Edexcel C34 2017 June — Question 10 9 marks

Exam BoardEdexcel
ModuleC34 (Core Mathematics 3 & 4)
Year2017
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHarmonic Form
TypeApplied context modeling
DifficultyStandard +0.3 This is a standard harmonic form question with routine application. Part (a) uses the textbook method for converting to R sin(θ-α), part (b) requires reflecting negative portions above the x-axis (straightforward graph transformation), and part (c) applies the result to a context. All techniques are standard C3/C4 material with no novel insight required, making it slightly easier than average.
Spec1.02s Modulus graphs: sketch graph of |ax+b|1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc

  1. (a) Write \(2 \sin \theta - \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R\) and \(\alpha\) are constants, \(R > 0\) and \(0 < \alpha \leqslant 90 ^ { \circ }\). Give the exact value of \(R\) and give the value of \(\alpha\) to one decimal place.
    (3)
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-32_513_1194_404_374} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} Figure 3 shows a sketch of the graph with equation \(y = 2 \sin \theta - \cos \theta , \quad 0 \leqslant \theta < 360 ^ { \circ }\) (b) Sketch the graph with equation $$y = | 2 \sin \theta - \cos \theta | , \quad 0 \leqslant \theta < 360 ^ { \circ }$$ stating the coordinates of all points at which the graph meets or cuts the coordinate axes. The temperature of a warehouse is modelled by the equation $$f ( t ) = 5 + \left| 2 \sin ( 15 t ) ^ { \circ } - \cos ( 15 t ) ^ { \circ } \right| , \quad 0 \leqslant t < 24$$ where \(\mathrm { f } ( t )\) is the temperature of the warehouse in degrees Celsius and \(t\) is the time measured in hours from midnight. State
(c) (i) the maximum value of \(f ( t )\),
(ii) the largest value of \(t\), for \(0 \leqslant t < 24\), at which this maximum value occurs. Give your answer to one decimal place.

Question 10:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(R = \sqrt{5}\)B1
\(\tan\alpha = \frac{1}{2} \Rightarrow \alpha = 26.6°\)M1, A1 For \(\tan\alpha = \pm\frac{1}{2}\) or \(\pm\frac{2}{1}\) or \(\sin\alpha = \pm\frac{1}{\sqrt{5}}\) or \(\cos\alpha = \pm\frac{2}{\sqrt{5}}\); awrt \(\alpha = 26.6°\)
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Correct shape including cusps, starting downwards from positive \(y\)-axis with two maximaB1 Curve parts under \(x\)-axis reflected; cusps must look pointed not rounded
\((0,1)\)B1 May be on diagram or in script; allow \((1,0)\) if marked correctly on sketch
\((\text{"26.6"},0)\) and \((\text{"206.6"},0)\)B1ft Allow radians i.e. \(\alpha\) and \(\pi+\alpha\); awrt 26.6 and awrt 207 or ft values
Part (c)(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(5 + {'}R{'} = 5 + \sqrt{5}\)B1ft Follow through including decimal answers
Part (c)(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(15t - \text{'26.6'} = 270 \Rightarrow t = 19.8\)M1, A1 Attempts \(15t - \text{"26.6"} = 90\) or \(270\); \(t = 19.8\) only
# Question 10:

## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = \sqrt{5}$ | B1 | |
| $\tan\alpha = \frac{1}{2} \Rightarrow \alpha = 26.6°$ | M1, A1 | For $\tan\alpha = \pm\frac{1}{2}$ or $\pm\frac{2}{1}$ or $\sin\alpha = \pm\frac{1}{\sqrt{5}}$ or $\cos\alpha = \pm\frac{2}{\sqrt{5}}$; awrt $\alpha = 26.6°$ |

## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct shape including cusps, starting downwards from positive $y$-axis with two maxima | B1 | Curve parts under $x$-axis reflected; cusps must look pointed not rounded |
| $(0,1)$ | B1 | May be on diagram or in script; allow $(1,0)$ if marked correctly on sketch |
| $(\text{"26.6"},0)$ and $(\text{"206.6"},0)$ | B1ft | Allow radians i.e. $\alpha$ and $\pi+\alpha$; awrt 26.6 and awrt 207 or ft values |

## Part (c)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5 + {'}R{'} = 5 + \sqrt{5}$ | B1ft | Follow through including decimal answers |

## Part (c)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $15t - \text{'26.6'} = 270 \Rightarrow t = 19.8$ | M1, A1 | Attempts $15t - \text{"26.6"} = 90$ or $270$; $t = 19.8$ **only** |

---
\begin{enumerate}
  \item (a) Write $2 \sin \theta - \cos \theta$ in the form $R \sin ( \theta - \alpha )$, where $R$ and $\alpha$ are constants, $R > 0$ and $0 < \alpha \leqslant 90 ^ { \circ }$. Give the exact value of $R$ and give the value of $\alpha$ to one decimal place.\\
(3)
\end{enumerate}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-32_513_1194_404_374}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

Figure 3 shows a sketch of the graph with equation $y = 2 \sin \theta - \cos \theta , \quad 0 \leqslant \theta < 360 ^ { \circ }$\\
(b) Sketch the graph with equation

$$y = | 2 \sin \theta - \cos \theta | , \quad 0 \leqslant \theta < 360 ^ { \circ }$$

stating the coordinates of all points at which the graph meets or cuts the coordinate axes.

The temperature of a warehouse is modelled by the equation

$$f ( t ) = 5 + \left| 2 \sin ( 15 t ) ^ { \circ } - \cos ( 15 t ) ^ { \circ } \right| , \quad 0 \leqslant t < 24$$

where $\mathrm { f } ( t )$ is the temperature of the warehouse in degrees Celsius and $t$ is the time measured in hours from midnight.

State\\
(c) (i) the maximum value of $f ( t )$,\\
(ii) the largest value of $t$, for $0 \leqslant t < 24$, at which this maximum value occurs. Give your answer to one decimal place.

\hfill \mbox{\textit{Edexcel C34 2017 Q10 [9]}}