Question 5 - A-Level Maths

CAIE Further Paper 1 2023 June — Question 5 12 marks

Exam Board	CAIE
Module	Further Paper 1 (Further Paper 1)
Year	2023
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Polar coordinates
Type	Maximum/minimum distance from pole or line
Difficulty	Challenging +1.3 This is a multi-part Further Maths polar coordinates question requiring sketching, finding extrema, area calculation, and optimization using calculus. Part (a) is straightforward (maximum r occurs at θ=0). Part (b) is a standard polar area integral. Part (c) requires setting up dy/dθ=0 for perpendicular distance from initial line, involving product/chain rule and algebraic manipulation to reach the given form, then numerical verification. While it requires multiple techniques and careful algebra, each step follows standard Further Maths procedures without requiring exceptional insight.
Spec	1.09a Sign change methods: locate roots 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve

5 The curve $C$ has polar equation $r ^ { 2 } = \frac { 1 } { \theta ^ { 2 } + 1 }$, for $0 \leqslant \theta \leqslant \pi$.

Sketch $C$ and state the polar coordinates of the point of $C$ furthest from the pole.
Find the area of the region enclosed by $C$, the initial line, and the half-line $\theta = \pi$.
Show that, at the point of $C$ furthest from the initial line, $$\left( \theta + \frac { 1 } { \theta } \right) \cot \theta - 1 = 0$$ and verify that this equation has a root between 1.1 and 1.2.

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Question 5:

Part 5(a):

Answer	Marks	Guidance
[Polar curve sketch: correct shape, $r$ strictly decreasing, domain $0$ to $\pi$]	B1*	Correct shape and domain, polar graph with $r$ strictly decreasing but condone if not strictly decreasing close to $\theta=\pi$.
[Fully correct including shape at $\theta=0$ correct, shape at $\theta=\pi$ correct]	DB1
$(1,0)$	B1	Identified as point furthest from the pole and given as coordinates.

Part 5(b):

Answer	Marks	Guidance
$\frac{1}{2}\int_0^{\pi}\frac{1}{\theta^2+1}\,\mathrm{d}\theta$	M1	Uses correct formula with correct limits.
$\frac{1}{2}\left[\tan^{-1}\theta\right]_0^{\pi}$	M1 A1	Integrates $\frac{1}{\theta^2+1}$.
$\frac{1}{2}\tan^{-1}\pi = 0.631$	A1

Part 5(c):

Answer	Marks	Guidance
$y = \frac{\sin\theta}{\sqrt{\theta^2+1}}$	B1	Uses $y = r\sin\theta$
$\frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{(\theta^2+1)^{\frac{1}{2}}\cos\theta - \theta(\theta^2+1)^{-\frac{1}{2}}\sin\theta}{\theta^2+1} = 0$	M1 A1	Sets derivative equal to zero.
$\theta \neq 0 \Rightarrow \left(\theta+\frac{1}{\theta}\right)\cot\theta - 1 = 0$	A1	AG.
$\left(1.1+\frac{1}{1.1}\right)\cot 1.1 - 1 = 0.02\ldots$ and $\left(1.2+\frac{1}{1.2}\right)\cot 1.2 - 1 = -0.209\ldots$	B1	Shows sign change (1sf or better).

## Question 5:

**Part 5(a):**

[Polar curve sketch: correct shape, $r$ strictly decreasing, domain $0$ to $\pi$] | B1* | Correct shape and domain, polar graph with $r$ strictly decreasing but condone if not strictly decreasing close to $\theta=\pi$.

[Fully correct including shape at $\theta=0$ correct, shape at $\theta=\pi$ correct] | DB1 |

$(1,0)$ | B1 | Identified as point furthest from the pole and given as coordinates.

**Part 5(b):**

$\frac{1}{2}\int_0^{\pi}\frac{1}{\theta^2+1}\,\mathrm{d}\theta$ | M1 | Uses correct formula with correct limits.

$\frac{1}{2}\left[\tan^{-1}\theta\right]_0^{\pi}$ | M1 A1 | Integrates $\frac{1}{\theta^2+1}$.

$\frac{1}{2}\tan^{-1}\pi = 0.631$ | A1 |

**Part 5(c):**

$y = \frac{\sin\theta}{\sqrt{\theta^2+1}}$ | B1 | Uses $y = r\sin\theta$

$\frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{(\theta^2+1)^{\frac{1}{2}}\cos\theta - \theta(\theta^2+1)^{-\frac{1}{2}}\sin\theta}{\theta^2+1} = 0$ | M1 A1 | Sets derivative equal to zero.

$\theta \neq 0 \Rightarrow \left(\theta+\frac{1}{\theta}\right)\cot\theta - 1 = 0$ | A1 | AG.

$\left(1.1+\frac{1}{1.1}\right)\cot 1.1 - 1 = 0.02\ldots$ and $\left(1.2+\frac{1}{1.2}\right)\cot 1.2 - 1 = -0.209\ldots$ | B1 | Shows sign change (1sf or better).

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5 The curve $C$ has polar equation $r ^ { 2 } = \frac { 1 } { \theta ^ { 2 } + 1 }$, 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 Find the area of the region enclosed by $C$, the initial line, and the half-line $\theta = \pi$.
\item Show that, at the point of $C$ furthest from the initial line,

$$\left( \theta + \frac { 1 } { \theta } \right) \cot \theta - 1 = 0$$

and verify that this equation has a root between 1.1 and 1.2.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q5 [12]}}

This paper (7 questions)

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Q1 7 Q2 8 Q3 7 Q4 12 Q5 12 Q6 15 Q7 14

Answer	Marks	Guidance
[Polar curve sketch: correct shape, \(r\) strictly decreasing, domain \(0\) to \(\pi\)]	B1*	Correct shape and domain, polar graph with \(r\) strictly decreasing but condone if not strictly decreasing close to \(\theta=\pi\).
[Fully correct including shape at \(\theta=0\) correct, shape at \(\theta=\pi\) correct]	DB1
\((1,0)\)	B1	Identified as point furthest from the pole and given as coordinates.

Answer	Marks	Guidance
\(\frac{1}{2}\int_0^{\pi}\frac{1}{\theta^2+1}\,\mathrm{d}\theta\)	M1	Uses correct formula with correct limits.
\(\frac{1}{2}\left[\tan^{-1}\theta\right]_0^{\pi}\)	M1 A1	Integrates \(\frac{1}{\theta^2+1}\).
\(\frac{1}{2}\tan^{-1}\pi = 0.631\)	A1

Answer	Marks	Guidance
\(y = \frac{\sin\theta}{\sqrt{\theta^2+1}}\)	B1	Uses \(y = r\sin\theta\)
\(\frac{\mathrm{d}y}{\mathrm{d}\theta} = \frac{(\theta^2+1)^{\frac{1}{2}}\cos\theta - \theta(\theta^2+1)^{-\frac{1}{2}}\sin\theta}{\theta^2+1} = 0\)	M1 A1	Sets derivative equal to zero.
\(\theta \neq 0 \Rightarrow \left(\theta+\frac{1}{\theta}\right)\cot\theta - 1 = 0\)	A1	AG.
\(\left(1.1+\frac{1}{1.1}\right)\cot 1.1 - 1 = 0.02\ldots\) and \(\left(1.2+\frac{1}{1.2}\right)\cot 1.2 - 1 = -0.209\ldots\)	B1	Shows sign change (1sf or better).