| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Session | Specimen |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Stationary points of curves |
| Difficulty | Challenging +1.2 This is a multi-part FP1 question involving t-substitution for harmonic forms and differentiation. Part (a) requires systematic application of the t = tan(x/4) substitution formulas and algebraic manipulation to reach the given form - this is a standard FP1 technique but involves careful bookkeeping across multiple steps. Parts (b) and (c) are contextual interpretation requiring minimal calculation. While the algebra is lengthy, this is a routine application of a core FP1 method with no novel insight required, making it moderately above average difficulty for A-level but standard for Further Maths. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07k Differentiate trig: sin(kx), cos(kx), tan(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dh}{dx} = 15\cos x + \frac{21}{2}\cos\left(\frac{x}{2}\right) - \frac{25}{2}\sin\left(\frac{x}{2}\right)\) | M1 | Differentiates \(h(x)\) |
| \(\frac{dh}{dx} = \ldots + \ldots\frac{1-t^2}{1+t^2} - \ldots\frac{2t}{1+t^2}\) | M1 | Applies \(t\)-substitution to both \(\left(\frac{x}{2}\right)\) terms with their coefficients |
| e.g. \(\frac{dh}{dx} = \ldots\left(2\left(\frac{1-t^2}{1+t^2}\right)^2 - 1\right) + \ldots\) or \(\frac{dh}{dx} = \ldots\frac{1-\left(\frac{2t}{1-t^2}\right)^2}{1+\left(\frac{2t}{1-t^2}\right)^2} + \ldots\) | M1 | Forms correct expression in \(t\) for the \(\cos x\) term using double angle formula and \(t\)-substitution |
| e.g. \(\frac{dh}{dx} = 15\left(2\left(\frac{1-t^2}{1+t^2}\right)^2 - 1\right) + \frac{21}{2}\left(\frac{1-t^2}{1+t^2}\right) - \frac{25}{2}\left(\frac{2t}{1+t^2}\right)\) | A1 | Fully correct expression in \(t\) for \(\frac{dh}{dx}\) |
| \(\ldots = \frac{15[4(1-t^2)^2 - 2(1+t^2)^2] + 21(1-t^2)(1+t^2) - 50t(1+t^2)}{2(1+t^2)^2}\) | M1 | Gets all terms over correct common factor; numerators appropriate for their terms |
| \(\ldots = \frac{9t^4 - 50t^3 - 180t^2 - 50t + 51}{2(1+t^2)^2} = \frac{(t^2 - 6t - 17)(9t^2 + 4t - 3)}{2(1+t^2)^2}\) | A1* | Achieves correct answer via expression with correct quartic numerator before factorisation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Accept any value between \(\frac{1}{40} = 0.025\) and \(\frac{1}{60} \approx 0.167\) inclusive | B1 | Note: taking high peak as reference gives \(\frac{1}{50}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Suitable: graphs both oscillate bi-modally with about the same periodicity | B1 | Should mention both bimodal nature and periodicity matching the graph of \(h\) |
| Not suitable: heights of peaks vary over time, but \(h(x)\) has fixed peak height | B1 | Mentions that the heights of peaks vary in each oscillation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t = \frac{6 \pm \sqrt{36 - 4 \times 1 \times 17}}{2} = 3 \pm \sqrt{26}\) or \(t = \frac{-4 \pm \sqrt{16 - 4 \times 9 \times (-3)}}{18} = \frac{-2 \pm \sqrt{31}}{9}\) | M1 | Solves at least one of the quadratics |
| Finds corresponding \(x\) values, \(x = 4\tan^{-1}(t)\) for at least one value of \(t\) from the \(9t^2 + 4t - 3\) factor | M1 | Must attempt to solve quadratic factor from which solution comes; uses \(t = \tan\left(\frac{x}{4}\right)\) |
| One correct value e.g. \(x =\) awrt \(-2.797\) or \(9.770\), \(1.510\) | A1 | At least one correct \(x\) value from requisite quadratic: awrt any of \(-2.797\), \(1.510\), \(9.770\), \(14.076\), \(22.336\), \(26.642\), \(34.902\) or \(39.208\) |
| Maximum peak height occurs at smallest positive value of \(x\); third of these peaks needed; so \(t = 1.509\ldots + 8\pi = 26.642\) is the required time | M1 | Uses graph of \(h\) to pick out \(x = 26.642\) as time corresponding to third of the higher peaks on 4th January |
| \(x = 26.642\) corresponds to 26 hours and 39 minutes (nearest minute) after 08:00 on 3rd January | M1 | Finds time for one value of \(t\) corresponding to highest peaks; only follow through on smallest positive \(t\) solution \(+ 4k\pi\) |
| Time of greatest tide height is approximately 10:39 (am) (also allow 10:38 or 10:40) | A1 | Time of greatest tide height on 4th January is approximately 10:39 |
## Question 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dh}{dx} = 15\cos x + \frac{21}{2}\cos\left(\frac{x}{2}\right) - \frac{25}{2}\sin\left(\frac{x}{2}\right)$ | M1 | Differentiates $h(x)$ |
| $\frac{dh}{dx} = \ldots + \ldots\frac{1-t^2}{1+t^2} - \ldots\frac{2t}{1+t^2}$ | M1 | Applies $t$-substitution to both $\left(\frac{x}{2}\right)$ terms with their coefficients |
| e.g. $\frac{dh}{dx} = \ldots\left(2\left(\frac{1-t^2}{1+t^2}\right)^2 - 1\right) + \ldots$ or $\frac{dh}{dx} = \ldots\frac{1-\left(\frac{2t}{1-t^2}\right)^2}{1+\left(\frac{2t}{1-t^2}\right)^2} + \ldots$ | M1 | Forms correct expression in $t$ for the $\cos x$ term using double angle formula and $t$-substitution |
| e.g. $\frac{dh}{dx} = 15\left(2\left(\frac{1-t^2}{1+t^2}\right)^2 - 1\right) + \frac{21}{2}\left(\frac{1-t^2}{1+t^2}\right) - \frac{25}{2}\left(\frac{2t}{1+t^2}\right)$ | A1 | Fully correct expression in $t$ for $\frac{dh}{dx}$ |
| $\ldots = \frac{15[4(1-t^2)^2 - 2(1+t^2)^2] + 21(1-t^2)(1+t^2) - 50t(1+t^2)}{2(1+t^2)^2}$ | M1 | Gets all terms over correct common factor; numerators appropriate for their terms |
| $\ldots = \frac{9t^4 - 50t^3 - 180t^2 - 50t + 51}{2(1+t^2)^2} = \frac{(t^2 - 6t - 17)(9t^2 + 4t - 3)}{2(1+t^2)^2}$ | A1* | Achieves correct answer via expression with correct quartic numerator before factorisation |
---
## Question 8(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Accept any value between $\frac{1}{40} = 0.025$ and $\frac{1}{60} \approx 0.167$ inclusive | B1 | Note: taking high peak as reference gives $\frac{1}{50}$ |
---
## Question 8(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Suitable: graphs both oscillate bi-modally with about the same periodicity | B1 | Should mention both bimodal nature and periodicity matching the graph of $h$ |
| Not suitable: heights of peaks vary over time, but $h(x)$ has fixed peak height | B1 | Mentions that the heights of peaks vary in each oscillation |
---
## Question 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t = \frac{6 \pm \sqrt{36 - 4 \times 1 \times 17}}{2} = 3 \pm \sqrt{26}$ or $t = \frac{-4 \pm \sqrt{16 - 4 \times 9 \times (-3)}}{18} = \frac{-2 \pm \sqrt{31}}{9}$ | M1 | Solves at least one of the quadratics |
| Finds corresponding $x$ values, $x = 4\tan^{-1}(t)$ for at least one value of $t$ from the $9t^2 + 4t - 3$ factor | M1 | Must attempt to solve quadratic factor from which solution comes; uses $t = \tan\left(\frac{x}{4}\right)$ |
| One correct value e.g. $x =$ awrt $-2.797$ or $9.770$, $1.510$ | A1 | At least one correct $x$ value from requisite quadratic: awrt any of $-2.797$, $1.510$, $9.770$, $14.076$, $22.336$, $26.642$, $34.902$ or $39.208$ |
| Maximum peak height occurs at smallest positive value of $x$; third of these peaks needed; so $t = 1.509\ldots + 8\pi = 26.642$ is the required time | M1 | Uses graph of $h$ to pick out $x = 26.642$ as time corresponding to third of the higher peaks on 4th January |
| $x = 26.642$ corresponds to 26 hours and 39 minutes (nearest minute) after 08:00 on 3rd January | M1 | Finds time for one value of $t$ corresponding to highest peaks; only follow through on smallest positive $t$ solution $+ 4k\pi$ |
| Time of greatest tide height is approximately **10:39 (am)** (also allow 10:38 or 10:40) | A1 | Time of greatest tide height on 4th January is approximately 10:39 |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a52911da-4b69-4d86-975e-d10e3a481e1d-16_407_1100_201_484}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the graph of the function $\mathrm { h } ( x )$ with equation
$$h ( x ) = 45 + 15 \sin x + 21 \sin \left( \frac { x } { 2 } \right) + 25 \cos \left( \frac { x } { 2 } \right) \quad x \in [ 0,40 ]$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac { \mathrm { d } h } { \mathrm {~d} x } = \frac { \left( t ^ { 2 } - 6 t - 17 \right) \left( 9 t ^ { 2 } + 4 t - 3 \right) } { 2 \left( 1 + t ^ { 2 } \right) ^ { 2 } }$$
where $t = \tan \left( \frac { x } { 4 } \right)$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a52911da-4b69-4d86-975e-d10e3a481e1d-16_581_1403_1263_331}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Source: ${ } ^ { 1 }$ Data taken on 29th December 2016 from \href{http://www.ukho.gov.uk/easytide/EasyTide}{http://www.ukho.gov.uk/easytide/EasyTide}
Figure 2 shows a graph of predicted tide heights, in metres, for Portland harbour from 08:00 on the 3rd January 2017 to the end of the 4th January $2017 { } ^ { 1 }$.
The graph of $k \mathrm {~h} ( x )$, where $k$ is a constant and $x$ is the number of hours after 08:00 on 3rd of January, can be used to model the predicted tide heights, in metres, for this period of time.
\item \begin{enumerate}[label=(\roman*)]
\item Suggest a value of $k$ that could be used for the graph of $k \mathrm {~h} ( x )$ to form a suitable model.
\item Why may such a model be suitable to predict the times when the tide heights are at their peaks, but not to predict the heights of these peaks?
\end{enumerate}\item Use Figure 2 and the result of part (a) to estimate, to the nearest minute, the time of the highest tide height on the 4th January 2017.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 Q8 [15]}}