Question 8 - A-Level Maths

Edexcel F2 2023 June — Question 8 13 marks

Exam Board	Edexcel
Module	F2 (Further Pure Mathematics 2)
Year	2023
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Polar coordinates
Type	Region bounded by curve and tangent lines
Difficulty	Challenging +1.8 This is a substantial Further Maths polar coordinates question requiring: (a) simple coordinate identification, (b) finding tangent using dy/dx = 0 via polar calculus formulas, and (c) computing area bounded by curve and two tangent lines using polar integration with trigonometric identities. The multi-step nature, need for polar tangent calculus, and algebraic manipulation of the area integral (requiring trig identities and exact form) make this significantly harder than average, though it follows standard Further Pure techniques without requiring exceptional insight.
Spec	1.07s Parametric and implicit differentiation 1.08h Integration by substitution 4.09a Polar coordinates: convert to/from cartesian 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve

8. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{709ed2f1-f81c-4820-ac31-e1f86baae9d7-28_552_759_246_660} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows a sketch of the curve $C$ with equation $$r = 6 ( 1 + \cos \theta ) \quad 0 \leqslant \theta \leqslant \pi$$ Given that $C$ meets the initial line at the point $A$, as shown in Figure 1,

write down the polar coordinates of $A$. The line $l _ { 1 }$, also shown in Figure 1, is the tangent to $C$ at the point $B$ and is parallel to the initial line.
Use calculus to determine the polar coordinates of $B$. The line $l _ { 2 }$, also shown in Figure 1, is the tangent to $C$ at $A$ and is perpendicular to the initial line. The region $R$, shown shaded in Figure 1, is bounded by $C , l _ { 1 }$ and $l _ { 2 }$
Use algebraic integration to find the exact area of $R$, giving your answer in the form $p \sqrt { 3 } + q \pi$ where $p$ and $q$ are constants to be determined.

Show mark scheme Show mark scheme source

Question 8:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\theta = 0$, $r = 6(1 + \cos 0) \Rightarrow 12$ or $(12, 0)$	B1	Correct values for $\theta$ and $r$ or correct coordinates. Condone $(0, 12)$

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\frac{d}{d\theta}(r\sin\theta)$: $= 6\sin\theta(1+\cos\theta) \Rightarrow 6\sin\theta + 6\sin\theta\cos\theta \Rightarrow 6\sin\theta + 3\sin 2\theta$	M1	Differentiates $r\sin\theta$. Allow sign errors only. The "6" may be missing
$\Rightarrow 2\cos^2\theta + \cos\theta - 1 = 0$	dM1	Dependent on previous M mark. Uses correct identity/identities to reach a 3TQ in $\cos\theta$ and solves. Must obtain at least one real consistent solution where $-1 \leq \cos\theta \leq 1$
$\theta = \frac{\pi}{3}$, $r = 6\left(1 + \cos\frac{\pi}{3}\right) = 9$ or $\left(9, \frac{\pi}{3}\right)$	A1	Either $r$ or $\theta$ correct
A1	Both coordinates correct. Only accept $\frac{\pi}{3}$ and $9$. Withhold last mark if additional answers offered and not rejected

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\left[\frac{1}{2}\right]\int r^2\, d\theta = \left[\frac{1}{2}\right]\int 36(1+\cos\theta)^2\,[d\theta]$	M1	Attempts $\int r^2\,d\theta$. The $\frac{1}{2}$ may be missing or wrong. Condone poor squaring if only two terms
$\int r^2\,d\theta = 36\int(1 + 2\cos\theta + \cos^2\theta)[d\theta] = 36\int\left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right)[d\theta]$	M1	Uses $\cos^2\theta = \pm\frac{1}{2}\cos 2\theta \pm \frac{1}{2}$ and obtains integrand of form $a + b\cos\theta + c\cos 2\theta$
$\int(a + b\cos\theta + c\cos 2\theta)[d\theta] = a\theta \pm b\sin\theta \pm \frac{c}{2}\sin 2\theta$	M1	Integrates; sign errors on trig terms only. May be two terms in $\theta$
$18\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]$	A1	Fully correct expression. Ignore limits but $\frac{1}{2}$ must have been used
$18\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{3}} = 18\left(\frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8}\right) = 9\pi + \frac{81}{4}\sqrt{3}$	dM1	Dependent on previous M. Substitution of "correct" limits; upper limit $\frac{\pi}{3}$ (provided $0 < \theta < \frac{\pi}{2}$), lower limit 0
E.g. $OB = 6\left(1+\cos\frac{\pi}{3}\right) \Rightarrow BQ = 6\left(1+\cos\frac{\pi}{3}\right)\sin\frac{\pi}{3} = \frac{9\sqrt{3}}{2}$	M1	Correct method for perpendicular distance between $l_1$ and initial line
$BP = 12 - 9\cos\frac{\pi}{3} = 12 - \frac{9}{2} = \frac{15}{2}$ $\Rightarrow$ area $OBPA = \frac{1}{2}\left(12 + \frac{15}{2}\right)\left(\frac{9\sqrt{3}}{2}\right) = \frac{351}{8}\sqrt{3}$	A1	Correct expression for area of trapezium $OBPA$
area of region $R = \frac{351}{8}\sqrt{3} - \frac{81}{4}\sqrt{3} - 9\pi = \frac{189}{8}\sqrt{3} - 9\pi$	A1	Correct answer: $p = \frac{189}{8}$, $q = -9$

# Question 8:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\theta = 0$, $r = 6(1 + \cos 0) \Rightarrow 12$ or $(12, 0)$ | B1 | Correct values for $\theta$ and $r$ or correct coordinates. Condone $(0, 12)$ |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{d}{d\theta}(r\sin\theta)$: $= 6\sin\theta(1+\cos\theta) \Rightarrow 6\sin\theta + 6\sin\theta\cos\theta \Rightarrow 6\sin\theta + 3\sin 2\theta$ | M1 | Differentiates $r\sin\theta$. Allow sign errors only. The "6" may be missing |
| $\Rightarrow 2\cos^2\theta + \cos\theta - 1 = 0$ | dM1 | Dependent on previous M mark. Uses correct identity/identities to reach a 3TQ in $\cos\theta$ and solves. Must obtain at least one real consistent solution where $-1 \leq \cos\theta \leq 1$ |
| $\theta = \frac{\pi}{3}$, $r = 6\left(1 + \cos\frac{\pi}{3}\right) = 9$ or $\left(9, \frac{\pi}{3}\right)$ | A1 | Either $r$ or $\theta$ correct |
| | A1 | Both coordinates correct. Only accept $\frac{\pi}{3}$ and $9$. Withhold last mark if additional answers offered and not rejected |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[\frac{1}{2}\right]\int r^2\, d\theta = \left[\frac{1}{2}\right]\int 36(1+\cos\theta)^2\,[d\theta]$ | M1 | Attempts $\int r^2\,d\theta$. The $\frac{1}{2}$ may be missing or wrong. Condone poor squaring if only two terms |
| $\int r^2\,d\theta = 36\int(1 + 2\cos\theta + \cos^2\theta)[d\theta] = 36\int\left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right)[d\theta]$ | M1 | Uses $\cos^2\theta = \pm\frac{1}{2}\cos 2\theta \pm \frac{1}{2}$ and obtains integrand of form $a + b\cos\theta + c\cos 2\theta$ |
| $\int(a + b\cos\theta + c\cos 2\theta)[d\theta] = a\theta \pm b\sin\theta \pm \frac{c}{2}\sin 2\theta$ | M1 | Integrates; sign errors on trig terms only. May be two terms in $\theta$ |
| $18\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]$ | A1 | Fully correct expression. Ignore limits but $\frac{1}{2}$ must have been used |
| $18\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{3}} = 18\left(\frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8}\right) = 9\pi + \frac{81}{4}\sqrt{3}$ | dM1 | Dependent on previous M. Substitution of "correct" limits; upper limit $\frac{\pi}{3}$ (provided $0 < \theta < \frac{\pi}{2}$), lower limit 0 |
| E.g. $OB = 6\left(1+\cos\frac{\pi}{3}\right) \Rightarrow BQ = 6\left(1+\cos\frac{\pi}{3}\right)\sin\frac{\pi}{3} = \frac{9\sqrt{3}}{2}$ | M1 | Correct method for perpendicular distance between $l_1$ and initial line |
| $BP = 12 - 9\cos\frac{\pi}{3} = 12 - \frac{9}{2} = \frac{15}{2}$ $\Rightarrow$ area $OBPA = \frac{1}{2}\left(12 + \frac{15}{2}\right)\left(\frac{9\sqrt{3}}{2}\right) = \frac{351}{8}\sqrt{3}$ | A1 | Correct expression for area of trapezium $OBPA$ |
| area of region $R = \frac{351}{8}\sqrt{3} - \frac{81}{4}\sqrt{3} - 9\pi = \frac{189}{8}\sqrt{3} - 9\pi$ | A1 | Correct answer: $p = \frac{189}{8}$, $q = -9$ |

Show LaTeX source

8.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{709ed2f1-f81c-4820-ac31-e1f86baae9d7-28_552_759_246_660}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

Figure 1 shows a sketch of the curve $C$ with equation

$$r = 6 ( 1 + \cos \theta ) \quad 0 \leqslant \theta \leqslant \pi$$

Given that $C$ meets the initial line at the point $A$, as shown in Figure 1,
\begin{enumerate}[label=(\alph*)]
\item write down the polar coordinates of $A$.

The line $l _ { 1 }$, also shown in Figure 1, is the tangent to $C$ at the point $B$ and is parallel to the initial line.
\item Use calculus to determine the polar coordinates of $B$.

The line $l _ { 2 }$, also shown in Figure 1, is the tangent to $C$ at $A$ and is perpendicular to the initial line.

The region $R$, shown shaded in Figure 1, is bounded by $C , l _ { 1 }$ and $l _ { 2 }$
\item Use algebraic integration to find the exact area of $R$, giving your answer in the form $p \sqrt { 3 } + q \pi$ where $p$ and $q$ are constants to be determined.
\end{enumerate}

\hfill \mbox{\textit{Edexcel F2 2023 Q8 [13]}}

This paper (8 questions)

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Q1 7 Q2 10 Q3 7 Q4 11 Q5 7 Q6 9 Q7 11 Q8 13

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(\left[\frac{1}{2}\right]\int r^2\, d\theta = \left[\frac{1}{2}\right]\int 36(1+\cos\theta)^2\,[d\theta]\)	M1	Attempts \(\int r^2\,d\theta\). The \(\frac{1}{2}\) may be missing or wrong. Condone poor squaring if only two terms
\(\int r^2\,d\theta = 36\int(1 + 2\cos\theta + \cos^2\theta)[d\theta] = 36\int\left(\frac{3}{2} + 2\cos\theta + \frac{1}{2}\cos 2\theta\right)[d\theta]\)	M1	Uses \(\cos^2\theta = \pm\frac{1}{2}\cos 2\theta \pm \frac{1}{2}\) and obtains integrand of form \(a + b\cos\theta + c\cos 2\theta\)
\(\int(a + b\cos\theta + c\cos 2\theta)[d\theta] = a\theta \pm b\sin\theta \pm \frac{c}{2}\sin 2\theta\)	M1	Integrates; sign errors on trig terms only. May be two terms in \(\theta\)
\(18\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]\)	A1	Fully correct expression. Ignore limits but \(\frac{1}{2}\) must have been used
\(18\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{3}} = 18\left(\frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8}\right) = 9\pi + \frac{81}{4}\sqrt{3}\)	dM1	Dependent on previous M. Substitution of "correct" limits; upper limit \(\frac{\pi}{3}\) (provided \(0 < \theta < \frac{\pi}{2}\)), lower limit 0
E.g. \(OB = 6\left(1+\cos\frac{\pi}{3}\right) \Rightarrow BQ = 6\left(1+\cos\frac{\pi}{3}\right)\sin\frac{\pi}{3} = \frac{9\sqrt{3}}{2}\)	M1	Correct method for perpendicular distance between \(l_1\) and initial line
\(BP = 12 - 9\cos\frac{\pi}{3} = 12 - \frac{9}{2} = \frac{15}{2}\) \(\Rightarrow\) area \(OBPA = \frac{1}{2}\left(12 + \frac{15}{2}\right)\left(\frac{9\sqrt{3}}{2}\right) = \frac{351}{8}\sqrt{3}\)	A1	Correct expression for area of trapezium \(OBPA\)
area of region \(R = \frac{351}{8}\sqrt{3} - \frac{81}{4}\sqrt{3} - 9\pi = \frac{189}{8}\sqrt{3} - 9\pi\)	A1	Correct answer: \(p = \frac{189}{8}\), \(q = -9\)