Question 12 - A-Level Maths

AQA Paper 1 2023 June — Question 12 8 marks

Exam Board	AQA
Module	Paper 1 (Paper 1)
Year	2023
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Trig Graphs & Exact Values
Type	Real-world modelling (tides, daylight, etc.)
Difficulty	Easy -1.2 This is a straightforward substitution question requiring only evaluation of cos(0) and sin(0), then basic arithmetic. It's simpler than average A-level questions as it involves no problem-solving, just direct calculation with exact values at a standard angle.
Spec	1.05a Sine, cosine, tangent: definitions for all arguments 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^2

12 One of the rides at a theme park is a room where the floor and ceiling both move up and down for $10 \pi$ seconds. At time $t$ seconds after the ride begins, the distance $f$ metres of the floor above the ground is $$f = 1 - \cos t$$ At time $t$ seconds after the ride begins, the distance $c$ metres of the ceiling above the ground is $$c = 8 - 4 \sin t$$ The ride is shown in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-16_448_766_932_635} 12

Show that the initial distance between the floor and ceiling is 8 metres.
[0pt] [1 mark]
\includegraphics[max width=\textwidth, alt={}]{6a03a035-ff32-4734-864b-a076aa9cbec0-17_2500_1721_214_148}

Show mark scheme Show mark scheme source

Question 12(a):

Answer	Marks	Guidance
$t = 0$	B1	Substitutes $t=0$ into both models; $8 - 4\sin 0 - (1-\cos 0) = 8$ metres; condone missing units

Question 12(b):

Answer	Marks	Guidance
$d = c - f$	M1	Models distance between ceiling and floor as $c - f$; condone sign error when expressions substituted
$= 8 - 4\sin t - (1 - \cos t) = 7 + \cos t - 4\sin t$	M1	Uses compound angle formula to obtain $R\cos\alpha = \pm 1$ or $\pm 4$, or $R\sin\alpha = \pm 4$ or $\pm 1$, or $\tan\alpha = \pm 4$ or $\pm\frac{1}{4}$
$R = \sqrt{17}$	A1	Obtains $R = \sqrt{17} \approx 4.1$; condone correct answer from $\sqrt{(\pm 1)^2 + (\pm 4)^2}$
$R\cos\alpha = 1$, $R\sin\alpha = 4$	A1	Obtains $\alpha = 1.33^c$ or $76°$; no incorrect working in finding $\alpha$
$\tan\alpha = 4,\ \alpha = 1.33$; $d = 7 + \sqrt{17}\cos(t + 1.33)$	R1	Completes argument; accept AWRT 4.1 for $\sqrt{17}$ and $\alpha = 1.33^c + 2n\pi$; do not award if $\sin\alpha = \pm 4$ or $\pm 1$ or $\cos\alpha = \pm 1$ or $\pm 4$ used leading to wrong $\tan\alpha$

Question 12(c):

Answer	Marks	Guidance
$7 - \sqrt{17} = 2.88\text{ m}$	M1	Subtracts their $R$ from 7 provided their $R < 7$
$2.88$ metres or $288$ cm	A1	Correct units must be seen

## Question 12(a):

$t = 0$ | B1 | Substitutes $t=0$ into both models; $8 - 4\sin 0 - (1-\cos 0) = 8$ metres; condone missing units

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## Question 12(b):

$d = c - f$ | M1 | Models distance between ceiling and floor as $c - f$; condone sign error when expressions substituted

$= 8 - 4\sin t - (1 - \cos t) = 7 + \cos t - 4\sin t$ | M1 | Uses compound angle formula to obtain $R\cos\alpha = \pm 1$ or $\pm 4$, or $R\sin\alpha = \pm 4$ or $\pm 1$, or $\tan\alpha = \pm 4$ or $\pm\frac{1}{4}$

$R = \sqrt{17}$ | A1 | Obtains $R = \sqrt{17} \approx 4.1$; condone correct answer from $\sqrt{(\pm 1)^2 + (\pm 4)^2}$

$R\cos\alpha = 1$, $R\sin\alpha = 4$ | A1 | Obtains $\alpha = 1.33^c$ or $76°$; no incorrect working in finding $\alpha$

$\tan\alpha = 4,\ \alpha = 1.33$; $d = 7 + \sqrt{17}\cos(t + 1.33)$ | R1 | Completes argument; accept AWRT 4.1 for $\sqrt{17}$ and $\alpha = 1.33^c + 2n\pi$; do not award if $\sin\alpha = \pm 4$ or $\pm 1$ or $\cos\alpha = \pm 1$ or $\pm 4$ used leading to wrong $\tan\alpha$

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## Question 12(c):

$7 - \sqrt{17} = 2.88\text{ m}$ | M1 | Subtracts their $R$ from 7 provided their $R < 7$

$2.88$ metres or $288$ cm | A1 | Correct units must be seen

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Show LaTeX source

12 One of the rides at a theme park is a room where the floor and ceiling both move up and down for $10 \pi$ seconds.

At time $t$ seconds after the ride begins, the distance $f$ metres of the floor above the ground is

$$f = 1 - \cos t$$

At time $t$ seconds after the ride begins, the distance $c$ metres of the ceiling above the ground is

$$c = 8 - 4 \sin t$$

The ride is shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-16_448_766_932_635}

12 (a) Show that the initial distance between the floor and ceiling is 8 metres.\\[0pt]
[1 mark]

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{6a03a035-ff32-4734-864b-a076aa9cbec0-17_2500_1721_214_148}
\end{center}

\hfill \mbox{\textit{AQA Paper 1 2023 Q12 [8]}}

This paper (16 questions)

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Q1 1 Q2 1 Q3 1 Q4 1 Q5 4 Q6 5 Q7 4 Q8 6 Q9 9 Q10 8 Q11 9 Q12 8 Q13 9 Q14 13 Q15 9 Q16 14

Answer	Marks	Guidance
\(d = c - f\)	M1	Models distance between ceiling and floor as \(c - f\); condone sign error when expressions substituted
\(= 8 - 4\sin t - (1 - \cos t) = 7 + \cos t - 4\sin t\)	M1	Uses compound angle formula to obtain \(R\cos\alpha = \pm 1\) or \(\pm 4\), or \(R\sin\alpha = \pm 4\) or \(\pm 1\), or \(\tan\alpha = \pm 4\) or \(\pm\frac{1}{4}\)
\(R = \sqrt{17}\)	A1	Obtains \(R = \sqrt{17} \approx 4.1\); condone correct answer from \(\sqrt{(\pm 1)^2 + (\pm 4)^2}\)
\(R\cos\alpha = 1\), \(R\sin\alpha = 4\)	A1	Obtains \(\alpha = 1.33^c\) or \(76°\); no incorrect working in finding \(\alpha\)
\(\tan\alpha = 4,\ \alpha = 1.33\); \(d = 7 + \sqrt{17}\cos(t + 1.33)\)	R1	Completes argument; accept AWRT 4.1 for \(\sqrt{17}\) and \(\alpha = 1.33^c + 2n\pi\); do not award if \(\sin\alpha = \pm 4\) or \(\pm 1\) or \(\cos\alpha = \pm 1\) or \(\pm 4\) used leading to wrong \(\tan\alpha\)

Answer	Marks	Guidance
\(7 - \sqrt{17} = 2.88\text{ m}\)	M1	Subtracts their \(R\) from 7 provided their \(R < 7\)
\(2.88\) metres or \(288\) cm	A1	Correct units must be seen