Question 3 - A-Level Maths

Edexcel F3 2023 June — Question 3 11 marks

Exam Board	Edexcel
Module	F3 (Further Pure Mathematics 3)
Year	2023
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Trig Graphs & Exact Values
Type	Prove or show algebraic identity
Difficulty	Challenging +1.8 This is a Further Maths arc length question requiring differentiation of tan/cot, algebraic manipulation to simplify the arc length integrand, and integration using trig identities. While the calculus is standard, the algebraic simplification (showing √(tan²x + cot²x + 2) = tan²x + cot²x) requires insight, and part (b) needs exact value evaluation at π/6 and π/3. The multi-step nature and Further Maths context place it above average difficulty.
Spec	4.08d Volumes of revolution: about x and y axes

3. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{f1efd9b3-d604-4088-a4b5-8680711aa8f1-08_353_474_301_781} \captionsetup{labelformat=empty} \caption{Figure 1}

\end{figure} Figure 1 shows a sketch of the curve $C$ with equation $$y = \frac { 1 } { 2 } ( \tan x + \cot x ) \quad \frac { \pi } { 6 } \leqslant x \leqslant \frac { \pi } { 3 }$$

Show that the length of $C$ is given by $$\frac { 1 } { 2 } \int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 3 } } \left( \tan ^ { 2 } x + \cot ^ { 2 } x \right) d x$$
Hence determine the exact length of $C$, giving your answer in simplest form.

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Question 3(a):

Answer	Marks	Guidance
$y=\frac{1}{2}(\tan x+\cot x)\Rightarrow \frac{dy}{dx}=\frac{1}{2}(\sec^2 x - \mathrm{cosec}^2 x)$	B1	Correct derivative, any equivalent
Applies $\sec^2 x = \pm\tan^2 x \pm 1$ and $\mathrm{cosec}^2 x = \pm\cot^2 x \pm 1$ to derivative	M1	As described
$\left(\frac{dy}{dx}\right)^2=\frac{1}{4}(\tan^4 x+\cot^4 x-2\tan^2 x\cot^2 x)$	dM1	Squares to a 3 (or 4 if middle terms uncollected) term expression; requires previous M mark
$\left\{1+\left(\frac{dy}{dx}\right)^2\right\}=1+\frac{1}{4}(\tan^4 x+\cot^4 x-2)$ leading to $\frac{1}{4}(\tan^4 x+\cot^4 x+2)$ or $\frac{1}{4}\tan^4 x+\frac{1}{4}\cot^4 x+\frac{1}{2}$	A1	Adds 1 and achieves expression shown; constant must be multiplied by $\tan^2 x\cot^2 x$; not implied, must be seen
$s=\int_{\pi/6}^{\pi/3}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \frac{1}{2}\int_{\pi/6}^{\pi/3}(\tan^2 x+\cot^2 x)\,dx$	M1 A1*	M1: applies arc length formula with their $\frac{dy}{dx}$; A1: correct result with no clear errors; condone omission of "dx" and/or limits and occasional missing arguments

Question 3(b):

Answer	Marks	Guidance
$\frac{1}{2}\int_{\pi/6}^{\pi/3}(\tan^2 x+\cot^2 x)\,dx = \frac{1}{2}\int_{\pi/6}^{\pi/3}(\sec^2 x-1+\mathrm{cosec}^2 x-1)\,dx$	M1	Applies $\tan^2 x=\pm\sec^2 x\pm 1$ and $\cot^2 x=\pm\mathrm{cosec}^2 x\pm 1$
$=\frac{1}{2}\left[\tan x-\cot x-2x\right]_{\pi/6}^{\pi/3}$	dM1 A1	M1: $\pm\sec^2 x\to\pm\tan x$ and $\pm\mathrm{cosec}^2 x\to\pm\cot x$; requires previous M; A1: correct integration, limits not required
$\frac{1}{2}\left(\tan\frac{\pi}{3}-\cot\frac{\pi}{3}-\frac{2\pi}{3}-\left(\tan\frac{\pi}{6}-\cot\frac{\pi}{6}-\frac{2\pi}{6}\right)\right)$	M1	Applies limits following completed integration; clear attempt at $f\!\left(\frac{\pi}{3}\right)-f\!\left(\frac{\pi}{6}\right)$
$\frac{1}{2}\left(\frac{4\sqrt{3}}{3}-\frac{\pi}{3}\right),\quad \frac{2\sqrt{3}}{3}-\frac{\pi}{6},\quad \frac{2}{\sqrt{3}}-\frac{\pi}{6},\quad \frac{1}{3}\left(2\sqrt{3}-\frac{\pi}{2}\right),\quad \frac{4\sqrt{3}-\pi}{6}$	A1	Correct answer in any exact simplified form with 2 terms

# Question 3(a):

$y=\frac{1}{2}(\tan x+\cot x)\Rightarrow \frac{dy}{dx}=\frac{1}{2}(\sec^2 x - \mathrm{cosec}^2 x)$ | B1 | Correct derivative, any equivalent

Applies $\sec^2 x = \pm\tan^2 x \pm 1$ and $\mathrm{cosec}^2 x = \pm\cot^2 x \pm 1$ to derivative | M1 | As described

$\left(\frac{dy}{dx}\right)^2=\frac{1}{4}(\tan^4 x+\cot^4 x-2\tan^2 x\cot^2 x)$ | dM1 | Squares to a 3 (or 4 if middle terms uncollected) term expression; requires previous M mark

$\left\{1+\left(\frac{dy}{dx}\right)^2\right\}=1+\frac{1}{4}(\tan^4 x+\cot^4 x-2)$ leading to $\frac{1}{4}(\tan^4 x+\cot^4 x+2)$ or $\frac{1}{4}\tan^4 x+\frac{1}{4}\cot^4 x+\frac{1}{2}$ | A1 | Adds 1 and achieves expression shown; constant must be multiplied by $\tan^2 x\cot^2 x$; not implied, must be seen

$s=\int_{\pi/6}^{\pi/3}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \frac{1}{2}\int_{\pi/6}^{\pi/3}(\tan^2 x+\cot^2 x)\,dx$ | M1 A1* | M1: applies arc length formula with their $\frac{dy}{dx}$; A1: correct result with no clear errors; condone omission of "dx" and/or limits and occasional missing arguments

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# Question 3(b):

$\frac{1}{2}\int_{\pi/6}^{\pi/3}(\tan^2 x+\cot^2 x)\,dx = \frac{1}{2}\int_{\pi/6}^{\pi/3}(\sec^2 x-1+\mathrm{cosec}^2 x-1)\,dx$ | M1 | Applies $\tan^2 x=\pm\sec^2 x\pm 1$ and $\cot^2 x=\pm\mathrm{cosec}^2 x\pm 1$

$=\frac{1}{2}\left[\tan x-\cot x-2x\right]_{\pi/6}^{\pi/3}$ | dM1 A1 | M1: $\pm\sec^2 x\to\pm\tan x$ and $\pm\mathrm{cosec}^2 x\to\pm\cot x$; requires previous M; A1: correct integration, limits not required

$\frac{1}{2}\left(\tan\frac{\pi}{3}-\cot\frac{\pi}{3}-\frac{2\pi}{3}-\left(\tan\frac{\pi}{6}-\cot\frac{\pi}{6}-\frac{2\pi}{6}\right)\right)$ | M1 | Applies limits following completed integration; clear attempt at $f\!\left(\frac{\pi}{3}\right)-f\!\left(\frac{\pi}{6}\right)$

$\frac{1}{2}\left(\frac{4\sqrt{3}}{3}-\frac{\pi}{3}\right),\quad \frac{2\sqrt{3}}{3}-\frac{\pi}{6},\quad \frac{2}{\sqrt{3}}-\frac{\pi}{6},\quad \frac{1}{3}\left(2\sqrt{3}-\frac{\pi}{2}\right),\quad \frac{4\sqrt{3}-\pi}{6}$ | A1 | Correct answer in any exact simplified form with 2 terms

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3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f1efd9b3-d604-4088-a4b5-8680711aa8f1-08_353_474_301_781}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

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

$$y = \frac { 1 } { 2 } ( \tan x + \cot x ) \quad \frac { \pi } { 6 } \leqslant x \leqslant \frac { \pi } { 3 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that the length of $C$ is given by

$$\frac { 1 } { 2 } \int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 3 } } \left( \tan ^ { 2 } x + \cot ^ { 2 } x \right) d x$$
\item Hence determine the exact length of $C$, giving your answer in simplest form.
\end{enumerate}

\hfill \mbox{\textit{Edexcel F3 2023 Q3 [11]}}

This paper (8 questions)

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Q1 5 Q2 8 Q3 11 Q4 12 Q5 7 Q6 13 Q7 9 Q8 10

Answer	Marks	Guidance
\(y=\frac{1}{2}(\tan x+\cot x)\Rightarrow \frac{dy}{dx}=\frac{1}{2}(\sec^2 x - \mathrm{cosec}^2 x)\)	B1	Correct derivative, any equivalent
Applies \(\sec^2 x = \pm\tan^2 x \pm 1\) and \(\mathrm{cosec}^2 x = \pm\cot^2 x \pm 1\) to derivative	M1	As described
\(\left(\frac{dy}{dx}\right)^2=\frac{1}{4}(\tan^4 x+\cot^4 x-2\tan^2 x\cot^2 x)\)	dM1	Squares to a 3 (or 4 if middle terms uncollected) term expression; requires previous M mark
\(\left\{1+\left(\frac{dy}{dx}\right)^2\right\}=1+\frac{1}{4}(\tan^4 x+\cot^4 x-2)\) leading to \(\frac{1}{4}(\tan^4 x+\cot^4 x+2)\) or \(\frac{1}{4}\tan^4 x+\frac{1}{4}\cot^4 x+\frac{1}{2}\)	A1	Adds 1 and achieves expression shown; constant must be multiplied by \(\tan^2 x\cot^2 x\); not implied, must be seen
\(s=\int_{\pi/6}^{\pi/3}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \frac{1}{2}\int_{\pi/6}^{\pi/3}(\tan^2 x+\cot^2 x)\,dx\)	M1 A1*	M1: applies arc length formula with their \(\frac{dy}{dx}\); A1: correct result with no clear errors; condone omission of "dx" and/or limits and occasional missing arguments

Answer	Marks	Guidance
\(\frac{1}{2}\int_{\pi/6}^{\pi/3}(\tan^2 x+\cot^2 x)\,dx = \frac{1}{2}\int_{\pi/6}^{\pi/3}(\sec^2 x-1+\mathrm{cosec}^2 x-1)\,dx\)	M1	Applies \(\tan^2 x=\pm\sec^2 x\pm 1\) and \(\cot^2 x=\pm\mathrm{cosec}^2 x\pm 1\)
\(=\frac{1}{2}\left[\tan x-\cot x-2x\right]_{\pi/6}^{\pi/3}\)	dM1 A1	M1: \(\pm\sec^2 x\to\pm\tan x\) and \(\pm\mathrm{cosec}^2 x\to\pm\cot x\); requires previous M; A1: correct integration, limits not required
\(\frac{1}{2}\left(\tan\frac{\pi}{3}-\cot\frac{\pi}{3}-\frac{2\pi}{3}-\left(\tan\frac{\pi}{6}-\cot\frac{\pi}{6}-\frac{2\pi}{6}\right)\right)\)	M1	Applies limits following completed integration; clear attempt at \(f\!\left(\frac{\pi}{3}\right)-f\!\left(\frac{\pi}{6}\right)\)
\(\frac{1}{2}\left(\frac{4\sqrt{3}}{3}-\frac{\pi}{3}\right),\quad \frac{2\sqrt{3}}{3}-\frac{\pi}{6},\quad \frac{2}{\sqrt{3}}-\frac{\pi}{6},\quad \frac{1}{3}\left(2\sqrt{3}-\frac{\pi}{2}\right),\quad \frac{4\sqrt{3}-\pi}{6}\)	A1	Correct answer in any exact simplified form with 2 terms