac{1}{2} - \frac{\sqrt{2}}{3}\) or \(0.0103\)	A1	(6)

# Question 10:

## Part (stationary points):
| $\frac{dr}{d\theta} = 2\cos\theta - 2\cos 2\theta$ | M1 | |
|---|---|---|
| $2c - 2(2c^2 - 1) = 0 \Rightarrow 2c^2 - c - 1 = 0$ | A1 | |
| $\Rightarrow (2c+1)(c-1) = 0$ | M1 | |
| $\Rightarrow c = -\frac{1}{2}$ or $1$ | A1 | |
| Required points on $C$: $\left(\frac{3}{2}\sqrt{3}, \frac{2}{3}\pi\right)$ | A1 | **(5)** |

## Part (sketch):
| Approximate shape and location | B1 | |
|---|---|---|
| Accurate scaling | B1 | **(2)** |

## Part (area):
| Area $= \frac{1}{2}\int_0^{\pi/4} 4\sin^2\theta(1 - 2\cos\theta + \cos^2\theta)d\theta$ | M1 M1 | |
|---|---|---|
| $= \int_0^{\pi/4}\left(1 - \cos 2\theta - 4\cos\theta\sin^2\theta + \frac{1}{4}[1-\cos 4\theta]\right)d\theta$ | A1A1 | |
| $= \left[\frac{5\theta}{4} - \frac{\sin 2\theta}{2} - 4\frac{\sin^3\theta}{3} - \frac{\sin 4\theta}{16}\right]_0^{\pi/4}$ | M1 | |
| $= \frac{5}{16}\pi - \frac{1}{2} - \frac{\sqrt{2}}{3}$ or $0.0103$ | A1 | **(6)** |

---

Show LaTeX source

10 The curve $C$ has polar equation $r = 2 \sin \theta ( 1 - \cos \theta )$, for $0 \leqslant \theta \leqslant \pi$. Find $\frac { \mathrm { d } r } { \mathrm {~d} \theta }$ and hence find the polar coordinates of the point of $C$ that is furthest from the pole.

Sketch $C$.

Find the exact area of the sector from $\theta = 0$ to $\theta = \frac { 1 } { 4 } \pi$.

\hfill \mbox{\textit{CAIE FP1 2013 Q10 [13]}}

This paper (12 questions)

View full paper

Q1 6 Q2 6 Q3 7 Q4 7 Q5 8 Q6 8 Q7 9 Q8 11 Q9 11 Q10 13 Q11 EITHER Q11 OR