| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Real-world modelling (tides, daylight, etc.) |
| Difficulty | Standard +0.3 This is a straightforward application of cosine graph transformations requiring students to find maximum values (when cos = 1) and solve a trigonometric equation. The steps are routine: identify amplitude/vertical shift for part (a), solve a basic cosine equation for part (b), though the degree mode and time conversion add minor complexity. Slightly above average due to the two-part structure and real-world context requiring careful interpretation, but all techniques are standard C2 material. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h = 3.7 + 2.5\cos(30t - 40)°\), \(0 \leq t < 24\) | ||
| Max \(= 3.7 + 2.5 = 6.2\) m | B1 | \(h_{\max} = 6.2\,(m)\). Units not important |
| \(30t - 40 = 0 \Rightarrow t = \frac{40}{30} = 1:20am\,(01:20)\) | M1 | Solves \(30t - 40 = 0 \Rightarrow t = ..\) or \(30t - 40 = 360 \Rightarrow t = ..\) May be implied by \(t = \frac{40}{30}, \frac{400}{30}\) |
| \(t = \frac{40}{30}\) | A1 | Or exact equivalents like \(\frac{4}{3}, 1.\dot{3}\) |
| \(1:20\)am or \((01:20)\) | A1 | Exact time of day required. \(1:20\) or \(1:20\)pm is incorrect |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3.7 + 2.5\cos(30t-40)° = 3 \Rightarrow \cos(30t-40)° = -\frac{0.7}{2.5} = (-0.28)\) | M1 | Sets \(h=3\) and proceeds to \(\cos(30t-40)° = ..\) Accept inequalities in place of \(=\) sign |
| \(30t - 40 = 106.3,\,(253.7)\) | A1 | Proceeds by taking invcos to reach either value. Accept inequalities |
| \(t =\) awrt \(4.9\) or \(9.8\) | A1 | One value for \(t\) correct. Accept either awrt 1 dp |
| \(2^{\text{nd}}\) Value: \(30t - 40 = 253.7 \Rightarrow t = ..\) | M1 | For correct method of finding second value of \(t\). Accept \(30t - 40 = (360 - \alpha) \Rightarrow t = ..\) where \(\alpha\) is principal value |
| \(t =\) awrt \(4.9\) and \(9.8\) | A1 | Both values correct awrt 1 dp. Ignore values where \(t \geq 12\) but withhold for extra values in range |
| Boat cannot enter harbour between \(04:53\) and \(09:47\) | A1 | cso. Accept both \(0453\) and \(0947\). Accept \(4:53\) and \(9:47\) without am. Accept \(293\) and \(587\) minutes |
## Question 13:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h = 3.7 + 2.5\cos(30t - 40)°$, $0 \leq t < 24$ | | |
| Max $= 3.7 + 2.5 = 6.2$ m | B1 | $h_{\max} = 6.2\,(m)$. Units not important |
| $30t - 40 = 0 \Rightarrow t = \frac{40}{30} = 1:20am\,(01:20)$ | M1 | Solves $30t - 40 = 0 \Rightarrow t = ..$ or $30t - 40 = 360 \Rightarrow t = ..$ May be implied by $t = \frac{40}{30}, \frac{400}{30}$ |
| $t = \frac{40}{30}$ | A1 | Or exact equivalents like $\frac{4}{3}, 1.\dot{3}$ |
| $1:20$am or $(01:20)$ | A1 | Exact time of day required. $1:20$ or $1:20$pm is incorrect |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3.7 + 2.5\cos(30t-40)° = 3 \Rightarrow \cos(30t-40)° = -\frac{0.7}{2.5} = (-0.28)$ | M1 | Sets $h=3$ and proceeds to $\cos(30t-40)° = ..$ Accept inequalities in place of $=$ sign |
| $30t - 40 = 106.3,\,(253.7)$ | A1 | Proceeds by taking invcos to reach either value. Accept inequalities |
| $t =$ awrt $4.9$ or $9.8$ | A1 | One value for $t$ correct. Accept either awrt 1 dp |
| $2^{\text{nd}}$ Value: $30t - 40 = 253.7 \Rightarrow t = ..$ | M1 | For correct method of finding second value of $t$. Accept $30t - 40 = (360 - \alpha) \Rightarrow t = ..$ where $\alpha$ is principal value |
| $t =$ awrt $4.9$ and $9.8$ | A1 | Both values correct awrt 1 dp. Ignore values where $t \geq 12$ but withhold for extra values in range |
| Boat cannot enter harbour between $04:53$ and $09:47$ | A1 | cso. Accept both $0453$ and $0947$. Accept $4:53$ and $9:47$ without am. Accept $293$ and $587$ minutes |
---13. The height of sea water, $h$ metres, on a harbour wall at time $t$ hours after midnight is given by
$$h = 3.7 + 2.5 \cos ( 30 t - 40 ) ^ { \circ } , \quad 0 \leqslant t < 24$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the maximum value of $h$ and the exact time of day when this maximum first occurs.
Fishing boats cannot enter the harbour if $h$ is less than 3
\item Find the times during the morning between which fishing boats cannot enter the harbour.\\
Give these times to the nearest minute.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q13 [10]}}