| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Applied context modeling |
| Difficulty | Standard +0.8 This is a Further Maths FP1 question requiring harmonic form manipulation with t-substitution (a non-standard technique), followed by solving a rational equation. Part (a) is routine substitution, but parts (b) and (c) require careful algebraic manipulation of trigonometric identities and solving a quadratic equation in context. The t-substitution method is less familiar than standard R cos(x-α) form, making this moderately challenging for Further Maths students. |
| Spec | 1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((H=)0.3\sin\left(\frac{30}{60}\right) - 4\cos\left(\frac{30}{60}\right) + 11.5 = 8.13\) {hours} | B1* | Uses \(x=30\) to show \(H=8.13\). Accept \(x=30\) seen substituted followed by 8.13, or \(x=30\) identified followed by 8.133… before rounding |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitutes \(\sin\left(\frac{x}{60}\right) = \frac{2t}{1+t^2}\) and \(\cos\left(\frac{x}{60}\right) = \frac{1-t^2}{1+t^2}\) into \(H\); \((H=)0.3\left(\frac{2t}{1+t^2}\right) - 4\left(\frac{1-t^2}{1+t^2}\right) + 11.5\) | M1 | Uses correct \(t\)-formulae, attempts to substitute into \(H\) |
| \((H=)\frac{0.6t - 4 + 4t^2 + 11.5(1+t^2)}{1+t^2} = \frac{15.5t^2 + 0.6t + 7.5}{1+t^2}\) | A1 | Fully correct method, expresses as single fraction with denominator \(1+t^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H = \frac{15.5t^2 + 0.6t + 7.5}{1+t^2} = 12 \Rightarrow 3.5t^2 + 0.6t - 4.5 = 0\) | M1 | Sets \(H=12\) (or any inequality in between) and rearranges to quadratic in \(t\) |
| \(t = \frac{-0.6 \pm \sqrt{0.6^2 - 4(3.5)(-4.5)}}{7}\); \(=\ldots(1.051\ldots, -1.222\ldots) \Rightarrow x = 120\tan^{-1}(``1.051\ldots") = \ldots(97.254\ldots)\) | dM1 | Dependent on M1. Solves quadratic by any means and uses to find value for \(x\) |
| \(x =\) awrt \(97\) | A1 | Correct value for \(x\) = awrt 97 or accept 98 following correct \(t\) |
| \(8^{\text{th}}\) or \(9^{\text{th}}\) April | A1 | Correct day of year. Accept \(8^{\text{th}}\) or \(9^{\text{th}}\) April following awrt 97 from correct method |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(H=)0.3\sin\left(\frac{30}{60}\right) - 4\cos\left(\frac{30}{60}\right) + 11.5 = 8.13$ {hours} | B1* | Uses $x=30$ to show $H=8.13$. Accept $x=30$ seen substituted followed by 8.13, or $x=30$ identified followed by 8.133… before rounding |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $\sin\left(\frac{x}{60}\right) = \frac{2t}{1+t^2}$ and $\cos\left(\frac{x}{60}\right) = \frac{1-t^2}{1+t^2}$ into $H$; $(H=)0.3\left(\frac{2t}{1+t^2}\right) - 4\left(\frac{1-t^2}{1+t^2}\right) + 11.5$ | M1 | Uses correct $t$-formulae, attempts to substitute into $H$ |
| $(H=)\frac{0.6t - 4 + 4t^2 + 11.5(1+t^2)}{1+t^2} = \frac{15.5t^2 + 0.6t + 7.5}{1+t^2}$ | A1 | Fully correct method, expresses as single fraction with denominator $1+t^2$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H = \frac{15.5t^2 + 0.6t + 7.5}{1+t^2} = 12 \Rightarrow 3.5t^2 + 0.6t - 4.5 = 0$ | M1 | Sets $H=12$ (or any inequality in between) and rearranges to quadratic in $t$ |
| $t = \frac{-0.6 \pm \sqrt{0.6^2 - 4(3.5)(-4.5)}}{7}$; $=\ldots(1.051\ldots, -1.222\ldots) \Rightarrow x = 120\tan^{-1}(``1.051\ldots") = \ldots(97.254\ldots)$ | dM1 | Dependent on M1. Solves quadratic by any means and uses to find value for $x$ |
| $x =$ awrt $97$ | A1 | Correct value for $x$ = awrt 97 or accept 98 following correct $t$ |
| $8^{\text{th}}$ or $9^{\text{th}}$ April | A1 | Correct day of year. Accept $8^{\text{th}}$ or $9^{\text{th}}$ April following awrt 97 from correct method |
---\begin{enumerate}
\item During 2029, the number of hours of daylight per day in London, H, is modelled by the equation
\end{enumerate}
$$H = 0.3 \sin \left( \frac { x } { 60 } \right) - 4 \cos \left( \frac { x } { 60 } \right) + 11.5 \quad 0 \leqslant x < 365$$
where $x$ is the number of days after 1st January 2029 and the angle is in radians.\\
(a) Show that, according to the model, the number of hours of daylight in London on the 31st January 2029 will be 8.13 to 3 significant figures.\\
(b) Use the substitution $t = \tan \left( \frac { x } { 120 } \right)$ to show that $H$ can be written as
$$H = \frac { a t ^ { 2 } + b t + c } { 1 + t ^ { 2 } }$$
where $a$, $b$ and $c$ are constants to be determined.\\
(c) Hence determine, according to the model, the date of the first day of 2029 when there will be at least 12 hours of daylight in London.
\hfill \mbox{\textit{Edexcel FP1 2022 Q2 [7]}}