| Exam Board | OCR |
|---|---|
| Module | H240/01 (Pure Mathematics) |
| Year | 2018 |
| Session | March |
| Marks | 12 |
| Topic | Trig Graphs & Exact Values |
| Type | Real-world modelling (tides, daylight, etc.) |
| Difficulty | Moderate -0.3 This is a standard A-level Pure 1 trigonometric modelling question with routine calculations. Parts (i)-(ii) involve straightforward reading of amplitude/vertical shift and solving a basic sine equation. Part (iii) requires determining parameters from given max/min values and times, which is a standard textbook exercise. Part (iv) is a simple conceptual question about period. All techniques are routine for this level with no novel problem-solving required. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| \(2.5m\) | B1 | State 2.5m |
| Answer | Marks | Guidance |
|---|---|---|
| \(30r = 90\); \(t = 3\), hence time is 0300 | M1, A1 | Identify that \(30r = 90\) soi; Solve for \(t\), and deduce time is 0300 |
| Answer | Marks | Guidance |
|---|---|---|
| \(1.7 + 0.8\sin(30t) = 1.2\); \(\sin(30t) = -0.625\); \(30t = 218.68\); \(t = 7.289\); hence time is 0717 | M1, M1, A1, A1FT | Equate to 1.3 and rearrange to \(\sin(30t) = k\); Attempt to find value for \(t\) from angle in third quadrant; Obtain 7.29, or better; Obtain time as 0717; FT their decimal value for \(t\), as long as 2dp or better |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 1.9, b = 1.2\); \(c = 30\); \(1.9 + 1.2\sin(150 + d) = 3.1\); \(150 + d = 90\); \(d = -60\) | B1, B1, M1, A1 | Correct values for \(a\) and \(b\); Correct value for \(c\); Attempt to find \(d\) eg use \(f(2) = 3.1\); Obtain correct value for \(d\) |
| Answer | Marks | Guidance |
|---|---|---|
| Identify decrease with reason | B1 | Eg There is a longer time between maximum heights, so the value of \(c\) will need decrease to give longer period for the sine model |
**(i) (a)**
| $2.5m$ | B1 | State 2.5m |
**(i) (b)**
| $30r = 90$; $t = 3$, hence time is 0300 | M1, A1 | Identify that $30r = 90$ soi; Solve for $t$, and deduce time is 0300 |
**(ii)**
| $1.7 + 0.8\sin(30t) = 1.2$; $\sin(30t) = -0.625$; $30t = 218.68$; $t = 7.289$; hence time is 0717 | M1, M1, A1, A1FT | Equate to 1.3 and rearrange to $\sin(30t) = k$; Attempt to find value for $t$ from angle in third quadrant; Obtain 7.29, or better; Obtain time as 0717; FT their decimal value for $t$, as long as 2dp or better |
**(iii)**
| $a = 1.9, b = 1.2$; $c = 30$; $1.9 + 1.2\sin(150 + d) = 3.1$; $150 + d = 90$; $d = -60$ | B1, B1, M1, A1 | Correct values for $a$ and $b$; Correct value for $c$; Attempt to find $d$ eg use $f(2) = 3.1$; Obtain correct value for $d$ |
**(iv)**
| Identify decrease with reason | B1 | Eg There is a longer time between maximum heights, so the value of $c$ will need decrease to give longer period for the sine model |11 The height, in metres, of the sea at a coastal town during a day may be modelled by the function
$$\mathrm { f } ( t ) = 1.7 + 0.8 \sin ( 30 t ) ^ { \circ }$$
where $t$ is the number of hours after midnight.
\begin{enumerate}[label=(\roman*)]
\item (a) Find the maximum height of the sea as given by this model.\\
(b) Find the time of day at which this maximum height first occurs.
\item Determine the time when, according to this model, the height of the sea will first be 1.2 m .
The height, in metres, at a different coastal town during a day may be modelled by the function
$$\mathrm { g } ( t ) = a + b \sin ( c t + d ) ^ { \circ }$$
where $t$ is the number of hours after midnight.
\item It is given that at this different coastal town the maximum height of the sea is 3.1 m , and this height occurs at 0500 and 1700. The minimum height of the sea is 0.7 m , and this height occurs at 1100 and 2300 . Find the values of the constants $a , b , c$ and $d$.
\item It is instead given that the maximum height of the sea actually occurs at 0500 and 1709 . State, with a reason, how this will affect the value of $c$ found in part (iii).\\
\includegraphics[max width=\textwidth, alt={}, center]{74a37bca-0b28-4c48-bd21-a9304f31b8f8-6_563_568_322_751}
The diagram shows the curve $y = \mathrm { e } ^ { \sqrt { x + 1 } }$ for $x \geqslant 0$.
\item Use the substitution $u ^ { 2 } = x + 1$ to find $\int \mathrm { e } ^ { \sqrt { x + 1 } } \mathrm {~d} x$.
\item Make $x$ the subject of the equation $y = \mathrm { e } ^ { \sqrt { x + 1 } }$.
\item Hence show that $\int _ { \mathrm { e } } ^ { \mathrm { e } ^ { 4 } } \left( ( \ln y ) ^ { 2 } - 1 \right) \mathrm { d } y = 9 \mathrm { e } ^ { 4 }$.
\section*{END OF QUESTION PAPER}
\section*{OCR}
Oxford Cambridge and RSA
\end{enumerate}
\hfill \mbox{\textit{OCR H240/01 2018 Q11 [12]}}